Learner support Grade 12 for 2021

CURRICULUM GRADE 10 -12 DIRECTORATE NCS (CAPS) LEARNER SUPPORT DOCUMENT GRADE 12 PHYSICAL SCIENCES STEP AHEAD PROGRAMME 2021 1

PREFACE This support document serves to assist Physical Sciences learners on how to deal with curriculum gaps and learning losses as a result of the impact of COVID-19 in 2020. It also captures the challenging topics in the Grade 10 -12 work. Activities should serve as a guide on how various topics are assessed at different cognitive levels and also preparing learners for informal and formal tasks in Physical Sciences. It will cover the following topics: TOPIC PAGE NUMBER 1. Work, Energy and Power 3 – 23 2. Reaction Rates 24 – 45 3. Chemical Equilibrium 46 – 57 4. Acids and Bases 58 – 72 Chemical Systems 5. 73 - 85 (Fertilizer Industry) 2

LEARNER DOCUMENT WORK  Define the work done on an object by a constant force F as F∙Δx cosθ, where F is the magnitude of the force, Δx the magnitude of the displacement and θ the angle between the force and the displacement.  The force doing the work and the motion must be in the same direction.  If the force F doing the work acts at an angle Θ, first determine the horizontal component of the force acting at an angle i.e. the force has to be resolved into its components.  Use the horizontal component of the force to determine the amount of work.  A force must act on the object on which work is done, all the time.  The object must move in the direction of the force that is applied i.e Displacement must be in the same direction.  Only the component of the applied force that is parallel to the motion does work on an object.  Positive work is done by a force that acts in the same direction as the motion of the object.  Negative work is done by a force that acts in the opposite direction of the motion of the object. (e.g friction).  Zero work (no work) is done by a force that acts perpendicular to the motion of the object ( e.g normal). For the force acting at an angle:-  Calculate the horizontal and the vertical components.  If an object is pulled , then F g= FN+Fv, therefore FN= Fg- Fv  If an object is pushed then Fg= FN- Fv, therefore FN= Fg+Fv  This is used when given the coefficients Worked examples A 6kg box is pulled, to the right on a horizontal surface, by a force of 20N. The box experiences a constant frictional force of 3N during its motion. The box covers a distance of 5m. calculate the work done by: a) The frictional force b) Applied force c) Normal force d) Weight e) Net work done Solutions a) 𝑊𝑓 = 𝑓𝛥𝑥𝑐𝑜𝑠𝜃 = (3)(5)cos1800 3

= -15J b) 𝑊𝐹 = 𝐹𝛥𝑥𝑐𝑜𝑠𝜃 =(20)(5)cos00 = 100J c) 𝑊𝑁 = 𝐹𝑁𝛥𝑥𝑐𝑜𝑠𝜃 = (6)(9.8)(5)cos900 = 0J d) 𝑊𝐹𝑔 = 𝐹𝑔𝛥𝑥𝑐𝑜𝑠𝜃 = (6)(9.8)(5)cos2700 =0J f) Wnet = Wf +WF+WFg+ WN = -15J + 100J + 0J +0J = 85J Learner’s notes  State the work-energy theorem: The net work done on an object is equal to the change in the object’s kinetic energy. OR The work done on an object by a net force is equal to the change in the object’s kinetic energy. 1  In symbols: Wnet = ∆Ek = Ekf – Eki OR Wnet = 2 m(vf 2 − vi 2 ), where W net is Work net measured in Joules (J), Ek is Kinetic energy measured in Joules(J), m is mass measured in kilograms (kg), v is velocity measured in m·s-1 .  Firstly take out the given data. METHOD ONE:  Draw a force / free-body diagram of all the forces acting on the object.  Determine the angle θ between each force and the direction of motion / displacement.  Use W = FΔxcosθ to determine the work done by each force 1  Then use W net = ∆EK OR W net = EKf – EKi OR Wnet = 2 m(vf 2 − vi 2 ), according to the given data and the variable you are calculating.  Calculate the sum of the work done by all the forces (on the left hand side): W net = W Fa +W f 1 equals to change in kinetic energy(on the right hand side): ∆EK = EKf – EKi OR ∆EK = 2 𝑚(𝑣𝑓 2 − 𝑣𝑖 2 ) METHOD TWO:  Draw a force / free-body diagram of all the forces acting along the plane of motion of the object.  Ignore the forces that act perpendicular to the plane, because they do zero work on the object.  Calculate the net force / the sum of the forces acting parallel to the plane, i.e.: Fnet = F a +Ffk  Calculate the net work done on the object using: W net = Fnet Δxcosθ (on the left hand side) equals 1 to ∆EK = EKf – EKi OR = 2 𝑚(𝑣𝑓 2 − 𝑣𝑖 2 ) on the right hand side. WORKED EXAMPLE (horizontal surface) 4

A formula 1 racing car of mass 640kg is travelling at 30m.s -1. It then accelerated in a straight line down the main straight. The engine exerts an average forward force of 12000N and the racing car experiences an average frictional force of 3000N. 1. State the work-energy theorem in words. 2. Draw a labelled free- body diagram showing ALL the forces acting on a racing car. 3. Use work-energy theorem to calculate the speed of the racing car after it has travelled 30m. Solutions Data: m = 640 kg, Vi = 30m.s-1, Fa= 12000N, fk = 3000N 1. The net work done on an object is equal to the change in the object’s kinetic energy. 2. FN FF FA Fg Firstly calculate the net work done on the car by each force : W a = Fa Δxcosθ = (12000)(30)cos 0° = +360 000J Wf = Ff Δxcosθ = (3000)(30)cos 180° = -90000J Find the net work done on the car by adding the work done by each force W net = +360000+ (-90000) = + 270 000J Secondly find the initial kinetic energy of the car: 1 1 Eki = 2 𝑚𝑣𝑖 2 = 2 (640)(30)2 = 288000J Use WORK-ENERGY theorem to find the final speed of the car: W net = Ekf – Eki 1 270 000 = 2 (640)𝑣𝑓 2 - 288000 1 270000+288000 = (640)𝑣𝑓 2 2 Vf = 41.76m.s-1 Second method Firstly calculate the net force on the car : Fnet = Fa +Ffk = 12000-3000 = 9000N Find the net work done on the car using the equation: W net = Fnet Δxcosθ = (9000)(30)cos0° =270000J Secondly find the initial kinetic energy of the car: 1 1 Eki = 2 𝑚𝑣𝑖 2 = (640)(30)2 = 288000J 2 Use WORK-ENERGY theorem to find the final speed of the car: W net = Ekf – Eki 5

1 270 000 = 2 (640)𝑣𝑓 2 - 288000 1 270000+288000 = 2 (640)𝑣𝑓 2 Vf = 41.76m.s-1 WORKED EXAMPLE (Incline plane) A 3kg remote-controlled car is driven up a plane inclined at 25° to the horizontal. The car motor exerts an average forward force of 60N. The car experiences a frictional force of 15N as it moves up the incline plane. The speed of the car at the bottom of the inclined plane is 5 m∙s-1. 1. State work-energy theorem in words. 2. Draw a free-body diagram showing all the forces acting on an object. 3. Use work-energy theorem to calculate the speed of the car after it has travelled 4m up the inclined plane. Solutions 1. The net/total work done on an object is equal to the change in the object's kinetic energy. OR The work done on an object by a net force is equal to the change in the object's kinetic energy. 2. FN Fa F g FF 3. Wnet = ΔK W F + W g// + W f = ½ mvf2 - ½ mvi2 F∆xcos𝜃°+mgsin𝜃∆xcos𝜃°+f∆xcos𝜃°+ =½ mvf2 - ½ mvi2 (60)(4)cos0°+(3)(9,8)sin25°(4)cos180°+ (15)(4)cos180° = ½(3) vf2 - ½(3)(5)2 Vf = 10,58 m.s-1 Second method Calculate the net force / the sum of the forces acting parallel to the plane: Fnet = F a +Ffk+Fg// = 60 - 15 - (3)(9,8)sin25° = 32,56N Find the net work done on the car using the equation: W net = Fnet Δxcosθ= (32,56)(4)cos0° =130,30J Secondly find the initial kinetic energy of the car: 1 1 Eki = 2 𝑚𝑣𝑖 2 = 2 (3)(5)2 = 37,5J 6

Use WORK-ENERGY theorem to find the final speed of the car: W net = Ekf – Eki 1 130,30 = 2 (3)𝑣𝑓 2 - 37,5 1 130,30+37,5 = 2 3𝑣𝑓 2 Vf = 10,58 m.s-1 GRAVITATIONAL POTENTIAL ENERGY: Gravitational potential energy (Ep) of an object is defined as the energy it has because of its position in the gravitational field relative to some reference level. 𝐄𝐩 = 𝐦𝐠𝐡. Where:  m= mass of object, measured in kilograms(kg)  g= acceleration due to gravity, measured in metres per second per second (m ∙ s −2 ). On earth g=9.8m ∙ s −2 .  h=vertical height of the object above some reference point, measured in metres (m) KINETIC ENERGY: Kinetic energy (Ek) is defined as the energy an object possess due to its motion 𝟏 𝐄𝐤 = 𝟐 𝐦𝐯 𝟐 . Where:  m=mass of the object, measured in kilograms(kg)  v=the magnitude of the velocity of the object, measured in metres per second(m/s) MECHANICAL ENERGY: The mechanical energy (Em) of an object is defined as the sum of its gravitational potential energy (Ep) and kinetic energy (Ek). 𝐄𝐦 = 𝐄𝐤 + 𝐄𝐩 . Energy is a scalar quantity and is measured in joules (J). LAW OF CONSERVATION OF MECHANICAL ENERGY: Principle of conservation of mechanical energy: The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. A system is isolated when the resultant/net external force acting on the system is zero. Worked Examples: Question 1 Given a 0.1kg bead at A sliding along a frictionless surface. Determine its speed at B if it slides up the frictionless ramp 7

Solution: Emech (A) = Emech (B). Ep(A) + Ek(A) = Ep(B) + Ek(B). 1 1 0 + 2 (0.1)(62 ) = (0.1)(9.8)(1) + 2 (0.1)(v 2 ). 𝐯 = 𝟒. 𝟎𝟓𝐦 ∙ 𝐬 −𝟏 . Question 2 A pendulum of mass 800g swings to a maximum height of 50cm a) Calculate the potential energy of the pendulum at point A. (3) b) What is the kinetic energy of the pendulum at point A. (1) c) Calculate the velocity of the pendulum at the lowest point of its swing. (4) Solution a) Ep(A) = mgh. 800 50 Ep(A) = (1000)(9.8)(100). 𝐄𝐩(𝐀) = 𝟑. 𝟗𝟐𝐉 b) Ek(A) = 0J c) Emech(A) = Emech(C) . Ep(A) + Ek(A) = Ep(C) + Ek(C) 8

