4.6 The rate and extent of chemical change — coverage pack

11 specification leaves · notes, questions, answers and worked methods

4.6.1.1 · Calculating rates of reactions

Mean rate is the quantity of reactant used or product formed divided by the time taken; suitable units include $\text{g s}^{-1}$ and $\text{cm}^3\text{ s}^{-1}$ .
On a quantity-time graph, a steeper gradient means a faster reaction. A curve becoming horizontal shows that the measured quantity is no longer changing.
To find a mean rate, subtract the two quantity readings and divide by the time interval; Higher Tier also uses $\text{mol s}^{-1}$ and calculates an instantaneous rate from a tangent gradient.
A common error is to divide the final reading by the clock time even when the graph or table starts from a non-zero quantity; always calculate the change first.

Tier 1 · Easy

1. A reaction produces $72\,\text{cm}^3$ of gas in $40\,\text{s}$ . Calculate its mean rate.[2 marks]
Answer
- $1.8\,\text{cm}^3\text{ s}^{-1}$
Method: Use $\text{mean rate}=\dfrac{\text{quantity formed}}{\text{time}}=\dfrac{72}{40}=1.8\,\text{cm}^3\text{ s}^{-1}$ .

Tier 2 · Standard

1. 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]
Answer
- $0.024\,\text{g s}^{-1}$
Method: The mass lost is $83.40-82.56=0.84\,\text{g}$ . Therefore the mean rate is $0.84/35=0.024\,\text{g s}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- R: $1.2\,\text{cm}^3\text{ s}^{-1}$
- S: $1.8\,\text{cm}^3\text{ s}^{-1}$
- S is $50\%$ faster than R.
Method: For R, $84/70=1.2\,\text{cm}^3\text{ s}^{-1}$ . For S, $99/55=1.8\,\text{cm}^3\text{ s}^{-1}$ . The increase relative to R is $1.8-1.2=0.6$ , so the percentage increase is $(0.6/1.2)\times100=50\%$ .

4.6.1.2 · Factors which affect the rates of chemical reactions

Reaction rate can be changed by concentration in solution, pressure for gases, surface area of a solid, temperature and the presence of a catalyst.
Increasing concentration, gas pressure, solid surface area or temperature usually increases rate; a suitable catalyst also increases rate.
In the required practical, concentration can be varied while gas volume is measured, or while a colour or turbidity change is timed; other variables must be controlled.
A common error is to change total volume as well as concentration without recognising the extra variable. Use measured volumes and keep temperature, quantities not under test and apparatus consistent.

Tier 1 · Easy

1. A solid reactant is crushed into smaller pieces without changing its mass. State the effect on the reaction rate.[1 mark]
Answer
- The reaction rate increases.
Method: Crushing exposes a greater surface area of the same solid, which increases the reaction rate.

Tier 2 · Standard

1. 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]
Answer
- Use measured acid concentrations with equal volumes, equal masses and sizes of marble chips, and the same temperature. Collect the gas in a gas syringe and record gas volume at regular times, repeating each concentration.
Method: Name the independent variable as acid concentration and a quantitative dependent variable such as gas volume over time. Keep acid volume, marble mass and surface area, temperature and apparatus constant. Start timing on mixing, take readings at fixed intervals and repeat so a mean or anomalies can be considered.

Tier 3 · Hard

1. 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]
Answer
- The time is inversely proportional to concentration for these data: doubling concentration halves the time, and tripling it divides the time by three. Judging disappearance is subjective, and a single end-point time does not show how rate changes during the reaction.
Method: Check concentration multiplied by time: $20(162)=3240$ , $40(81)=3240$ and $60(54)=3240$ , so time is inversely proportional to concentration here. Valid limitations include subjective visual judgement, inconsistent lighting or observer, delay in mixing and timing, and recording only one end point rather than a continuous quantity-time curve.

4.6.1.3 · Collision theory and activation energy

A reaction occurs only when reactant particles collide and the collision has at least the activation energy.
Greater concentration or gas pressure puts more particles into a given volume, so collisions occur more frequently and rate increases.
Higher temperature makes particles move faster, causing more frequent collisions and a larger fraction of collisions with energy at least equal to the activation energy.
Smaller solid pieces have a larger surface area to volume ratio. A common error is to say that temperature only increases collision frequency and omit the greater collision energy.

Tier 1 · Easy

1. State the two conditions needed for a collision between reactant particles to lead to a reaction.[2 marks]
Answer
- The particles must collide, and the collision must have energy equal to or greater than the activation energy.
Method: Collision alone is insufficient. Award one condition for a collision occurring and one for sufficient collision energy.

