4.6 The rate and extent of chemical change

4.6.1.1

Calculating rates of reactions

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Reaction rate measures how quickly a reactant is used or a product is formed.
Calculate mean rate using $\text{mean rate}=\dfrac{\text{change in quantity}}{\text{time taken}}$ ; quantity may be mass in grams or gas volume in $\text{cm}^3$ , giving units such as $\text{g s}^{-1}$ or $\text{cm}^3\text{ s}^{-1}$ .
On a quantity–time graph, gradient represents rate: a steeper section is faster and a horizontal section has zero rate.
A tangent gives the rate at one instant.
Higher tier: calculate the tangent's gradient and also express quantity in moles, with units $\text{mol s}^{-1}$ .

A product–time curve with a tangent whose gradient gives the instantaneous rate.

Worked example

The mass of a flask decreases from $92.80\,\text{g}$ to $91.96\,\text{g}$ in $35\,\text{s}$ . Calculate the mean rate of mass loss.

1.Find the mass used: $92.80-91.96=0.84\,\text{g}$ .
2.Substitute into the rate equation: $\text{rate}=0.84\div35$ .
3.Evaluate and include the correct compound unit: $0.024\,\text{g s}^{-1}$ .

Answer: $0.024\,\text{g s}^{-1}$

Common mistakes

Don't fall into the trap of dividing the final balance reading by the time instead of first finding the change in mass.
Don't fall into the trap of using the whole curve for an instantaneous rate instead of drawing a tangent at the stated time.

Exam tip

For a ‘calculate the rate’ question, show change ÷ time and give a quantity-per-second unit.

Tier 1 · Easy

ORIGINAL

A reaction produces

72\,\text{cm}^3

of gas in

40\,\text{s}

. Calculate its mean rate.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The mass of a reaction flask falls from

83.40\,\text{g}

to

82.56\,\text{g}

during the first

35\,\text{s}

. Calculate the mean rate of mass loss.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Reaction R forms

84\,\text{cm}^3

of gas in

70\,\text{s}

. Reaction S forms

99\,\text{cm}^3

in

55\,\text{s}

. Calculate both mean rates and determine the percentage by which S is faster than R.

[5 marks]

Total for this question: 5

4.6.1.2

Factors which affect the rates of chemical reactions

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Reaction rate is affected by reactant concentration in solution, gas pressure, the surface area of a solid, temperature and catalysts.
Increasing concentration, pressure, surface area or temperature increases rate; a suitable catalyst also increases it.
Required practical 5 investigates concentration in two ways: measuring gas volume produced over time and timing a colour or turbidity change.
Develop a hypothesis, vary only concentration and keep temperature, total liquid volume, amount and surface area of solid, and apparatus controlled.
Start timing when reactants mix, collect quantitative results and repeat trials so a mean can be calculated and anomalies identified.

Worked example

Describe how to investigate the effect of sodium thiosulfate concentration on reaction rate using a disappearing cross.

1.Place measured sodium thiosulfate solution over a marked cross and add a fixed volume and concentration of acid.
2.Start the timer on mixing and stop it when the cross is no longer visible through the sulfur precipitate.
3.Repeat for several thiosulfate concentrations while keeping total volume, acid, temperature, viewing position and cross constant.
4.Repeat each concentration and compare mean $1\div\text{time}$ values.

Answer: A controlled concentration series with repeated disappearance times provides comparable relative-rate data.

Common mistakes

Don't fall into the trap of changing acid concentration by changing its volume, so concentration and total volume both become independent variables.
Don't fall into the trap of calling disappearance time the reaction rate without using a comparable measure such as $1\div\text{time}$ .

Exam tip

In a practical-method answer, name the independent, dependent and at least two controlled variables before describing repeats.

Tier 1 · Easy

ORIGINAL

A solid reactant is crushed into smaller pieces without changing its mass. State the effect on the reaction rate.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student investigates how acid concentration affects the rate of gas production from marble chips. Describe how the student should collect suitable results while changing only the acid concentration.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

In a turbidity experiment, relative acid concentrations of

20\%

,

40\%

and

60\%

give disappearance times of

162

,

81

and

54\,\text{s}

. Describe the relationship shown and give two limitations of using the disappearance time as a rate measurement.