1 3.92 + 0 = 0 + 2 (0.8)(v 2 ) 𝐯 = 𝟑. 𝟏𝟑𝐦 ∙ 𝐬 −𝟏 LAW OF CONSERVATION OF ENERGY: The law of conservation of energy states that energy is neither created nor destroyed; it is ONLY converted from one form to another. CONSERVATIVE FORCE AND NON-CONSERVATIVE FORCES: Conservative force: Conservative force is a force for which the work done in moving an object between two points is independent of the path taken. Examples are gravitational force, the elastic force in a spring and electrostatic forces (coulomb forces). Worked Example: The diagram shows a 70kg skateboarder who skates down a slope while experiencing a frictional force of 190N. The slope forms an angle of 30 0 with the horizontal. The skateboarder covers a distance of 10m between points A and B. The speed of the skateboarder at point A is 6m ∙ s−1 . Calculate the work done by gravity. . Solution: Option 1: W = F ∙ ∆xCosθ. Wg =mg∙ ∆xCosθ. Wg = 70(9.8)(10)Cos60. 𝐖𝐠 = 𝟑𝟒𝟑𝟎𝐉. Option 2: opposite h Sin30° = hypotenuse = 10. h = 5m. W = F ∙ ∆xCosθ. 9

Wg = mg ∙ hCosθ. Wg = 70(9.8)(5)Cos0. 𝐖𝐠 = 𝟑𝟒𝟑𝟎𝐉. Therefore, irrespective of the path taken, work done by gravitational force does not change. Non-conservative force: Non-conservative force is a force for which the work done in moving an object between two points depends on the path taken. Examples are frictional force, air resistance, tension in a chord, etc. Worked Example: In the diagram below, a dynamics trolley of mass 1kg moving at 3.5𝑚 ∙ 𝑠 −1 , rolls from point X to Y along a frictionless runway. The length of the runway is 1.5m. the vertical height between points X and Y is 0.5m 1) Is the mechanical energy of the trolley conserved as it moves from X to Y? Explain your answer. 2) Calculate the speed of the trolley at point Y. 3) Now suppose the surface XY is rough and the trolley experiences a constant frictional force of 5N as it moves from X to Y. Will the trolley reach point Y? Justify your answer by means of a calculation. Solution: 1. Yes, the system is isolated/there are no non-conservative forces present. 2. Wnc = ∆Ek + ∆Ep . Wnc = Ek(f) − Ek(i) + Ep(f) − Ep(i) . 1 1 0 = 2 (1)v 2 − 2 (1)(3.52 ) + 1(9.8)(0.5) − 0. 𝐯 = 𝟏. 𝟓𝟕𝐦/𝐬. 3. Wnc = ∆Ek + ∆Ep . Wf = Ek(f) − Ek(i) + Ep(f) − Ep(i) . fk ∆xCosθ = Ek(f) − Ek(i) + Ep(f) − Ep(i). 1 5∆xCos180 = 0 − 2 (1)(3.52 ). ∆x = 1.225m. The trolley will not reach Y. 10

Power is a rate at which work is done or energy is expended. 𝑊 In symbols: 𝑃 = ∆𝑡 , where P is power, the unit of power is the joule per second which is known as the watt (W). 1 W = 1 J∙s–1 W is the work done, measured in joules (J) t is time taken to do work, measured in seconds(s) WORKED EXAMPLE 1 During a canoe race each of two different teams, A and B, exerts a net force of 80 N to pull their canoes across a distance of 70 m. Team A takes 100 s and Team B takes 95 s. 1. Calculate the net work done by both teams. 2. Determine the power of each team. Solutions 1. W net = Fnet∆xcos𝜃 = (80)(70)cos0° = 5600J 𝑊 5600 2. 𝑃 = ∆𝑡 = 95 = 58,95W WORKED EXAMPLE 2 A power lifter is able to lift a large weight through a vertical height of 1,5m in 3s. The weight lifter applies an average upwards force of 2700N. 1. Calculate the work done by an average upwards force. 2. Calculate the power output of the weightlifter. Solutions 1. W Fa = Fa∆xcos𝜃 = (2700)(1,5)cos0° = 4050J 𝑊 4050 2. 𝑃 = ∆𝑡 = = 1350W 3 AVERAGE POWER  Suppose a car moves along an even horizontal road at a constant velocity. The driving force of the engine is equal in size and opposite in direction to the frictional force and the car does not experience any acceleration.  Distance that the car moves in time, t: Δx = vΔt  Work done by the driving force: W = FΔxcos0° = FvΔt 𝑊 FvΔt  𝑃 = ∆𝑡 = = Fv P = Fv ∆𝑡 Use P = Fv to calculate the average power when a force F (the component of the force in the direction of the motion) acts on an object moving at an average speed v. P ave = Fvave WORKED EXAMPLE 1(horizontal surface) A woman, shopping in a hurry, applies a force of 60N and moves her trolley at a constant speed of 3m.s-1. Calculate the average power of a woman. Solution 11

Pave = Fvave = (60)(3) = 180 W WORKED EXAMPLES (inclined plane) A cyclist rides up the rough inclined plane at a constant speed of 3m.s -1, as shown in figure 4.96. The combined mass of the cyclist and bicycle is 90kg. The cycle experiences a frictional force of 250N.  Firstly identify all the forces acting parallel to the plan. There are two forces acting on down the slope(frictional force and the parallel weight component to the slope) Fg// = mgsin𝜃 ° = (90)(9,8) sin20° = 301.7N Fg// + f = 301,7 +250 = 551,7 N In order for the cyclist to maintain the constant speed up the slope, he must exert an equal forward force of 551,7N up the slope. Pave = Fvave = (551,7)(3) = 1655,1W WORKED EXAMPLE 2 A vehicle of mass 2000kg travels up a plane inclined at 25° to the horizontal, at a constant speed of 8m.s-1. The vehicle experiences a frictional force of 1500N. 1. Draw a labelled free body diagram showing all the forces acting on a vehicle on an inclined plane. 2. Calculate the force exerted by the vehicles engine. 3. Calculate the average power developed by the vehicles engine. Solution 1. 2. F = Fg// + f = mgsin𝜃 ° +f = (2000)(9,8)sin 25° + 1500 = 9783,3N 3. Pave = Fvave = (9783,3)(8) = 78267 W 12

ACTIVITIES Question 1 A 10kg box is pulled, to the right on a horizontal surface, by a force of 50N for 10m. The box experiences a constant frictional force of 5N during its motion. 1.1 Draw a free body diagram, show all the forces acting on the object 1.2 Name two physical quantities needed for the work to be done. 1.3 Calculate work done by the 1.3.1 Applied force 1.3.2 Frictional force 1.3.3 Net work done by the two forces on an object QUESTION 2 13

A 3kg block is pulled 8m to the right, by a 15N force, applied at an angle of 30 o to the horizontal. The surface has a coefficient of kinetic friction of 0.2. 15N o 3kg 30 2.1.1 Drawing a free-body diagram of all the forces acting on the block 2.1.2 Calculate the work done by each force 2.1.3 Calculate the net work done by all the forces 2.1.1 2.1.2 2.1.3 14

LEARNER’S WORKSHEET ACTIVITY 1 In the diagram below, a 4 kg block lying on a rough horizontal surface is connected to a 6kg block by a light inextensible string passing over a light frictionless pulley. Initially the blocks are HELD AT REST. When the blocks are released, the 6 kg block falls through a vertical distance of 1,6 m. 1. State the work-energy theorem in words. (2) 2. Draw a labelled free-body diagram for the 6 kg block. (2) 3. Calculate the work done by the gravitational force on the 6 kg block. (3) The coefficient of kinetic friction between the 4 kg block and the horizontal surface is 0,4. Ignore the effects of air resistance. The coefficient of kinetic friction between the 4 kg block and the horizontal surface is 0,4. Ignore the effects of air resistance. 4. Use WORK-ENERGY THEOREM to calculate the speed of the 6 kg (5) block when it falls through 1,6 m while still attached to the 4 kg block. 15

ACTIVITY 2 A rescue helicopter is stationary (hovers) above a soldier. The soldier of mass 80 kg is lifted vertically through a height of 20 m by a cable at a CONSTANT SPEED of 4 m·s-1. The tension in the cable is 960 N. Assume that there is no sideways motion during the lift. Air friction is not to be ignored. 1. State the work-energy theorem in words. (2) 2. Draw a labelled free body diagram showing ALL the forces acting on (3) the soldier while being lifted upwards. 3. Write down the name of a non-constant force that acts on the soldier (1) during the upward lift. 4. Use the WORK-ENERGY THEOREM to calculate the work done on (5) 16

the soldier by friction after moving through the height of 20 m. 5. Identify TWO forces which do negative work. (2) ACTIVITY 3 The block moves from point B at a velocity of 4,95 m·s-1 up a rough inclined plane to point C. The speed of the block at point C is 2 m·s-1. Point C is 0,5 m above the horizontal, as shown in the diagram below. During its motion from B to C a uniform frictional force acts on the block. The total work done by force F that moves the vehicle from point A to point B is 4,80 x 106 J. 4,95 m·s-1 C 2 kg 0,5 m B 1. State the work-energy theorem in words. (2) 2. Draw a labelled free- body diagram showing ALL the forces acting on (3) an object. 3. Use work-energy theorem to calculate the work done by the frictional (5) force when the 2 kg block moves from point B to point C. 17

Activity 4 The diagram below shows a boy skate boarding on a ramp which is inclined 20° to the horizontal. A constant frictional force of 50N acts on the skate board as it moves from point P to Q. consider the boy and the skateboard as a single unit of mass 60kg. ignore the effect of air friction 1. Draw a labelled free- body diagram, showing ALL the forces acting (3) on the boy-skateboard unit while moving down the ramp from P to Q. Point P and Q on the ramp are 25m apart. The skate board passes point P at a speed of vi and passes point Q at a speed of 15 m·s-1 . 2. State work-energy theorem in words. (2) 3. Use work-energy theorem to calculate the speed v i of the (5) skateboarder at point P. 18

WORK SHEET: ACTIVITY 1 Question 1 During a flood a tree trunk of mass 100 kg falls down a waterfall. The waterfall is 5 m high. If air resistance is ignored, calculate: 1.1 the potential energy of the tree trunk at the top of the waterfall. (3) 1.2 the kinetic energy of the tree trunk at the bottom of the waterfall. (4) 1.3 the magnitude of the velocity of the tree trunk at the bottom of the waterfall. (4) Question 2 A mountain climber who is climbing a mountain in the Drakensberg during winter, by mistake drops her water bottle which then slides 100 m down the side of a steep icy slope to a point which is 10 m lower than the climber's position. The mass of the climber is 60 kg and her water bottle has a mass of 500 g 19