Tier 2 · Standard

1. Explain, using collision theory, why increasing the pressure of two reacting gases increases their reaction rate at constant temperature.[3 marks]
Answer
- The same number of gas particles occupies a smaller volume, so particles are closer together and collide more frequently. This produces more successful collisions per second.
Method: Link the macroscopic change to particles: higher pressure at constant temperature means a greater particle concentration. Collision frequency therefore rises, so the number of collisions with sufficient energy per second rises and the reaction is faster.

Tier 3 · Hard

1. 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]
Answer
- $192\,\text{cm}^2$ for the $2.0\,\text{cm}$ cubes
- $384\,\text{cm}^2$ for the $1.0\,\text{cm}$ cubes
- The $1.0\,\text{cm}$ cubes react faster because their greater exposed area gives more frequent collisions at the solid surface.
Method: A $2.0\,\text{cm}$ cube has volume $8.0\,\text{cm}^3$ , so there are $64/8=8$ cubes. Each has area $6(2.0^2)=24\,\text{cm}^2$ , giving $8(24)=192\,\text{cm}^2$ . There are $64$ cubes of side $1.0\,\text{cm}$ , each with area $6\,\text{cm}^2$ , giving $64(6)=384\,\text{cm}^2$ . The doubled exposed area permits more acid-particle collisions per second, increasing the rate.

4.6.1.4 · Catalysts

A catalyst increases reaction rate without being used up overall; different reactions may require different catalysts, and enzymes are biological catalysts.
Catalysts provide a different reaction pathway with a lower activation energy, so a greater fraction of collisions can lead to reaction at the same temperature.
On a reaction profile, the catalysed curve has a lower peak but the reactant and product energy levels, and therefore the overall energy change, stay the same.
A common error is to say a catalyst gives particles more energy or increases product yield. It changes the pathway and rate, not the energy levels or final amount fixed by a limiting reactant.

Tier 1 · Easy

1. Complete the statement: a catalyst increases reaction rate by providing a different pathway with a lower what?[1 mark]
Answer
- Activation energy.
Method: The defining energy change is a lower activation energy for the alternative pathway.

Tier 2 · Standard

1. 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]
Answer
- The activation energy is reduced by $36\,\text{kJ mol}^{-1}$ ; the overall energy change remains $-24\,\text{kJ mol}^{-1}$ .
Method: The reduction is $79-43=36\,\text{kJ mol}^{-1}$ . A catalyst changes the pathway but not the reactant or product energy levels, so the overall change stays $-24\,\text{kJ mol}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- X raises the initial mean rate from $0.70$ to $1.6\,\text{cm}^3\text{ s}^{-1}$ , does not change the final gas volume, and is recovered unchanged, so it increases rate without being used up.
Method: Without X the first- $30$ -second mean rate is $21/30=0.70\,\text{cm}^3\text{ s}^{-1}$ ; with X it is $48/30=1.6\,\text{cm}^3\text{ s}^{-1}$ . The equal final volume shows that X changes how quickly product forms rather than the final quantity. Its unchanged mass shows it was not used up overall. Together these observations identify X as a catalyst.

4.6.2.1 · Reversible reactions

In a reversible reaction, products can react to form the original reactants; the equation uses the reversible arrow $\rightleftharpoons$ .
Changing conditions can favour one direction, so the same chemical system may be driven towards products or back towards reactants.
For a general reaction $\mathrm{A+B\rightleftharpoons C+D}$ , the forward process forms C and D while the reverse process consumes C and D to reform A and B.
A common error is to treat the double arrow as meaning the reaction must have reached equilibrium. Reversibility is possible even before the two rates become equal.

Tier 1 · Easy

1. State what is meant by a reversible reaction.[1 mark]
Answer
- The products can react to form the original reactants.
Method: The essential idea is that the reaction can proceed in the reverse direction, turning products back into reactants.

Tier 2 · Standard

1. 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]
Answer
- L and M react to form J and K.
Method: Read the equation from right to left for the reverse direction: the right-hand substances L and M are its reactants, and J and K are its products.

Tier 3 · Hard

1. 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]
Answer
- Heating drives the hydrated salt towards anhydrous copper sulfate and water; adding water drives the products back to hydrated copper sulfate. Products reform the original reactant, so the reaction is reversible.
Method: Identify the forward observation under heating, then the reverse observation when water is supplied. The recovery of the original blue hydrated substance from the products is the evidence of reversibility; the conditions are heating in one direction and addition of water in the other.