[5 marks]

Total for this question: 5

4.6.1.3

Collision theory and activation energy

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Collision theory states that reacting particles must collide with sufficient energy. The minimum energy needed is the activation energy.
Greater solution concentration or gas pressure places more particles in a given volume, so collisions occur more frequently.
Breaking a solid into smaller pieces increases its surface-area-to-volume ratio, exposing more particles for collisions.
Raising temperature has two effects: particles collide more often and collisions are more energetic, so a greater proportion meet or exceed the activation energy.
Examiners expect a complete chain from the changed condition to collision frequency or energy, then to more successful collisions per second and a faster rate.

Worked example

A $27\,\text{cm}^3$ cube is cut into $1\,\text{cm}$ cubes. Compare the total surface area before and after cutting and explain the rate change.

1.The original cube has side $3\,\text{cm}$ , so area $=6(3^2)=54\,\text{cm}^2$ .
2.There are $27$ small cubes, each with area $6(1^2)=6\,\text{cm}^2$ .
3.Total area after cutting $=27\times6=162\,\text{cm}^2$ , three times larger.
4.The greater exposed area gives more frequent collisions at the solid surface, so reaction is faster.

Answer: Surface area increases from $54$ to $162\,\text{cm}^2$ , so the smaller cubes react faster.

Common mistakes

Don't fall into the trap of saying smaller pieces contain more particles even though the mass and number of particles are unchanged.
Don't fall into the trap of explaining temperature only through more frequent collisions and omitting that collisions are more energetic.

Exam tip

For an ‘explain using collision theory’ question, finish with ‘more successful collisions per second’.

Tier 1 · Easy

ORIGINAL

State the two conditions needed for a collision between reactant particles to lead to a reaction.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain, using collision theory, why increasing the pressure of two reacting gases increases their reaction rate at constant temperature.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A fixed

64\,\text{cm}^3

of solid is cut either into cubes of side

2.0\,\text{cm}

or cubes of side

1.0\,\text{cm}

. Calculate the total surface area for each set and explain which reacts faster with an acid.

[6 marks]

Total for this question: 6

4.6.1.4

Catalysts

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A catalyst changes the rate of a reaction but is not used up overall; different reactions need different catalysts, and enzymes are biological catalysts. It provides an alternative pathway with a lower activation energy.
At the same temperature, a greater fraction of collisions therefore has enough energy to react, so the reaction is faster.
On a reaction profile the catalysed pathway has a lower peak, while the reactant and product energy levels remain fixed.
A catalyst can be identified because rate increases and it is not included in the overall chemical equation.
It does not give particles extra energy or change the reaction’s overall energy transfer.

Catalysed and uncatalysed pathways have different activation energies but the same energy change.

Worked example

A substance is recovered with unchanged mass after making a reaction faster. Explain why it may be a catalyst.

1.State the rate evidence: the reaction is faster when the substance is present.
2.Use the recovery evidence: unchanged mass shows it was not used up overall.
3.Explain the action: it provides an alternative pathway with lower activation energy.

Answer: It fits the definition of a catalyst because it increases rate without being used up overall.

Common mistakes

Don't fall into the trap of writing that a catalyst supplies energy to reactant particles.
Don't fall into the trap of drawing different reactant or product energy levels for the catalysed pathway.

Exam tip

To explain catalytic action, state ‘alternative pathway’ and ‘lower activation energy’ for both marks.

Tier 1 · Easy

ORIGINAL

Complete the statement: a catalyst increases reaction rate by providing a different pathway with a lower what?

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

An uncatalysed reaction has an activation energy of

79\,\text{kJ mol}^{-1}

and an overall energy change of

-24\,\text{kJ mol}^{-1}

. A catalyst lowers the activation energy to

43\,\text{kJ mol}^{-1}

. State the reduction in activation energy and the catalysed reaction's overall energy change.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In the first

30\,\text{s}

, a reaction forms

48\,\text{cm}^3

of gas with solid X and

21\,\text{cm}^3

without X. Both tests eventually form

72\,\text{cm}^3

, and the dry mass of X is unchanged. Use the data to explain why X is a catalyst.

[5 marks]

Total for this question: 5

4.6.2.1

Reversible reactions

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

In a reversible reaction, products can react to produce the original reactants.
The equation uses $\rightleftharpoons$ : for $\mathrm{A+B\rightleftharpoons C+D}$ , A and B form C and D in the forward direction, while C and D reform A and B in the reverse direction.
Changing the conditions can change which direction is favoured.
A familiar example is hydrated copper sulfate: heating blue hydrated copper sulfate produces white anhydrous copper sulfate and water, while adding water reverses the change.
Reversible describes the ability to proceed both ways; it does not by itself mean that equilibrium has been reached.