2.1 If the bottle starts from rest, how fast is it travelling by the time it reaches the bottom of (4) the slope? (Neglect friction.) 2.2 What is the total change in the climber's potential energy as she climbs down the (5) mountain to fetch her fallen water bottle? i.e. what is the difference between her potential energy at the top of the slope and the bottom of the slope? ACTIVITY 2 Question 1 1.1 Sate the law of conservation of energy. (2) 1.2 State the law of conservation of mechanical energy (2) 1.3 Define a conservative force and give an example of such a force. (3) 1.4 A child of mass 38kg slides from rest down a slide at the playground. The height of the slide is 2m. the speed of the child at the bottom of the slide is 3m/s a) Is mechanical energy of the child conserved? Explain your answer. (2) b) Calculate the work done by gravity on the child. (4) 20

ACTIVITY 3 Question 1 In the diagram below, a 10kg crate is pushed up an inclined ramp with a force of 150N being applied parallel to the ramp. The ramp is inclined at 350 to the horizontal. The crate experiences a constant frictional of 80N. (a) Draw a labelled free-body diagram showing all the forces acting on the crate as it (4) moves up the ramp. (b) From your free-body diagram above, identify conservative and non-conservative (3) forces. 21

Question 2 Ronald, a cyclist, is free-wheeling (moving without peddling) along a horizontal surface at a constant speed of 𝟏𝟎𝐦 ∙ 𝐬 −𝟏 . He reaches the bottom of the ramp (position A). The ramp has a height of 1.2m and a length of 8m. While free-wheeling up the ramp, he experiences a frictional force of 8N. The total mass of the cyclist and cycle is 55kg 2.1 Is the cyclist’s mechanical energy conserved as he moves from point A to point B? (2) Give a reason for your answer. 2.2.1 What is a conservative force? (2) 2.2.2 Name the conservative force acting on the cyclist as he moves. (1) 2.3 Calculate the kinetic energy of the cyclist: 2.3.1 at position A (3) 2.3.2 at the top of the ramp, using 𝑊𝑛𝑐 =∆𝐸𝑘 + ∆𝐸𝑝 (6) 2.3.3 Verify your answer 1.3.2 above using the WORK-ENERGY THEOREM. (6) 22

ACTIVITY Question 1 Define power: (2) Question 2 What are the two possible units for measuring power. (2) Question 3 A girl pushes a 30kg box with a horizontal force of 120N along a horizontal floor a distance of 3m in 5s. the force of friction between the box and the floor in 20N 3.1 Calculate the power of the girl (4) 3.2 Calculate the rate at which energy id dissipated as heat. (4) Question 4 A car maintains a constant speed of 110𝐾𝑚 ∙ ℎ−1 on a straight, level road. The total resistive force acting on the car is 750N. Calculate: 4.1 The work done against friction over a distance of of 10Km. (3) 4.2 The power output of the car. (4) 23

Reaction mechanisms, Collision Theory and the Rates of Reaction Reaction mechanisms Key Concepts In this session we will focus on summarising what you need to know about:  Activation energy and activation complex  Energy profile of a reaction  Enthalpy (ΔH)  Endothermic and exothermic reactions  Energy changes during a reaction Most reactions do not begin until an amount of energy (activation energy) has been added to the reaction mixture. The activation energy is often called the ‘energy hill’ which must be ‘overcome’ by the addition of this amount of energy before a reaction can take place. When activation energy is added to the reactants, a so-called activated complex is formed. Activated complex is a temporary, unstable, high-energy composition of atoms, which represents a transition state between reactants and the products. When the activated complex is formed during a reaction, this complex can lead either to the formation of new bonds, i.e. molecules of the products, or to re-formation of the old bonds, thereby returning to being reactants. The peak of the energy hill indicates the energy of the activated complex. When an activated complex is formed during a reaction, this complex can lead either to the formation of new bonds, i.e. molecules of the products, or to the re-formation of the old bonds thereby returning to being the reactants. This is reversibility for the reaction. 24

ENTHALPY (HEAT OF REACTION)  Enthalpy (∆H) is the difference between the energy of the products and the energy of the reactants. ∆H = Energy of products – Energy of reactants For an endothermic reaction, Energy of products > Energy of reactants, therefore, ∆H is positive. For an exothermic reaction, Energy of products < Energy of reactants, ∆H is negative. A catalyst mechanism: the function of a catalyst is to provide an alternate route for the reaction to take place. This route has a lower activation energy and the rate of the reaction increases. A catalyst forms part of the activated complex and when this decomposes, the catalyst is released unchanged. 25

Collision Theory Collision theory provides an explanation for how particles interact to cause a reaction and the formation of new products. According to the collision theory, the following criteria must be met in order for a chemical reaction to occur:  Molecules must collide with sufficient energy, known as the activation energy, so that chemical bonds can break.  Molecules must collide with the proper orientation.  A collision that meets these two criteria, and that results in a chemical reaction, is known as a successful collision or an effective collision. Factors affecting Reaction Rates Reaction rate, in chemistry is the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time. Alternatively, it may be defined in terms of the amounts of the reactants consumed or products formed in a unit of time. Nature of Reactants Certain substances are more reactive than others based on their chemical properties. Example: Lithium is more reactive than Gold. The more bonds that need to be broken the slower the reaction rate will be. Concentration Increasing the concentration of the reactants will result in more effective collisions because there are more particles. More effective collisions = a faster reaction rate (shorter time) Surface area Increasing the surface area of a reactant increases the number of sites where a collision can take place. Greater surface area = more effective collisions = faster reaction rate (shorter time) Temperature Increasing temperature causes an increase in the average kinetic energy of molecules. Higher kinetic energy means the particles move faster. This results in more effective collisions and a faster reaction rate. Pressure A change in pressure only affects gases! Increase pressure by decreasing the volume. Less space between particles results in more effective collisions per unit time. More collisions gives a faster reaction rate. Catalyst Catalyst: a substance that speeds up a reaction but IS NOT A PART of the reaction. The catalyst lowers the activation energy. (Less energy needed for an effective collision) 26

The Maxwell-Boltzmann distribution Curve 27

a. The effect of increasing the temperature of the reaction mixture If the temperature increases: • The particles in the sample have a higher average kinetic energy and the most probable energy is higher than at a low temperature. • The graph is wider and the peak is lower. • The total number of particles is the same, so the total area below the graph is the same. • More particles have kinetic energy greater than the activation energy (the shaded area is bigger) • More effective collisions take place per second and the collisions are more energetic the reaction rate increases. b. The effect of increasing the concentration of the reactants. If the concentration of the reactants increases: • The particles in the sample have the same average kinetic energy and the most probable energy is the same the graph has the same curve/shape 28

• The graph is higher as there are now more particles per unit volume; • The total number of particles is greater, so the total area below the graph is bigger. • More particles have energy greater than the activation energy (the shaded area is bigger) • More effective collisions take place per second • The reaction rate increases. c. The effect of adding a catalyst 29

Worked examples 1.1 Define each of the following terms 1.1.1 Activation energy (2) 1.1.2 Exothermic reaction (2) 1.1.3 Endothermic reaction (2) 1.1.4 Activated complex (2) 1.2 In a limited supply of oxygen, such as in a car which is not tuned properly, octane burns incompletely to produce, amongst others, carbon monoxide. The following balanced chemical equation represents the reaction during which carbon monoxide forms: 30

2C8H18(ℓ) + 17O2(g) → 16CO(g) + 18H2O(g) ∆H < 0 Activated complex Potential energy reactants products Course of reaction 1.2.1 Is this graph representing an exothermic or endothermic reaction? Explain your (3) answer 1.2.2 Is the reaction homogeneous or heterogeneous (1) 1.2.3 On the same set of axis draw a catalyzed pathway (2) QUESTION 2 2.1 State TWO requirements for effective collision according to the collision theory (2) QUESTION 3 31

3. Consider the following two ways of producing hydrogen (H 2) gas in the laboratory: Test tube A Test tube B HCl (1mol.dm-3) HCl (1,5mol.dm-3) Zinc powder Zinc granules 3.1 Define rate of reaction (2) 3.2 In test tube B, H2 (g) is produced at a faster rate than in test tube A. Give three reasons why H2(g) is produced at a faster rate in test-tube B. (6) 3.3 Mention 2 ways, other than those mentioned in 3.2 above, of increasing the rate of the (2) reaction in test-tube B 4. The rate of the reaction between methanol and hydrochloric acid is investigated. The concentration of HCl(aq) was measured at different time intervals. The following results were obtained: TIME (minutes) HCl CONCENTRATION (𝒎𝒐𝒍 ∙ 𝒅𝒎−𝟑) 0 1,90 15 1,45 55 1,10 100 0,85 215 0,60 Calculate the average rate, in (𝑚𝑜𝑙 ∙ 𝑑𝑚−3) ∙ 𝑚𝑖𝑛−1 during the first 15 minutes. (3) 5. In general, a teaspoonful of sugar dissolves much quicker in hot water than in the same amount of cold water. Use the graph below and your knowledge of the collision theory to explain this observation. 32

6. A learner investigates the relationship between the mass of a metal and the volume of (5) the gas produced when the metal reacts with dilute hydrochloric acid. During the investigation the learner adds the metal in amounts of 0,4 g to a certain volume of acid in a container. After the complete reaction between the metal and the acid, the learner measures the volume of gas that forms after each addition of the metal. (2) 6.1 State the hypothesis of this investigation (2) 400 P 350 300 Volume of gas (cm3) 250 200 150 100 50 0 0 0,4 0,8 1,2 1,6 2,0 2,4 Mass of metal (g) 6.2 Name TWO variables that must be controlled during this investigation (2) 6.3 What conclusion can be drawn from the section PQ on this investigation (2) 6.4 Use the graph to predict the volume of the gas that will be produced when 0,4g of the (2) metal results with the acid 33

SOLUTIONS QUESTION 1 1.1 1.1.1 Minimum energy needed for the chemical reaction to start (2) 1.1.2 Reaction that releases energy during chemical reaction (2) 1.1.3 Reaction that absorbs energy during the chemical reaction (2) 1.1.4 Unstable transition state from reactants to products (2) 1.2 1.2.1 Exothermic reaction, ∆H < 0 or energy of products is less tan energy or reactants (3) 1.2.2 Heterogeneous (1) 1.2.3 (2) Activated complex Potential energy reactants products Course of reaction QUESTION 2 2.1 Sufficient kinetic energy of particles (2) Correct orientation of particles 34

QUESTION 3 3. 3.1 Change in concentration of reactants or products (2) 3.2 HCl has a higher concentration Reaction mixture is heated thus temperature is higher (6) Zinc powder has been used resulting to a larger surface area 3.3 Addition of positive catalyst (2) Nature of a reactant 4. Average rate = ∆c ∆t = 1,45 – 1,90 15 – 0 = -0,03 𝑚𝑜𝑙 ∙ 𝑑𝑚−3 = 0,03 𝑚𝑜𝑙 ∙ 𝑑𝑚−3 5. The more water gets hotter the higher the temperature, according to collision theory,  The speeds of the particles increase  The average kinetic energy of the particles increases  More particles have sufficient kinetic energy  Which increases the number of effective collisions taking place per unit time  Thus, rate of reaction increases 6. 6.1 The higher the amount of the metal, the higher the volume of the hydrogen gas (2) produced 6.2 Concentration of HCl (2) Volume of HCl Temperature 6.3 The reaction has reached completion/ reaction has stopped/ reactants has been used (2) up 6.4 125 cm3 (2) 35