4.6.2.2 · Energy changes and reversible reactions

If the forward direction of a reversible reaction is exothermic, the reverse direction is endothermic.
The two directions transfer the same amount of energy with opposite signs because one direction exactly reverses the energy change of the other.
A forward energy change of $-q$ corresponds to a reverse energy change of $+q$ ; reactant and product energy levels exchange roles.
A common error is to change only the sign but not recognise the change in energy flow: the exothermic direction releases energy and the endothermic direction takes it in.

Tier 1 · Easy

1. The forward direction of a reversible reaction is exothermic. State the energy-change type of the reverse direction.[1 mark]
Answer
- Endothermic.
Method: Reversing an exothermic process reverses the direction of energy transfer, so the reverse process is endothermic.

Tier 2 · Standard

1. 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]
Answer
- $+67\,\text{kJ mol}^{-1}$
Method: Transferring energy to the surroundings makes the forward change $-67\,\text{kJ mol}^{-1}$ . The reverse reaction transfers the same amount in the opposite direction, so its change is $+67\,\text{kJ mol}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- Reverse activation energy $=183\,\text{kJ mol}^{-1}$ ; reverse overall energy change $=+38\,\text{kJ mol}^{-1}$ .
Method: The products are $38\,\text{kJ mol}^{-1}$ below the reactants. The peak is $145\,\text{kJ mol}^{-1}$ above the reactants, so it is $145+38=183\,\text{kJ mol}^{-1}$ above the products. Reversing the reaction changes the sign of the overall change, giving $+38\,\text{kJ mol}^{-1}$ .

4.6.2.3 · Equilibrium

A reversible reaction can reach dynamic equilibrium only in a closed system that prevents reactants and products from escaping.
At equilibrium, the forward and reverse reactions continue at exactly the same rate; neither reaction has stopped.
Because the two rates are equal, the macroscopic amounts or concentrations remain constant even though particles continue to react in both directions.
A common error is to say equilibrium means equal amounts of reactants and products. The amounts are constant, but they do not have to be equal.

Tier 1 · Easy

1. State the relationship between the forward and reverse reaction rates at equilibrium.[1 mark]
Answer
- The forward and reverse reactions occur at the same rate.
Method: Dynamic equilibrium is defined by equality of the forward and reverse rates.

Tier 2 · Standard

1. A student says, 'The concentrations stay constant at equilibrium because both reactions have stopped.' Explain why this statement is incorrect.[3 marks]
Answer
- Both reactions continue. Their rates are equal, so each substance is formed at the same rate as it is used and its concentration stays constant.
Method: Correct the word 'stopped': equilibrium is dynamic. Then connect equal opposing rates to no net change in concentration.

Tier 3 · Hard

1. 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]
Answer
- Equilibrium is first reached at $40\,\text{s}$ . The equal, non-zero rates at $40$ and $60\,\text{s}$ show dynamic equilibrium continues with no net change.
Method: Find the first row in which the two rates are equal: both are $2.8$ at $40\,\text{s}$ . Their remaining equal and non-zero at $60\,\text{s}$ shows that both directions continue at a shared rate rather than stopping.

4.6.2.4 · The effect of changing conditions on equilibrium (HT only)

The relative amounts of reactants and products at equilibrium depend on the reaction conditions.
If an equilibrium condition changes, the system responds in the direction that counteracts that change; this is Le Chatelier's principle.
Predict a shift by identifying the imposed change first, then choosing the direction that consumes an addition or replaces a removal, or that opposes a temperature or pressure change.
A common error is to assume every change moves equilibrium towards products. Some changes favour reactants, and simultaneous changes can have opposing effects.

Tier 1 · Easy

1. State Le Chatelier's principle for a system at equilibrium when a condition is changed.[1 mark]
Answer
- The system responds so as to counteract the change.
Method: The required principle is that the equilibrium shifts in the direction that opposes the imposed change.

Tier 2 · Standard

1. 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]
Answer
- The equilibrium shifts to the right. More R reacts to replace some of the removed P, counteracting the decrease in P concentration.
Method: Identify the change as a decrease in product concentration. The counteracting response is the forward reaction, which forms more P, so the position moves right.

Tier 3 · Hard

1. 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]
Answer
- Increasing B favours the forward reaction and more C, but increasing temperature favours the endothermic reverse reaction and less C. The effects oppose one another, and their relative sizes are not given.
Method: Treat each change separately. Extra B is counteracted by consuming B, shifting right and increasing C. Extra thermal energy is counteracted by the endothermic reverse direction, shifting left and decreasing C. Because no data compare the magnitudes of these shifts, the net change cannot be decided.