Worked example

Heating blue hydrated copper sulfate makes a white solid. Adding water makes it blue again. Explain what this shows.

1.Heating changes hydrated copper sulfate into anhydrous copper sulfate and water.
2.Adding water makes the products reform the original hydrated copper sulfate.
3.Products reforming the original reactant is evidence that the reaction is reversible.

Answer: The observations show a reversible reaction whose direction changes with the conditions.

Common mistakes

Don't fall into the trap of reading the right-hand substances as products in both directions instead of reactants for the reverse reaction.
Don't fall into the trap of claiming the reversible arrow means the forward and reverse rates are already equal.

Exam tip

When asked for evidence of reversibility, state that the products reform the original reactants under changed conditions.

Tier 1 · Easy

ORIGINAL

State what is meant by a reversible reaction.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

The reaction

\mathrm{J+K\rightleftharpoons L+M}

is reversible. State which substances react in the reverse reaction and which substances they form.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Blue hydrated copper sulfate is heated and forms white anhydrous copper sulfate and water. Adding water to the white solid reforms the blue substance. Explain how these observations show a reversible reaction and identify the change of condition used in each direction.

[4 marks]

Total for this question: 4

4.6.2.2

Energy changes and reversible reactions

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

The two directions of a reversible reaction have opposite energy transfers.
If the forward reaction is exothermic, it transfers energy to the surroundings; the reverse reaction must take the same amount of energy from the surroundings and is endothermic.
Conversely, an endothermic forward reaction has an exothermic reverse.
On an energy-level diagram the reactant and product levels simply exchange roles when the direction is reversed, so an energy change of $-q$ forwards becomes $+q$ backwards.
Examiners expect both the opposite classification and the fact that the magnitude of the energy transfer is unchanged.

Worked example

The forward direction of a reversible reaction transfers $38\,\text{kJ}$ to the surroundings. Describe the reverse reaction.

1.Transferring energy to the surroundings makes the forward reaction exothermic.
2.The reverse reaction has the opposite energy transfer, so it is endothermic.
3.The same energy magnitude is involved, so the reverse takes in $38\,\text{kJ}$ .

Answer: The reverse reaction is endothermic and takes in $38\,\text{kJ}$ .

Common mistakes

Don't fall into the trap of writing that the reverse reaction transfers a different amount of energy.
Don't fall into the trap of changing the sign of the energy change without stating whether energy enters or leaves the surroundings.

Exam tip

For a reverse-reaction energy question, keep the numerical magnitude, reverse the sign and name the opposite energy-transfer type.

Tier 1 · Easy

ORIGINAL

The forward direction of a reversible reaction is exothermic. State the energy-change type of the reverse direction.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

The forward direction of a reversible reaction transfers

67\,\text{kJ mol}^{-1}

to the surroundings. State the energy change for the reverse direction, including its sign.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A reversible reaction has a forward activation energy of

145\,\text{kJ mol}^{-1}

and a forward overall energy change of

-38\,\text{kJ mol}^{-1}

. Calculate the reverse activation energy and state the reverse overall energy change.

[4 marks]

Total for this question: 4

4.6.2.3

Equilibrium

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Dynamic equilibrium is reached when a reversible reaction occurs in a closed system, so reactants and products cannot escape, and the forward and reverse reactions happen at exactly the same rate. Both reactions continue: ‘dynamic’ means particles still react in each direction.
Equal rates make the observable amounts or concentrations of every substance remain constant.
Those amounts need not be equal; an equilibrium mixture may contain much more reactant than product.
If material escapes from the apparatus, a stable equilibrium cannot be maintained.
In definitions, AQA expects the closed-system condition and equality of forward and reverse rates.

Inside one closed container, separate A and B particles and joined AB pairs are intermixed. Equal forward and reverse arrows show A and B joining as AB and AB separating back into A and B.

Worked example

A sealed reaction mixture has constant concentrations, but particles continue changing between reactants and products. Explain this observation.

1.The sealed apparatus provides the required closed system.
2.Constant concentrations are consistent with equal forward and reverse reaction rates.
3.Particles still reacting in both directions shows that the equilibrium is dynamic.

Answer: The mixture is at dynamic equilibrium: forward and reverse reactions continue at equal rates.