Worksheet: Reaction rate QUESTION 1 1.1 Define each of the following terms 1.1.1 Activation energy (2) 1.1.2 Exothermic reaction (2) 1.1.3 Endothermic reaction (2) 1.1.4 Activated complex (2) 1.2 In a limited supply of oxygen, such as in a car which is not tuned properly, octane burns incompletely to produce, amongst others, carbon monoxide. The following balanced chemical equation represents the reaction during which carbon monoxide forms: 2C8H18(ℓ) + 17O2(g) → 16CO(g) + 18H2O(g) ∆H < 0 Activated complex Potential energy reactants products Course of reaction 36

1.2.1 Is this graph representing an exothermic or endothermic reaction? Explain your (3) answer 1.2.2 Is the reaction homogeneous or heterogeneous (1) 1.2.3 On the same set of axis draw a catalyzed pathway (2) QUESTION 2 2.1 State TWO requirements for effective collision according to the collision theory (2) QUESTION 3 37

3. Consider the following two ways of producing hydrogen (H 2) gas in the laboratory: 3.1 Define rate of reaction (2) 3.2 In test tube B, H2 (g) is produced at a faster rate than in test tube A. (6) Give three reasons why H2(g) is produced at a faster rate in test-tube B. 3.3 Mention 2 ways, other than those mentioned in 3.2 above, of increasing the rate of the (2) reaction in test-tube B 4. The rate of the reaction between methanol and hydrochloric acid is investigated. The concentration of HCl(aq) was measured at different time intervals. The following results were obtained: TIME (minutes) HCl CONCENTRATION (𝒎𝒐𝒍 ∙ 𝒅𝒎−𝟑) 0 1,90 15 1,45 55 1,10 100 0,85 215 0,60 Calculate the average rate, in (𝑚𝑜𝑙 ∙ 𝑑𝑚−3) ∙ 𝑚𝑖𝑛−1 during the first 15 minutes. (3) 5. In general, a teaspoonful of sugar dissolves much quicker in hot water than in the same amount of cold water. Use the graph below and your knowledge of the collision theory to explain this observation. 38

6. A learner investigates the relationship between the mass of a metal and the volume of the gas produced when the metal reacts with dilute hydrochloric acid. During the investigation the learner adds the metal in amounts of 0,4 g to a certain volume of acid in a container. After the complete reaction between the metal and the acid, the learner measures the volume of gas that forms after each addition of the metal. (2) 6.1 State the hypothesis of this investigation (2) 400 P Q 350 300 Volume of gas (cm3) 250 200 150 100 50 0 0 0,4 0,8 1,2 1,6 2,0 2,4 Mass of metal (g) 6.2 Name TWO variables that must be controlled during this investigation (2) 6.3 What conclusion can be drawn from the section PQ on this investigation (2) 6.4 Use the graph to predict the volume of the gas that will be produced when 0,4g of the (2) metal results with the acid 39

CHEMICAL EQUILIBRIUM SUMMARY NOTES Definitions Open system – reactants or products can escape from reaction vessel (An open system continuously interacts with its environment) Closed system – reactants or products can escape from reaction vessel (A closed system is isolated from its surroundings) Macroscopic changes – measurable or visible changes, eg. changes in colour, temperature, pressure, volume, concentration Yield – the amount of product formed during a chemical reaction Endothermic reaction – a reaction which absorbs more energy than is released Exothermic reaction – a reaction which releases more energy than is absorbed Reversible reaction – reaction that does not go to completion and occurs in both the the forward and reverse directions (products can be converted back to reactants) Dynamic chemical equilibrium – the rate of the forward reaction equals the rate of reverse reaction and they occur simultaneously Homogeneous equilibrium – all the substances in the system are in the same phase Heterogeneous equilibrium – substances of different phases occur in the system Equilibrium constant – ratio of the concentration of products to the concentration of reactants FACTORS AFFECTING EQUILIBRIUM  concentration  temperature  pressure (in the case of gases) If any of the conditions (factors) are changed, the forward or reverse reaction will be favoured (will occur faster) until a new equilibrium is established. NB: *Adding a catalyst has NO EFFECT on the equilibrium position, a catalyst only increases the rate of both the forward and the reverse reactions equally. *If a catalyst is added initially, then the equilibrium position is reached much quicker. *The addition of an inert gas does not alter the equilibrium position. Examples: 1. Equilibrium mixtures of solutions (temperature and concentration play a role) Equation: CoCℓ42- (aq) + 6H2O ⇌ Co(H2O)62+ + 4Cℓ- ∆H < 0 40

Disturbance Effect on equilibrium Colour change 1. Temperature o increase o favours endothermic o turns blue o decrease (reverse) reaction o favours exothermic o turns pink (forward) reaction 2. Concentration o [H2O] is increased by o favours forward reaction o turns pink adding H2O (which uses up water) o [Cℓ] is increased by o favours reverse reaction o turns blue adding HCℓ/NaCℓ (with (which will decrease Cℓ-) common Cℓ- ion) o [Cℓ-] is decreased by o favours forward reaction o turns pink adding Ag+ (which produces more Cℓ-) (Ag+ + Cℓ- → AgCℓ(s)) o favours reverse reaction o [H2O] is decreased by (which will decrease Cℓ-) o turns blue adding a dehydrating agent, eg. H2SO4 2. Gaseous equilibrium mixtures (temperature and pressure play a role) Equation: NO2(g) ⇌ N2O4(g) ∆H < 0 Disturbance Effect on equilibrium Colour change o increase o favours endothermic o colourless to 1. Temperature (reverse) reaction brown o decrease o favours exothermic o brown to (forward) reaction colourless o increase o favours forward o brown to 2. Pressure reaction (less moles of colourless gas molecules) o colourless to o favours reverse brown o decrease reaction (more moles of gas molecules) 41

FACTORS AFFECTING EQUILIBRIUM CONCENTRATION If the concentration of a reactant increases, more of that reactant reacts, and so the concentration of the reactants on the other side of the equation increases. Removing a reactant lowers the concentration of the reactant and so less of that reactant reacts and so t The rates of the forward and reverse reactions will not be equal for a short while. The concentrations are restored to the original values. The reaction rates once again become equal and equilibrium is restored. he concentration of the reactants on the other side of the equation decreases The value of Kequ is unaffected by adding or removing reactants, providing the temperature does not change. Worked examples: 1. Consider the following reaction that is at equilibrium in a closed container N2O4(g)  2 NO2 (g) CHANGE OF EFFECT REACTION CHANGE IN CHANGE IN CHANGE FACTOR ON FAVOURED THE THE IN KC REACTION AMOUNT AMOUNT OF RATE OF REACTANTS PRODUCTS Increase in Overall Forward Amount of Amount of Remains concentration reaction reaction product reactants the same of a reactant rate (N2O4) (NO2) [NO2] increases increases decreases BUT rate of forward 42

reaction is faster Increase in Overall Reverse Amount of Amount of Remains concentration reaction reaction product reactants the same of a product rate (N2O4) (NO2) [N2O4] increases decreases increases BUT rate of reverse reaction is faster Decrease in Overall Reverse Amount of Amount of Remains concentration reaction reaction product reactants the same of a reactant rate (N2O4) (NO2) [NO2] decreases decreases increases BUT rate of reverse reaction is faster Decrease in Overall Forward Amount of Amount of Remains concentration reaction reaction product reactants the same of a product rate (N2O4) (NO2) [N2O4] decreases increases decreases BUT rate of forward reaction is faster PRESSURE Changes in pressure only affect the equilibrium if the reactants or products are in the gas phase. A change in pressure can be brought about by changing the volume of the container. Adding or removing one of the reacting gases changes the partial pressure of that gas and disturbs the equilibrium Adding or removing a non-reacting gas has no effect on the equilibrium. TEMPERATURE Increasing the temperature of an equilibrium system increases the rate of both the forward and the reverse reactions as more particles have sufficient energy to make effective collisions. However, in order to counteract the increase in temperature, the endothermic reaction rate increases even more. After a time the rates adjust and equilibrium is restored. However, as both rates are higher, the equilibrium has a new value for Eequ. ΔH values given with the equation for a reversible reaction refer to the forward reaction. Worked example: The following reversible reaction reaches equilibrium in a closed container: 43

2SO2(g) + O2(g) ↔ 2SO3(g) ∆H<0 1.1 What is meant by “ dynamic equilibrium” 1.2 State Le Chatelier’s principle 1.3 How will the equilibrium concentration of the sulphur trioxide be affected by: 1.3.1 an increase in temperature 1.3.2 an increase in pressure 1.3.3 an increase in [SO2] SUGGESTED SOLUTIONS 1.1 In a closed system dynamic equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. 1.2 Le Chatelier’s principle state that in a closed system when equilibrium is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance. 1.3.1 DECREASE 1.3.2 INCREASE 1.3.3 INCREASE A CATALYST A catalyst affects the rate of both the forward and the reverse reactions equally and so has no effect on the position of the equilibrium or the value of Kequ These observations and explanations can be summed up in le Chatelier’s principle If a system in chemical equilibrium is subjected to a change of temperature, pressure or concentration, processes occur within the system which tend to counteract the change imposed. Imposed change System response Change in Kequ Increase Favour forward reaction to none concentration of increase concentration of products reactants Decrease Favour reverse reaction to None concentration of decrease concentration of reactants products 44

Increase pressure Favour the reaction that produces none fewer particles Decrease pressure Favour the reaction that produces none more particles Increase temperature Favour the endothermic reaction Yes Decrease temperature Favour the exothermic reaction Yes Worked example 2: For the reaction: N2(g) + O2(g) ⇌ 2NO(g) H = 180 kJ state the effect on the reaction equilibrium if you use 1.1. Higher concentration of O2. Increase in [O2 ], increases the overall reaction rates but forward reaction will be favoured, amount of reactants increase amounts of product decreases 1.2. Lower concentration of NO. Decrease in [NO], overall reaction rates decrease, forward reaction is favoured, amount of reactants decrease, amount of products decreases 1.3. Presence of a catalyst. The rate of both the forward reaction and reverse reaction increases equally. Equilibrium is not disturbed, amounts remain constant Worked example 3: 1. State Le Chatelier’s principle. When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance. 2. Study the following equation : 2CO (g) + O2 (g) ⇌ 2CO2 (g) Use Le Chatelier’s principle to predict what effect the following changes will have on the equilibrium. 2.1 increase in the concentration of O2 overall rates of reaction increases, forward reaction is favoured, amount of reactants decreases , amount of products decreases until new equilibrium is re- instated 2.2 Besides concentration state TWO ways in which the rate of the forward reaction can be increased. Increase pressure, increase temperature 45