4.6.2.5 · The effect of changing concentration (HT only)

Increasing a reactant concentration shifts equilibrium towards products until a new equilibrium is established.
Decreasing a product concentration also shifts equilibrium towards products because the forward reaction replaces some of the removed product.
For any concentration change, identify which side contains the changed substance and choose the direction that consumes an addition or replaces a removal.
A common error is to say the imposed concentration is completely restored. The system only counteracts the change, and all concentrations settle at new constant values.

Tier 1 · Easy

1. A reactant is added to a mixture at equilibrium. State the direction in which the equilibrium shifts.[1 mark]
Answer
- Towards the products.
Method: The system counteracts the added reactant by consuming some of it in the forward reaction, so the position shifts towards products.

Tier 2 · Standard

1. 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]
Answer
- The equilibrium shifts to the right, so more D and E react and more F is formed to oppose its removal.
Method: Removing F lowers a product concentration. The forward reaction replaces some F, so reactants are consumed and the equilibrium position shifts right.

Tier 3 · Hard

1. 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]
Answer
- The equilibrium shifts right. Some of the added hydrogen and some nitrogen are consumed, while more ammonia is formed; the hydrogen increase is partly, not completely, opposed.
Method: The imposed change is increased $\mathrm{H_2}$ concentration. The forward reaction consumes $\mathrm{H_2}$ , so it is favoured. Consequently $\mathrm{N_2}$ is also used and $\mathrm{NH_3}$ increases until the opposing rates are equal again. Do not claim that all added hydrogen disappears.

4.6.2.6 · The effect of temperature changes on equilibrium (HT only)

Increasing temperature favours the endothermic direction because that direction takes in energy and counteracts the heating.
Decreasing temperature favours the exothermic direction because that direction releases energy and counteracts the cooling.
First label the forward reaction as exothermic or endothermic, then apply the temperature change to predict whether the relative amount of products rises or falls.
A common error is to use the rule that higher temperature increases rate and conclude that product yield must rise. Both directions speed up; equilibrium position depends on energy transfer.

Tier 1 · Easy

1. The forward reaction is endothermic. State the effect of increasing temperature on the relative amount of products at equilibrium.[1 mark]
Answer
- The relative amount of products increases.
Method: Heating favours the direction that takes in energy. Here that is the endothermic forward reaction, so the equilibrium shifts towards products.

Tier 2 · Standard

1. The forward direction of $\mathrm{X\rightleftharpoons Y}$ is exothermic. Explain the effect of decreasing temperature on the equilibrium yield of Y.[3 marks]
Answer
- The yield of Y increases because the equilibrium shifts in the exothermic forward direction, which releases energy and counteracts the cooling.
Method: A temperature decrease favours the energy-releasing direction. Since the forward direction is exothermic, the position moves right and the relative amount of Y rises.

Tier 3 · Hard

1. 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]
Answer
- The forward reaction is exothermic because increasing temperature lowers the product yield. An intermediate temperature compromises between a higher equilibrium yield at low temperature and a faster production rate at high temperature.
Method: The decreasing product yield as temperature rises shows that heating favours the reverse direction. The reverse direction is therefore endothermic and the forward direction exothermic. Low temperature favours products but gives a slow reaction; high temperature gives a faster reaction but a poorer equilibrium yield. An intermediate value balances these competing effects.

4.6.2.7 · The effect of pressure changes on equilibrium (HT only)

For gaseous equilibria, increasing pressure shifts the position towards the side with fewer gas molecules in the balanced symbol equation.
Decreasing pressure shifts the position towards the side with more gas molecules; only gaseous species are counted for this rule.
Count the stoichiometric coefficients of gases on both sides before predicting a shift. If the totals are equal, pressure does not change the equilibrium position.
A common error is to count different chemical formulae rather than molecules, or to include solids and liquids when applying the pressure rule.

Tier 1 · Easy

1. 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]
Answer
- To the right.
Method: Increasing pressure favours the side with fewer gas molecules. One on the right is fewer than three on the left.

Tier 2 · Standard

1. 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]
Answer
- The ammonia yield increases because the equilibrium shifts right, from four gas molecules to two, counteracting the pressure increase.
Method: Count gaseous coefficients: $1+3=4$ on the left and $2$ on the right. Higher pressure favours the side with fewer gas molecules, so the position shifts right and produces relatively more ammonia.

Tier 3 · Hard

1. 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]
Answer
- For (1), the equilibrium shifts left and the relative amount of $\mathrm{SO_3}$ decreases. For (2), there is no shift because each side has two gas molecules.
Method: A pressure decrease favours more gas molecules. In (1), the left has $2+1=3$ gas molecules and the right has $2$ , so the position shifts left and product amount falls. In (2), the left has $1+1=2$ and the right has $2$ ; because the gas totals are equal, changing pressure does not alter the equilibrium position.