Common mistakes

Don't fall into the trap of stating that both reactions stop when equilibrium is reached.
Don't fall into the trap of stating that equilibrium requires equal amounts of reactants and products rather than equal reaction rates.

Exam tip

A two-mark definition normally needs both ‘closed system’ and ‘forward and reverse reactions at the same rate’.

Tier 1 · Easy

ORIGINAL

State the relationship between the forward and reverse reaction rates at equilibrium.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student says, 'The concentrations stay constant at equilibrium because both reactions have stopped.' Explain why this statement is incorrect.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a sealed vessel, the measured forward and reverse rates in arbitrary units are: at

0\,\text{s}

,

5.2

and

0.0

; at

20\,\text{s}

,

3.6

and

1.7

; at

40\,\text{s}

,

2.8

and

2.8

; at

60\,\text{s}

,

2.8

and

2.8

. Determine when equilibrium is first reached and explain what the later readings show.

[4 marks]

Total for this question: 4

4.6.2.4

The effect of changing conditions on equilibrium (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: the relative amounts of reactants and products at equilibrium depend on the conditions.
Le Chatelier’s principle states that when a condition changes, the equilibrium system responds in the direction that counteracts that change.
A prediction therefore needs three parts: identify the imposed change, decide which direction opposes it, then state whether the relative amount of product increases or decreases.
Concentration, temperature and gas pressure can change equilibrium position, but their rules differ.
The response only partially counteracts the disturbance and produces a new equilibrium; it does not necessarily restore the original composition.

Worked example

Higher tier: $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ is at equilibrium. Predict and explain what happens to the equilibrium position and the relative amount of ammonia when the pressure is increased at constant temperature.

1.Count the gaseous molecules from the balanced equation: there are $1+3=4$ on the left and $2$ on the right.
2.Increasing pressure favours the side with fewer gas molecules, which counteracts the pressure increase.
3.The equilibrium therefore shifts to the right, so the relative amount of ammonia increases as a new equilibrium is reached.

Answer: The equilibrium shifts to the right and the relative amount of $\mathrm{NH_3}$ increases because the product side has fewer gas molecules: $2$ compared with $4$ on the reactant side.

Common mistakes

Don't fall into the trap of writing that every condition change shifts equilibrium towards products.
Don't fall into the trap of describing a faster reaction rate without stating the new relative amount of product at equilibrium.

Exam tip

For a ‘predict and explain’ question, name the direction of shift, the counteracted change and the effect on product amount.

Tier 1 · Easy

ORIGINAL

State Le Chatelier's principle for a system at equilibrium when a condition is changed.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

For

\mathrm{R\rightleftharpoons2P}

at equilibrium, some P is removed. Predict the direction of shift and explain it using Le Chatelier's principle.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

The forward reaction

\mathrm{A+2B\rightleftharpoons C}

is exothermic. At equilibrium, the concentration of B and the temperature are both increased. Explain why the information given is insufficient to predict the final change in the amount of C.

[4 marks]

Total for this question: 4

4.6.2.5

The effect of changing concentration (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: changing the concentration of a reactant or product disturbs equilibrium, so all concentrations change until a new equilibrium is reached.
If a reactant concentration is increased, the system shifts towards products and uses some of the added reactant.
If a product concentration is decreased, the system also shifts towards products and replaces some of what was removed.
Apply the same counteracting logic to any named substance: an added substance is consumed and a removed substance is replaced.
The shift changes relative amounts but does not completely undo the imposed change, and the concentrations eventually become constant at new values.

Worked example

For $\mathrm{A+B\rightleftharpoons C+D}$ , some C is removed from an equilibrium mixture. Predict the effect.

1.C is a product, so removing it decreases a product concentration.
2.The system counteracts the removal by favouring the forward reaction.
3.More A and B react, producing more C and D until a new equilibrium is reached.

Answer: Equilibrium shifts to the right and the relative amounts of C and D increase.

Common mistakes

Don't fall into the trap of assuming increasing any concentration must increase the amount of product.
Don't fall into the trap of saying the system fully restores the original concentration rather than only counteracting the change.

Exam tip

Circle whether the changed substance is a reactant or product, then choose the direction that consumes an addition or replaces a removal.