CHEMICAL EQUILIBRIUM_ quantity and concentration of substances at equilibrium  Remember these equations from grade 10 and 11 𝑐 = 𝑛⁄𝑉 , 𝑛 = 𝑚⁄𝑀 ,𝑛 = 𝑁⁄𝑁 , 𝑛 = 𝑉⁄𝑉 these are mostly used to find or convert the amount of 𝐴 𝑚 substances at equilibrium  𝑐 = 𝑛⁄𝑉 _used when you have a solution C=concentration (mol/dm3) and V = volume of solution (dm3)  𝑛 = 𝑚⁄𝑀 _used when given the mass of a substance m=mass (g) and M=molar mass (g/mol)  𝑛 = 𝑉⁄𝑉 _used when given gases and specific temperatures and pressure most cases STP 𝑚 V= volume of a gas (dm3) and Vm= molar volume of a gas(dm3/mol) 22.4 ONLY at STP  𝑛 = 𝑁⁄𝑁 _used when amount is given as number of particles/atoms/molecules 𝐴 N= number of particles/atoms/molecules and NA =Avogadro’s number : 6.022x10 23  Equilibrium concentration is the amount of substance per unit volume the instant the reaction reaches equilibrium, the amount of substance at equilibrium can be achieved by :  𝑛𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = 𝑛𝐼𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑛𝐶ℎ𝑎𝑛𝑔𝑒 / 𝑛𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = 𝑛𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑛𝐶ℎ𝑎𝑛𝑔𝑒  Note: [ ] mean concentration of  Summarise the 1st column of the equilibrium table as R.I.C.E Reactant ⇌ Product Ratio nInitial (mol) nChange (mol) nEquilibrium (mol) Volume (dm3) [equilibrium] mol/dm3  Explain that the concentration of liquids and solids remains constant, for volume of liquids and solids on the table put a cross.  Use the following steps to fill in the table Step 1_Set up the RICE table as shown above Step 2_write each Change moles in terms of an unknown (eg. x, y, a, b ) in pencil using the stoichiometric ratio of each substance as the the coefficient of the chosen unknown Step 3_fill in quantities given on the table in ink if quantities are given as mass, concentration, volume convert to number of moles Step 4_ fill in the volume (dm3) of the reaction vessel on the volume row, put a cross for liquids and solids. Step 5_ put + in change in moles of substance formed and – for the substance that decreases or is used up 46

Step6_ Solve for the unknown and substitute actual values of the Change To do this find one substance in the reaction with all I.C.E values (including the unknown) Step 7_ find moles at equilibrium Step 8_ calculate equilibrium concentration using 𝑐 = 𝑛⁄𝑣 Examples 1. A mixture of 5 moles of H2 and 6 moles of I2(g) is placed in a sealed container of 2dm3 at a temperature of 430⁰C.The balanced chemical equation for this reaction is: H2(g) + I2(g) ⇌ 2HI(g) Equilibrium is reached in a certain period. At equilibrium, there are 4 moles of HI(g) in the container. Calculate the concentrations of H2(g) and I 2(g) at equilibrium. SOLUTION: Table H2(g) + I2(g) ⇌ 2HI(g) Ratio 1 1 2 nInitial (mol) 5 6 0 nChange (mol) (- (-1a) (+2a) 1a) -2 -2 +4 nEquilibrium 3 4 4 (mol) Volume (dm3) 2 2 2 [equilibrium] 1,5 2 2 mol/dm3 2. The following reaction takes place in a 250 cm3 container at 150°C 2NO(g) +Br2 (g) ⇌ 2NOBr (g) The reaction initially contains 0.25moles of NO and 0.1 moles Br 2. When equilibrium is reached, it is found that 80% of NO remained at, calculate the concentrations at equilibrium 2NO(g) +Br2 (g) ⇌ 2NOBr(g) Ratio 2 1 2 nInitial (mol) 0.25 0.1 0 nChange (mol) (-2y) -0.5 (-1y) - (+2y) +0.5 0.25 nEquilibrium (mol) 0.75 0.5 0.2 3 Volume (dm ) 0.25 0.25 0.25 47

[equilibrium] 0.8 0.3 0.2 mol/dm3 NO at equilibrium 𝟖𝟎 × 𝟎. 𝟐𝟓 = 𝟎. 𝟐 𝟏𝟎𝟎 QUESTION 1 1.1 1.1.1 (2) 48

1.1.2 N2O4(g) is placed in an evacuated, sealed container. c The following reaction takes place in the container at constant temperature: N2O4(g) ⇌ 2NO2(g) The concentration of the product is measured over time. Which ONE of the following graphs correctly illustrates the relationship between the nitrogen dioxide (NO2) concentration and time? 49

1.1.3 (2) 1.1.4 The graph below represents the change in concentration of a reactant against time for a chemical reaction Concentration Time In which ONE of the following graphs does the dotted line show the effect of catalyst on this reactant? 50

Concentration A Concentration C Time Time 51 Concentration Concentration D B Time Time

2.1 The substance that lowers the activation energy in a reaction. (1) 2.2 The measure of the average kinetic energy in a system. (1) 2.3 Rate of forward reaction equals rate of reverse reaction. (1) 2.4 A reaction in which products can be converted back to reactants (1) 2.5 The ratio of concentration of products to the ratio of concentration of (1) reactants at equilibrium. 3 A fertiliser company produces ammonia on a large scale at a temperature of 450 °C. The balanced equation below represents the reaction that takes place in a sealed container. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0 To meet an increased demand for fertiliser, the management of the company instructs their engineer to make the necessary adjustments to increase the yield of ammonia. In a trial run on a small scale in the laboratory, the engineer makes adjustments to the TEMPERATURE, PRESSURE and CONCENTRATION of the equilibrium mixture. The graphs below represent the results obtained. 3.1 Identify the changes made to the equilibrium mixture at each of the 52

following times: 3.1.1 t1 (2) 3.1.2. t2 (2) 3.1.3 t3 (2) 3.2 At which of the above time(s) did the change made to the reaction (2) mixture lead to a higher yield of ammonia? Write down only t 1 and/or t2 and/or t3 4 A sample of N 2 O 4 gas is sealed in a container and heated. The N 2 O 4 gas decomposes to NO 2 gas and the reaction reaches equilibrium according to the following balanced equation: N 2 O 4 (g) ⇌ 2NO 2 (g) ΔH > 0 The graph below shows how the concentrations of the two gases change as a result of changes made to the reaction conditions Concentration () 4.1 Define the term chemical equilibrium. (2) 4.2 How does the rate of the forward reaction compare to that of the reverse reaction at each of the following times? Only write down HIGHER THAN, LOWER THAN or EQUAL TO. 4.2.1 t1 (1) 4.2.2 t2 (1) 4.3 What change was made to the reaction conditions at each of the following times? In both instances, the equilibrium constant for the reaction did not change 53

4.3.1 t3 (1) 4.3.2 t4 (1) 4.4 How will an increase in temperature influence the yield of NO 2 (g)? (3) Write down INCREASES, DECREASES or REMAINS THE SAME. Use Le Chatelier's principle to explain the answer. 5 Hydrogen and iodine are injected into a closed container at constant temperature. The reaction reaches equilibrium according to the following equation: H2(g) + I2(g) ⇌ 2HI(g) The graph below illustrates changes that were then made to the equilibrium mixture. 5.1 What information about the reaction can be obtained from the graph (1) between times t0 and t1? 5.2 Describe all the changes that occurred in the system between t 1 and t2 (4) 6 The following equation represents a reversible reaction that has reached equilibrium at 470 °C in a closed container: N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0 A change was then made to the NH3 in the equilibrium mixture at t2. A graph showing the effect of this change is drawn below. (The graph is not drawn to scale.) 54

6.1 What is the meaning of the horizontal lines between t 1 and t2? (1) 6.2 State the change that was made to the NH3 in the mixture at time t2 (1) 6.3 Explain how the change mentioned in QUESTION 6.2 affected the (3) concentration of H2 and N2 gases as shown in the graph. 7 Pure hydrogen iodide, sealed in a 2 dm3 container at 721 K, decomposes according to the following balanced equation: 2HI(g) ⇌ H2(g) + I2(g) ∆H = + 26 kJ∙mol-1 The graph below show how reaction rate changes with time for this reversible reaction. 7.1 Write down the meaning of the term reversible reaction. (1) 55

7.2 How does the concentration of the reactant change between the 12th and (1) the 15th minute? Write down only INCREASES, DECREASES or NO CHANGE. 7.3 The rates of both the forward and the reverse reactions suddenly change at t = 15 minutes. 7.3.1 Give a reason for the sudden change in reaction rate (1) 7.3.2 Fully explain how you arrived at the answer to QUESTION 7.3.1 (3) 8 Hydrogen gas, H2(g), reacts with sulphur powder, S(s), according to the following balanced equation: H2(g) + S(s) ⇌ H2S(g) ∆H < 0 The system reaches equilibrium at 900C 8.1 Define the term chemical equilibrium. (2) 8.2 The sketch graph below was obtained for the equilibrium (2) A catalyst is added to the equilibrium mixture at time t1 Redraw the graph above in your ANSWER BOOK. On the same set of axes, complete the graph showing the effect of the catalyst on the reaction rates. 9 Hydrogen and iodine are sealed in a 2 dm3 container. The reaction is allowed to reach equilibrium at 700 K according to the following balanced equation: H2(g) + I2(g) ⇌ 2HI(g) 9.1 Give a reason why changes in pressure will have no effect on the (1) equilibrium position. The reaction rate versus time graph below represents different changes made to the equilibrium mixture. 56

9.2 What do the parallel lines in the first two minutes indicate? (1) 9.3 State TWO possible changes that could be made to the reaction conditions (2) at t = 2 minutes. 9.4 The temperature of the equilibrium mixture was changed at t = 4 minutes 9.4.1 Is the forward reaction EXOTHERMIC or ENDOTHERMIC? (3) Fully explain the answer 9.4.2 How will this change influence the Kc value? Choose from INCREASES, (1) DECREASES or REMAINS THE SAME 9.5 What change was made to the equilibrium mixture at t = 8 minutes? (1) 57

ACIDS AND BASES Properties of acids 1. sour to taste 2. temperature rises when reacting with water-shows that reaction is taking place 3. Ability to neutralise bases 4. Have a pH less than 7 5. Conduct electricity in solution Properties of bases 1. Bitter to taste 2. Have a soapy feel 3. temperature rises when reacting with water-shows that reaction is taking place 4. Ability to neutralise acids 5. Have a pH greater than 7 Acid Formula Strength Base Formula Strength Sodium Hydrochloric hydroxide NaOH strong acid HCl strong Potassium KOH strong Sulphuric hydroxide H2SO4 strong acid Lithium LiOH strong hydroxide Nitric acid HNO3 strong Calcium Ca(OH)2 Strong Phosphoric hydroxide H3PO4 Weak acid Magnesium Mg(OH)2 Strong Carbonic hydoxide H2CO3 weak acid Ammonia NH3 weak Oxalic acid (COOH)2 weak Potassium K2CO3 weak H2C2O4 carbonate Sodium Acetic acid CH3COOH weak NaHCO3 weak bicarbonate (etanoic acid) Sodium Na2CO3 weak carbonate 58

WORKED EXAMPLES Acid-Base reactions (Common acids and bases) Eg1) Complete and balance the following chemical equations a) 2 HCl + Ca(OH)2  __CaCl2____ + __2_H2O___ b) 2 HNO3 + K2O  __2 KNO3____ + __H2O__ c) MgCO3 + H2SO4  _ MgSO4__ + _ H2O _ + __CO2__ % Purity composition and Yield eg1) Determine the % composition of Na in 2,5 g of Na 3PO4 Mr (Na3PO4) = 3(23) +31 + 4(16) = 164 gmol -1 % of Na = 69 X 100 = 42,073 % 164 Mass of Na = 42,073 X 2,5 g = 1,052 g 100 eg2) A sample consisting of 3,25 g of Impure Potassium oxide reacts completely with excess nitric acid as shown by following balanced chemical equation. 2KOH + H2SO4  K2SO4 + 2H20 During the reaction, 2,56 g of Potassium sulphate was formed. Calculate the % purity of Potassium oxide Solution : 2 KOH : 1 K2SO4 112g : 174 g x : 2,56 Mass of pure KOH = 1,648 g % purity of KOH = 1,648 x 100 3,25 = 0,51 % DISCUSSION: Complete and balance the following chemical equations: a) 2 HCl + Ca(OH)2  _________ + _________ b) 2 HNO3 + K2O  __________ + __________ c) MgCO3 + H2SO4  ________+ _______ + _________ % Purity composition and Yield eg1) Determine the % composition of Na in 2,5 g of Na 3PO4 eg2) A sample consisting of 3,25 g of Impure Potassium oxide reacts completely with excess nitic acid as shown by following balanced chemical equation. 2KOH + H2SO4  K2SO4 + 2H20 During the reaction, 2,56 g of Potassium sulphate was formed. Calculate the % purity of Potassium oxide 59

NOTES Acids and Bases • Acids and bases are chemicals that exhibit their chemical behaviour when dissolved in water. Hence, when dealing with them the word aqueous will be repeatedly used, so is in the chemical equation for acids and bases. Acids and bases produces ions when dissolved in water; therefore, are called electrolytes. • An electrolyte is an ionic solution that is able to conduct electricity. • There are different types of acids and bases that occur naturally; however, there also others which are derived in the laboratory. • Acids and bases are defined based of two theories i.e. Lowry-Bronsted and Arrhenius Acids and bases are defined based of two theories i.e. Lowry-Bronsted and Arrhenius Table1: Illustrates acid and base definition using different theories. Definition Acid Base Lowry-Bronsted An acid is a proton donor (H+) A base is a proton acceptor (H+) theory HCl + H2O  H3O+(aq) + Cl-(aq) NH3 + H2O ⇌ NH4+(aq) + OH-(aq) Arrhenius theory An acid liberates hydrogen A base liberates hydroxyl ions ions (H+) in aqueous solutions (OH-) in aqueous solutions HCl  H+(aq) + Cl-(aq) NaOH  Na+(aq) + OH-(aq) Discuss Conjugate Acid- base pairs Conjugate acid- formed when a base accepts a proton Conjugate base- formed when an acid loses a proton(H+) .a1 b2 a2 b1 HCl + H2O  H3O+(aq) + Cl- (aq) Discuss Ampholyte/Amphiprotic substances a1 b2 a2 b1 H2O + HSO4-  H2SO4 + OH- 60

a1 b2 a2 b1 HSO4- + H2O  H3O+ + SO42- An ampholyte is a substance that can act as an acid and a base. HSO4- + H2O ⇌ H2SO4 + OH- HSO4- acts as a base HSO4- + H2O ⇌ SO42- + H3O+ HSO4- acts as an acid Activity 1 & 2 are educator facilitated: Activity 1 1.1 Which one of the following substances may be regarded as an ampholyte A. H2O B. CO32- C. SO42- D. CI- 1.2 Use the relevant Equations to show that the following substances are ampholytes. 1.2.1 H2PO4- 1.2.2 H2O Activity 2 2 For each of the following equations, write down the conjugate base: 2.1 NH3 + HCI NH4+ + CI- 2.2 HCN + H2O ⇌ H3O+ + CN- 2.2 For each of the following equations, write down the TWO acids: 2.2.1 H2SO4 + H2O H3O+ + SO42- 2.2.2 CO32- + H2O ⇌ HCO3- + OH- 61

NOTES Types of acids Strong Acid(HCl,H2SO4,HNO3) Weak Acid(CH3COOH) 1. Ionises completely in water to form 1. Ionises partially in water to form low high concentration of hydronium ions concentration of hydroxide ions 2. Has a lower pH value 2. Has pH value closer to 7 3. Has bigger Ka 3. Has smaller Ka 4. Stronger electrolytes, high conductivity 4. Weak electrolytes, lower conductivity Dilute acid – acid where the ratio of the Concentrated acid – acid in which the number of moles of acid to volume of ratio of the number of moles of the acid solution is small. to volume is large. This applies to both strong and weak acids . Strengths of acids : In decreasing order of strength HClO4, HCl ,HNO3, H2SO4, (COOH)2 , H2SO3 , HSO4-, H3PO4, HF, HNO2,CH3COOH, H2CO3, H2S ,NH4+ Strengths of bases : In decreasing order of strength OH- , HCO3- , CO32- , O2- Define pH as: pH is a measure of concentration of H+ ions in a solution.it is indication of the acidity or basicity of a solution. pH is mathematically defined as: pH = -log[H3O+] OR pH = -log[H+] [ H3O+][OH-] = 1 X 10 -14 Conversion for Volumes: 1 l = 1 dm3 1 ml = 1 cm3 1 cm3 = 1x10-3 dm3 1000ml = 1dm3 = 1l = 1000cm3 62

pH is measured with a pH meter; for grade 12, pH meter ranges from 0 to 14 ◦ Acidic pH is less than 7 ◦ Neutral pH is equal to 7 ◦ Basic pH is greater than 7 Auto-ionisation of water • Water has tendency to ionise itself by transferring a proton (H+) from one water molecule to the other. This ionisation is shown in the equation below: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) • Since, the above reaction shows an equilibrium nature, then one can write the equilibrium constant expression for water auto-ionisation as follows: Worked Examples 1. Calculate the concentration of H3O+ and OH- ions in H3PO4 of concentration 0,1 mol.dm-3 when it ionizes in water.(assume complete ionization) H3PO4 ⇌ 3H+ + PO43- Ratio: 1 3 1 0,1 0,3 0,1 [H3O+] = 3(0,1) = 0,3 mol.dm-3 [ H3O+][OH-] = 1 X 10 -14 (0,3) [OH-] = 1 X 10 -14 [OH-] = 3,33 X 10 -14 mol.dm-3 2. Calculate the pH of the solution of H3PO4 in question 1 above. pH = - log[H3O+] = - log(0,3) = 0,52 63

NOTES Hydrolysis- reaction of salt with water TEST Dissociate the following salts in water a) NaCl  _____ + _____ (2) b) NH4Cl  ______ + ______ (2) c) CH3OONH4  ______ + ______ (2) d) NH4NO3  ______ + ______ (2) (10 minutes) Reaction of salts with water Acidic salts - The hydrolysis of a salt derived from a strong acid and a weak base result in an acidic solution. eg ) NH4Cl + H2O NH4+ + Cl-   derived from NH3 HCl Basic salts - The hydrolysis of a salt derived from a strong base and weak acid results in a basic solution eg) Na2CO3 + H2O 2Na+ + CO32-   derived from NaOH H2CO3 64

Neutral salts - The hydrolysis of a salt derived from a strong acid and a strong base results in a neutral solution eg) NaCl + H2O Na+ + Cl-   derived from NaOH HCl The effect of salts on pH  The salt that is formed in a neutralization reaction should be neutral, with a pH equal to 7. But if the acid is much stronger than the base, or the base is much stronger than the acid, the pH is not 7, but just above or below 7.  If the acid is much stronger than the base, the salt is acidic, with a pH just lower than 7.  If the base is much stronger than the acid, the salt is alkaline/basic, with a pH just higher than 7. Acidic salt Alkaline salt NH4Cℓ ammonium Na2CO3 sodium chloride carbonate CH3COONa sodium ethanoate (COOH)2Na2 sodium oxalate pH of Example Hydrolysis of salt salt NaOH + HCℓ → NaCℓ + H2O none strong KOH + H2SO4 →Na2SO4 + none 7 a 2H2O (neutral) + LiOH + HNO3 → LiNO3 + H2O none strong b strong <7 a NH 3 + HCℓ →NH 4 Cℓ NH 4 + + H 2O →NH3 (acidic) + + + H 3O weak b NaOH + H2CO3 →Na2CO3 + CO32– + H2O H2O →HCO3– + OH– >7 weak – NaOH + CH 3COOH CH 3COO + H 2O (basic/ a + →CH3COONa + H2O →CH3COOH + OH– alkaline) – strong NaOH + (COOH) 2 → (COOH) 2Na 2 (COO ) 2 + 2H 2O→ – b + H 2O (COOH) 2 + 2OH 65

ACTIVITIES-LESSON 1 ACTIVITY 1 1.Complete the following table of salts formed during acid-base reactions 1. Acid/Base HCI HNO3 H2SO4 KOH NaOH Na2O MgO NaHCO3 CaCO3 2. NAME THE FOLLOWING ACIDS, BASES OR IONS 2.1 NaOH 2.2 CH3COOH 2.3 (COOH)2 2.4 Ca(OH)2 2.5 HNO3 2.6 H2SO4 2.7 NaHCO3 2.8 H3O+ 2.9 HSO4- 2.10 CO32- 2.11 NH4+ 2.12 HCO3- 2.13 NH3 3. DATA SHEET: 66

n=V Vm 3.1 Calculate the concentration of a 0,2 mol solution of Na 2CO3 with a volume of 150 cm3. (3) 3.2 Calculate the number of mol of NaOH in 250 cm3 solution of concentration of 0,25 moldm-3 (3) 3.3 Calculate the mass of NaCl(s) needed to prepare a solution of concentration 0,2 moldm-3 in a 150 ml flask. (3) 3.4 Calculate the concentration of H+ ions in 0,25 moldm-3 H2SO4. (3) 3.5 1,2 g of impure sodium hydrogen carbonate reacts with excess hydrochloric acid to form 44,8 dm3 of carbon dioxide gas at STP .The balanced equation for this reaction is given below. NaHCO3 + HCl  NaCl + CO2 + H2O Calculate the % purity of sodium hydrogen carbonate in this mass. (5) 3.6 Calcium carbonate reacts completely with excess nitic acid according to the following balanced chemical equation. CaCO3 + 2 HNO3  Ca(NO3)2 + CO2 + H2O Calculate the mass of CaCO3 needed to completely react with 30 cm3 solution of nitric acid of concentration 0,0125 moldm-3. (5) HOMEWORK ACTIVITY-LESSON 1 1 If 20g of impure Na2CO3 produces 14,625 g of pure NaCl according to the reaction: Na2CO3 + 2HCl → 2NaCl + H2O + CO2 What is the percentage purity of Na2CO3 used? 2 Sulphuric acid is a strong diprotic acid. 2.1 Define the term strong acid (2) 2.2 Give a reason why sulphuric acid is referred to as a diprotic acid. (1) 67