Tier 1 · Easy

ORIGINAL

A reactant is added to a mixture at equilibrium. State the direction in which the equilibrium shifts.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

For

\mathrm{D+E\rightleftharpoons F}

, some F is continuously removed from an equilibrium mixture. Explain the effect on the relative amount of F that is formed.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

For

\mathrm{N_2+3H_2\rightleftharpoons2NH_3}

at equilibrium, extra hydrogen is added while the temperature is kept constant. Predict the effect of this concentration change on the amounts of all three substances as a new equilibrium is reached.

[4 marks]

Total for this question: 4

4.6.2.6

The effect of temperature changes on equilibrium (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: treat energy as part of the equilibrium: the endothermic direction takes in energy and the exothermic direction releases it.
Increasing temperature shifts equilibrium in the endothermic direction, counteracting the added heat; decreasing temperature shifts it in the exothermic direction, counteracting the cooling.
Therefore, for an endothermic forward reaction, heating increases the relative amount of products and cooling decreases it.
For an exothermic forward reaction, the pattern is reversed.
Temperature also affects reaction rates, but an equilibrium-position answer must use the direction of energy transfer to predict the final relative amounts.

Worked example

The forward reaction in $\mathrm{X\rightleftharpoons Y}$ is exothermic. Predict the effect of increasing temperature.

1.The forward direction is exothermic, so the reverse direction is endothermic.
2.Increasing temperature favours the endothermic direction to counteract the heating.
3.The equilibrium shifts towards X, so the relative amount of Y decreases.

Answer: The equilibrium shifts left and the equilibrium yield of Y decreases.

Common mistakes

Don't fall into the trap of claiming higher temperature always increases product yield because reactions happen faster.
Don't fall into the trap of using the sign of the forward reaction without first identifying which direction is endothermic.

Exam tip

Annotate the equation with ‘exo’ and ‘endo’ before deciding which direction counteracts the temperature change.

Tier 1 · Easy

ORIGINAL

The forward reaction is endothermic. State the effect of increasing temperature on the relative amount of products at equilibrium.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

The forward direction of

\mathrm{X\rightleftharpoons Y}

is exothermic. Explain the effect of decreasing temperature on the equilibrium yield of Y.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

For the same equilibrium, product yields are

68\%

at

300\,\text{K}

,

49\%

at

400\,\text{K}

and

32\%

at

500\,\text{K}

. Reaction time falls as temperature rises. Deduce the energy-change type of the forward reaction and explain why an industrial process might use an intermediate temperature.

[5 marks]

Total for this question: 5

4.6.2.7

The effect of pressure changes on equilibrium (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: pressure changes affect equilibria involving gases.
Increasing pressure shifts equilibrium towards the side with the smaller number of gas molecules, which counteracts the increase; decreasing pressure favours the side with more gas molecules.
Use the balanced equation and add the coefficients of gaseous species only.
For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ , there are four gas molecules on the left and two on the right, so increasing pressure favours ammonia.
If both sides contain equal numbers of gas molecules, pressure does not change the equilibrium position, although it may still affect reaction rates.

Worked example

Predict the effect of increasing pressure on $\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}$ .

1.Count gaseous molecules from the coefficients: $2+1=3$ on the left and $2$ on the right.
2.Increasing pressure favours the side with fewer gas molecules.
3.The equilibrium shifts right, increasing the relative amount of sulfur trioxide.

Answer: The equilibrium shifts towards $\mathrm{SO_3}$ because the product side has fewer gas molecules.

Common mistakes

Don't fall into the trap of counting the number of different gas formulae instead of adding their balanced coefficients.
Don't fall into the trap of including solids or liquids when comparing the numbers of gas molecules.

Exam tip

Write the gas-molecule totals under both sides of the equation before stating the pressure shift.

Tier 1 · Easy

ORIGINAL

A gaseous equilibrium has three gas molecules on the left of its equation and one on the right. State the direction of shift when pressure is increased.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

For

\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}

, predict and explain the effect of increasing pressure on the equilibrium yield of ammonia.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Pressure is decreased for each equilibrium: (1)

\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}

; (2)

\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}

. Predict the effect on the product amount in each case and justify both predictions.

[5 marks]

Total for this question: 5

Calculating rates of reactions

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Factors which affect the rates of chemical reactions

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Collision theory and activation energy

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Catalysts

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Reversible reactions

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Energy changes and reversible reactions

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Equilibrium

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The effect of changing conditions on equilibrium (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The effect of changing concentration (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The effect of temperature changes on equilibrium (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The effect of pressure changes on equilibrium (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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