3 The hydrogen carbonate ion can act as both an acid and a base. It reacts with water according to the following balanced equation: - HCO3-(aq) + H2O(ℓ) ⇌ H2CO3 (aq) + OH (aq) 3.1 Write down ONE word for the underlined phrase. (1) 3.2 HCO-3(aq) acts as base in the above reaction. (1) Write down the formula of the conjugate acid of HCO -3(aq) 4 Ammonia can readily dissolve in water according to the equation below: NH3(g) + H2O(ℓ) → NH+4(aq) + OH-(aq) 4.1 Explain why a hydroxide ion is regarded as a Lowry-Brønsted base (2) 4.2 Write a balanced equation to show how the ampholyte in the above (2) equation will act as a base when it reacts with hydrochloric acid (HCℓ) -3 5 5 dm3of nitric acid (HNO33), with a concentration of 0,75 mol·dm , is spilled accidentally in a small pond of water. The acid and water has 3 a total volume of 1 000 dm . To neutralise the acid, calcium hydroxide is added to the water 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O 5.1 Define concentration of a solution (2) 5.2 Calculate the concentration of the acid AFTER it was spilled in the (4) pond. 5.3 Use calculations to determine if 120 g of calcium hydroxide will be (6) sufficient to react completely with ALL the acid in the pond. NOTES/ACTIVITIES-LESSON 2 Acids and Bases • Acids and bases are chemicals that exhibit their chemical behaviour when dissolved in water. Hence, when dealing with them the word aqueous will be repeatedly used, so is in the chemical equation for acids and bases. Acids and bases produces ions when dissolved in water; therefore, are called electrolytes. • An electrolyte is an ionic solution that is able to conduct electricity. • There are different types of acids and bases that occur naturally; however, there also others which are derived in the laboratory. 68

• Acids and bases are defined based of two theories i.e. Lowry-Bronsted and Arrhenius Acids and bases are defined based of two theories i.e. Lowry-Bronsted and Arrhenius Table1: Illustrates acid and base definition using different theories. Definition Acid Base Lowry-Bronsted An acid is a proton donor (H+) A base is a proton acceptor (H+) theory HCl + H2O  H3O+(aq) + Cl-(aq) NH3 + H2O ⇌ NH4+(aq) + OH-(aq) Arrhenius theory An acid liberates hydrogen A base liberates hydroxyl ions ions (H+) in aqueous solutions (OH-) in aqueous solutions HCl  H+(aq) + Cl-(aq) NaOH  Na+(aq) + OH-(aq) Discuss Conjugate Acid- base pairs Conjugate acid- formed when a base accepts a proton Conjugate base- formed when an acid loses a proton(H+) .a1 b2 a2 b1 HCl + H2O  H3O+(aq) + Cl- (aq) Discuss Ampholyte/Amphiprotic substances a1 b2 a2 b1 H2O + HSO4-  H2SO4 + OH- a1 b2 a2 b1 HSO4- + H2O  H3O+ + SO42- An ampholyte is a substance that can act as an acid and a base. 69

HSO4- + H2O ⇌ H2SO4 + OH- HSO4- acts as a base HSO4- + H2O ⇌ SO42- + H3O+ HSO4- acts as an acid Activity 1 & 2 are educator facilitated: Activity 1 1.1 Which one of the following substances may be regarded as an ampholyte A. H2O B. CO32- C. SO42- D. CI- 1.2 Use the relevant Equations to show that the following substances are ampholytes. 1.2.1 H2PO4- 1.2.2 H2O Activity 2 2 For each of the following equations, write down the conjugate base: 2.1 NH3 + HCI NH4+ + CI- 2.2 HCN + H2O ⇌ H3O+ + CN- 2.2 For each of the following equations, write down the TWO acids: 2.2.1 H2SO4 + H2O H3O+ + SO42- 2.2.2 CO32- + H2O ⇌ HCO3- + OH- ACTIVITIES 1. Define the following terms: 1.1 Strong acid (2) 1.2 Strong base (2) 1.3 Concentrated acid (2) 1.4 Dilute base (2) 70

2. Calculate the concentration of H3O+ and OH- ions in the following solutions: 2.1 NaOH of concentration 0,00001 mol.dm-3 (4) 2.2 HCI of concentration 0,5 mol.dm-3 (4) 3. Calculate the pH of: 3.1 0,2 mol.dm-3 of HNO3 (3) 3.2 0,04 mol.dm-3 of H2SO4 (4) 3.3 0,2 mol.dm-3 of Ba(OH)2 (5) 4. The pH of a solution of HCl is 3. Calculate the concentration of HCl. (3) 5. Calculate the concentration of phosphoric acid(H 3PO4) if the pH of the acid is 4.92. (Assume complete dissociation of the acid) (5) HOMEWORK/CLASSWORK ACTIVITY-LESSON 4 1 Define the term End point (2) 3 2 45 cm of sodium hydroxide solution is pipetted into a conical flask and titrated with a 0,12 mol·dm-3 oxalic acid (H2C2O4 ). Using a suitable indicator, it was found that 20,3 cm3 of acid was needed to neutralise the base. 2NaOH + H2C2O4 Na2C2O4 + 2H2O 2.1 Write down the name an indicator that would be suitable for the above (3) titration. Give a reason for your answer 2.2 How many grams of oxalic acid is necessary to make 150cm3 of (4) standard solution? 2.3 Calculate the concentration of the sodium hydroxide solution. (4) 3 A learner accidentally spills some sulphuric acid of concentration -3 6mol∙dm from a flask on the laboratory bench. Her teacher tells her to neutralise the spilled acid by sprinkling sodium hydrogen carbonate powder onto it. The reaction that takes place is: (Assume that the H2SO4 ionises completely.) H2SO4(aq) + 2NaHCO3(s) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g) The fizzing, due to the formation of carbon dioxide, stops after the learner has added 27 g sodium hydrogen carbonate to the spilled 71

acid. 3.1 Calculate the volume of sulphuric acid that spilled. Assume that all the (6) sodium hydrogen carbonate reacts with all the acid. -3 The learner now dilutes some of the 6 mol∙dm sulphuric acid solution -3 in the flask to 0,1 mol∙dm . 3.2 Calculate the volume of the 6 mol∙dm-3 sulphuric acid solution needed (2) 3 to prepare 1 dm of the dilute acid. 3 -3 4 During a titration 25 cm of the 0,1 mol∙dm sulphuric acid solution is -3 added to an Erlenmeyer flask and titrated with a 0,1 mol∙dm sodium hydroxide solution. Calculate the pH of the solution in the flask after the addition of 30 (8) 3 cm of sodium hydroxide. The endpoint of the titration is not yet reached at this point 72

THE CHEMICAL SYSTEM FERTILIZERS NOTES - LESSON 1 Fertilizer is a substance that is used to make the soil fertile. A fertilizer consists of an organic or chemical substance with the aim of improving the quality or amount of plant growth. When correctly applied, fertilizer makes plants more vigorous, helps leaves grow larger, helps to develop strong root system and makes plants stronger. TYPES OF FERTILIZERS ORGANIC FERTILIZER These are fertilizers that are derived mainly from plant remains and animal excretion. Examples: Plant compost; animal manure; guano and bone meal INORGANIC FERTILIZERS These are artificial fertilisers manufactured chemically in the laboratory. The sample of inorganic fertilizer Example of a bag of inorganic fertiliser 73

Examples: Urea; ammonia; ammonium phosphate; ammonium nitrate; superphosphate; and NPK fertilizers. WHY FERTILIZERS ARE NEEDED?  To produce more food  To fertilize soil faster than naturally  To increase the rate of food production in a given space of land  Land available for agriculture decrease. NUTRIENTS NEEDED BY PLANTS There are 16 nutrients. They are divided into two. Mineral and non-mineral nutrient. Mineral nutrients.  There are three primary nutrients: o Nitrogen (N) ; Phosphorus(P) and Potassium (K).  NPK ratio is the proportion of nitrogen, phosphorus and potassium.  Mineral nutrients are derived from organic sources and inorganic sources.  NITROGEN is derived from: o organic sources like guano and manure and o inorganic sources: urea; ammonium nitrate and ammonia  Functions:- o promotes growth of the leaves o stimulates growth of the plants  POTASSIUM is derived from: o organic sources:- potash o inorganic sources:- Potassium Chloride; Potassium Nitrate and Potassium Sulphate o Function improves quality of fruits and flowers.  PHOSPHORUS comes from: o organic sources :- guano and bone meal. o Inorganic sources:- superphosphate; triple superphosphate and ammonium phosphate o Function- promotes growth of roots and stems.  Non-mineral nutrients ( CHO ) Carbon (C) – from carbon dioxide in the atmosphere Hydrogen (H) – from rain 74

Oxygen ( O) – from rain and air in the atmosphere. LESSON 2 NOTES INTERPRETATION OF NPK RATIO  Interpret the N: P: K fertiliser ratio and perform calculations based on the ratio.  N P K is the ratio of Nitrogen(N),Phosphorus(P) and Potassium(K) in a certain fertilizer.  This can be expressed as 3:1:5 (36) 20kg  36 percentage fertilizer in the bag.(NB In this bag 64% will be the fillers)  3 parts of 9 parts is Nitrogen  1 part of 9 parts is Phosphorus  5 parts of 9 parts is Potassium  Calculate the total percentages and mass of Nitrogen, Phosphorus and Potassium. 3+1+5=9 % N =3/9 x36 =12% % P = 1/9x36 =4% %K= 5/9x36 =20% Mass of Nitrogen =12% x 20kg =2,4kg Mass of Phosphorus= 4% x 20kg =0,8 kg Mass of Potassium = 20% x 20kg =4kg LESSON 3 NOTES Fractional distillation of liquid air Industrial preparation of nitrogen from air. Steam reforming Preparation of hydrogen from earth gas (methane) 75

Reaction: CH4 + H2O → 3H2 + CO Haber process Industrial preparation of ammonia Reaction: N2 + 3H2 ⇌ 2NH3 Iron (Fe) catalyst Ostwald process Industrial preparation of nitric acid Reactions: 1. Catalytic oxidation of ammonia; catalyst: Pt 4NH3+ 5O2 ⇌ 4NO + 6H2O 2NO + O2 → 2NO2 3NO2 + H2O → 2HNO3 + NO OR 4NO2 + 2H2O + O2 → 4HNO3 Contact process Industrial preparation of sulphuric acid Reactions: 1. S + O2 → SO2 2. 2SO2 + O2 → 2SO3 Contact catalyst: vanadium pentoxide (V2O5) 3. SO3 + H2SO4 → H2S2O7 (H2S2O7: fuming sulphuric acid OR pyro sulphuric acid OR oleum) 4. H2S2O7 + H2O → 2H2SO4 Preparation of ammonium sulphate Ammonia + sulphuric acid → ammonium sulphate 2NH3 + H2SO4 → (NH4)2SO4 Preparation of ammonium nitrate Ammonia + nitric acid → ammonium nitrate NH3 + HNO3 → NH4NO3 LESSON 4 NOTES Excessive use of fertiliser and the environment o The correct application of fertiliser to crops is essential for high quality, fast growing crops but using too much or unnecessary fertiliser has a negative effect on the environment. o Fertiliser leaching (spreading by water) into the ground contaminates groundwater. o Soil becomes acidic (pH decreases). Many plants do not grow in acidic soil. 76

o Invasive plants grow excessively while indigenous plants die. Invasive plants are undesirable non-indigenous (foreign) plants that grow too fast and out-compete local plants. o Fertiliser in dams and rivers leads to “eutrophication”, defined below. o High nitrate concentrations in drinking water decreases the ability of haemoglobin in the blood to carry oxygen and leads to ‘blue baby syndrome’ o Eutrophication The process by which an ecosystem, e.g. a river or dam, becomes enriched with inorganic plant nutrients, especially phosphorus and nitrogen, resulting in excessive plant growth. As plant growth becomes excessive, the amount of dead and decaying plant material increases rapidly. Alternatives to inorganic fertilisers Although fertilisers are essential for the fast growth of high-quality crops, the negative effects of inorganic compounds on the environment must be taken into account. Alternative sources of organic nutrients that can be used to ensure good crops are:  Bone meal;  Animal manure;  Natural plant compost;  Bat guano (faeces);  Fish emulsions;  Kelp meal Advantages of organic fertilisers:  Break down and release nutrients more slowly than inorganic fertilisers, so there is less chance of the fertiliser leaching into the soil and causing contamination of groundwater;  Usually cost less and Are often available free. Disadvantages of organic fertilisers:  Not enough is available for large scale usage;  Provide less nutrients – more has to be used;  Slow release of nutrients sometimes harms plants;  Slow release may cause nutrients to be available too late in the plant’s growth cycle. 77

QUESTION 1 1.1 Which of the following is a primary mineral nutrient that is needed by plants 1.1.1 A. N B. C C. Mg D. Na (2) 1.2 The rapidly increasing human population is resulting in an ever- increasing demand for food. To meet this demand, farmers apply fertilizer to the same cultivated land EACH YEAR. 1.2.1 Explain why farmers have to apply fertilizers to the land each year. (1) 1.2.2 Write down one negative impact that over- fertilization can have on (1) humans. 1.2.3 Write down one positive impact of fertilizer on humans. (1) QUESTION 2( Lesson 2) 1.1 A 10 kg bag of NPK fertilizer is labeled 6:1:5 (22) 78

1.1.1 What is the meaning of NPK? (1) 1.1.2 What is the meaning of (22) on the label (1) 1.1.3 Calculate the mass of potassium in the bag (4) 1.2 A 2 kg bag of fertilizer is labeled as follows: 2:3:2 (22) . Calculate the mass of the : (3) 1.2.1 Phosphorus in the bag (3) 1.2.2 Filler in the bag 1.3 1.3.1 Two 50kg bags containing fertilisers P and Q respectively, are labeled as follows: Fertiliser P: 5:2:3 (25) Fertiliser Q: 1:3:4 (20) What do the numbers (25) and (20) on the labels represent? 1.3.2 Using calculations,determine which fertiser (P or Q) contains the greater (4) mass of potassium 1.4 1.4.1 The following substances are present in a bag of fertilizer: 20kg (5) Ammonium nitrate,12kg sodium phosphate,18 kg potassium chloride. Calculate the NPK ratio of the fertilizer. 79

QUESTION 1 1.1 The reaction represented below take place during one of the industrial processes used in the fertilizer industry. 1. 4NH3(g) + 5O2(g) ----Pt------ 4NO(g) + 6H2O(g) 11. NO(g) + O2(g) X 111.NO2 + H2O(l) --------------HNO3(aq) + -------- Write down: 1.1.1 The name of this industrial process 1.1.2 The function of Pt in reaction 1 1.1.3 The NAME of product X 1.1.4 A balanced equation for reaction 111 1.1.5 Two ways in which the yield of the NO(g) obtained in reaction 1 can be increased Without changing the amount of reactants and products. 1.2 The four steps in the manufacture of an inorganic fertilizer are listed below. These steps are NOT written in the order in which they occur. Step 1: Sulphuric acid reacts with ammonia to produce ammonium sulphate’ Step 2: Sulphur dioxide reacts with oxygen to produce Sulphur trioxide Step 3: Oleum is diluted with water to produce sulphuric acid Step 4: Sulphur trioxide is bubbled in concentrated sulphuric acid to produce oleum. Write down the : 1.2.1 Correct order in which the steps occur in the preparation of the inorganic fertilizer by using the numbers 1 to 4. 1.2.2 Balance chemical equation for step 2 1.2.3 NAME of the catalyst used in step 2 1.2.4 Balanced chemical equation for step 4 1.2.5 Reason why Sulphur trioxide is NOT dissolved in water in step 4. Worksheet 3 (Lesson 3) Activity 2 2.1 2.1.1 Ammonium Nitrate is often mixed with Potassium Chloride and Ammonium Phosphate. Give a reason why it is mixed with these compounds 80

2.2 2.2.1 The letters A to F below represent some fertilisers and raw materials used in the preparation of fertilisers. A Sulphur B Methane C Ammonium Sulphate D Air E Potassium Chloride F Ammonium Nitrate 2.2.1.1 LETTERS representing TWO raw materials used in the preparation of compound F .2.2.1.2 NAME or FORMULA of the acid needed to prepare compound F 2.2.1.3 LETTER representing the solid raw material used in the contact process 2.2.1.4 Balanced equation for the preparation of compound E 2.2.1.5 LETTER representing the raw material that supplies the primary nutrients needed for development of flowers 81

REVISION QUESTIONS Question 1 The industrial process for the preparation of sulphuric acid involves a series of stages. The second stage in this process involves the conversion of sulphur dioxide into sulphur trioxide in a converter as illustrated below. In the converter the gases are passed over vanadium pentoxide(V2O5) placed in layers as shown below. 1.1 Write down the: 1.1.1 Balanced equation for the reaction taking place in the converter (3) 1.1.2 Function of the vanadium pentoxide (1) The table below shows data obtained during the second stage 1.2 Is the reaction in the second stage EXOTHERMIC or ENDOTHERMIC? Refer to the data in the table to give a reason for the answer. (2) 1.3 After the conversion at each layer the gases are cooled down to 450 °C. Fully explain why the gases must be cooled to this temperature. (3) 1.4 During the third stage sulphur trioxide is dissolved in sulphuric acid rather than in water to produce oleum. 1.4.1 Write down the FORMULA of oleum. (1) 1.4.2 Give a reason why sulphur trioxide is not dissolved in water. (1) 1.5 Sulphuric acid reacts with ammonia to form a fertiliser. Write down a balanced equation for this reaction. (3) [14] 82

QUESTION 2 Ammonia is an important fertiliser. Large amounts are prepared from hydrogen and nitrogen in industry. 2.1 For the industrial preparation of ammonia, write down: 2.1.1 The name of the process used (1) 2.1.2 A balanced equation for the reaction that occurs (3) 2.1.3 The source of nitrogen (1) 2.2 The yield of ammonia changes with temperature and pressure during its industrial preparation. The graphs below show how the percentage of ammonia in the reaction mixture that leaves the reaction vessel varies under different conditions. GRAPHS OF THE PERCENTAGE OF AMMONIA IN THE REACTION MIXTURE VERSUS PRESSURE 2.2.1 Use the appropriate graph to estimate the percentage of ammonia present in the reaction mixture at 240 atmosphere and 400 °C (1) 2.2.2 State TWO advantages of using high pressure in the preparation of ammonia. (2) 2.2.3 The advantage of using a low temperature is the large percentage of ammonia formed. What is the disadvantage of using a low temperature? (1) 2.3 Ammonia is also used in the preparation of other fertilisers such as ammonium nitrate. Calculate the mass of nitrogen in a 50 kg bag of pure ammonium nitrate fertiliser. (3) [12] 83

Question 3 A chemical company produces ammonium sulphate, (NH 4)2SO4, starting from the raw materials P, Q and R, as shown in the flow diagram below. 3.1 Write down the NAME of raw material: 3.1.1 P (1) 3.1.2 Q (1) 3.1.3 R (1) 3.2 Write down the: 3.2.1 NAME of process 1 (1) 3.2.2 NAME of compound X (1) 3.2.3 FORMULA of compound Y (1) 3.2.4 Balanced equation for reaction 1 (3) 3.3 The company compares the nitrogen content of ammonium sulphate with that of ammonium nitrate, NH4NO3. 3.3.1 Determine, by performing the necessary calculations, which ONE of the two fertilisers has the higher percentage of nitrogen per mass. (4) 3.3.2 Write down the name of the process that should be included in the flow diagram above if the company wants to prepare ammonium nitrate instead of ammonium sulphate. (1) [14] 84

ACTIVITY 4 (FS September 2014) Thabo wants to start a vegetable garden in his community. The department of agriculture donates the four fertilisers described below. FERTILISER A: 7:1:3 (21) FERTILISER B: 3:2:1 (28) FERTILISER C: 2:3:4 (22) FERTILISER D: Ammonium sulphate 4.1 What is the meaning of (21) indicated at fertiliser A? (1) 4.2 From FERTILISERS A, B and C, choose a fertilizer which is most suitable for: 4.2.1 Spinach (1) 4.2.2 Tomatoes (1) 4.2.3 Maize (1) 4.3 Calculate the: 4.3.1 Percentage of nitrogen in fertiliser B (2) 4.3.2 Mass of potassium present in a 20 kg bag of fertiliser C. (3) 4.4 Write down a balanced equation for the preparation of fertiliser D from ammonia and a suitable acid. (3) [12] ACTIVITY 5 (Gauteng 2014) In the CONTACT process for the manufacture of sulphuric acid the second step of the process produces sulphur (VI) oxide as in the following equation: 2SO2(g) + O2 (g) ⇌ -197 kJ·mol-1 The rate of the formation of SO3 is increased in a number of ways using rate and equilibrium principles. 5.1 Name the catalyst used in the Contact Process. (1) 5.2 Ammonia compounds are necessary for the production of fertiliser. It is manufactured in the Haber process. Give a balanced equation for the Haber process (3) 5.3 Calculate the number of moles of nitrogen in a bag of fertiliser that has a ratio of 7:1:3 (22). The mass of the bag is 5 kg. (5) 5.4 The abuse of large amounts of fertiliser can result in the enrichment of inorganic plant nutrients in water sources. Explain how this leads to dead and decaying plant material. (1) [10] 85