1 A mixture of gases is heated in a closed container. The reaction rate increases.
Which statement explains why the rate increases?
A More molecules have an energy greater than the activation energy.
B The activation energy decreases.
C The activation energy increases.
D The concentration of the gases increases.
Your answer [1]
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, 2 Crystal violet (CV) is a purple dye. In the presence of an alkali, CV reacts to form a colourless product.
A student uses a colorimeter to investigate the rate of the reaction between CV and sodium hydroxide, NaOH.
• The student mixes 10.0 cm3 of 2.8 × 10–7 mol dm–3 CV with 10.0 cm3 of 0.016 mol dm–3 NaOH.
• A large excess of NaOH is used, so that the reaction is effectively zero-order with respect to OH– ions.
• The student places a sample of the reaction mixture in a colorimeter and measures the absorbance over
time.
The student uses the absorbance readings to calculate the concentration of CV and plots a graph of
concentration of CV against time, as shown below.
Using collision theory, explain why the gradient decreases over time.
[1]
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, 3 Which row in the table explains how a catalyst affects the activation energy (Ea) and the proportion of molecules
with energy > Ea?
How the activation energy Proportion of molecules with energy > Ea
changes
A decreases decreases
B decreases increases
C increases decreases
D increases increases
Your answer [1]
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, 4(a) This question is about energy changes.
Hydrogen peroxide decomposes as shown in Reaction 16.1.
Reaction 16.1
The table shows enthalpy changes of formation and entropies.
ΔHfe/kJ mol1 Se / J K–-1 mol–1
H2O2(l) –188 110
H2O(l) –286 70.0
O2(g) 0 205
i. Calculate the free-energy change, ΔG, in kJ mol–1, of Reaction 16.1 at 25 °C.
Give your answer to 3 significant figures.
ΔG = .............................................. kJ mol–1 [4]
ii. The decomposition of hydrogen peroxide shown in Reaction 16.1 is feasible.
Suggest why Reaction 16.1 does not take place at 25 °C despite being feasible.
[1]
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Which statement explains why the rate increases?
A More molecules have an energy greater than the activation energy.
B The activation energy decreases.
C The activation energy increases.
D The concentration of the gases increases.
Your answer [1]
© OCR 2025. You may photocopy this page. 1 of 65 Created in ExamBuilder
, 2 Crystal violet (CV) is a purple dye. In the presence of an alkali, CV reacts to form a colourless product.
A student uses a colorimeter to investigate the rate of the reaction between CV and sodium hydroxide, NaOH.
• The student mixes 10.0 cm3 of 2.8 × 10–7 mol dm–3 CV with 10.0 cm3 of 0.016 mol dm–3 NaOH.
• A large excess of NaOH is used, so that the reaction is effectively zero-order with respect to OH– ions.
• The student places a sample of the reaction mixture in a colorimeter and measures the absorbance over
time.
The student uses the absorbance readings to calculate the concentration of CV and plots a graph of
concentration of CV against time, as shown below.
Using collision theory, explain why the gradient decreases over time.
[1]
© OCR 2025. You may photocopy this page. 2 of 65 Created in ExamBuilder
, 3 Which row in the table explains how a catalyst affects the activation energy (Ea) and the proportion of molecules
with energy > Ea?
How the activation energy Proportion of molecules with energy > Ea
changes
A decreases decreases
B decreases increases
C increases decreases
D increases increases
Your answer [1]
© OCR 2025. You may photocopy this page. 3 of 65 Created in ExamBuilder
, 4(a) This question is about energy changes.
Hydrogen peroxide decomposes as shown in Reaction 16.1.
Reaction 16.1
The table shows enthalpy changes of formation and entropies.
ΔHfe/kJ mol1 Se / J K–-1 mol–1
H2O2(l) –188 110
H2O(l) –286 70.0
O2(g) 0 205
i. Calculate the free-energy change, ΔG, in kJ mol–1, of Reaction 16.1 at 25 °C.
Give your answer to 3 significant figures.
ΔG = .............................................. kJ mol–1 [4]
ii. The decomposition of hydrogen peroxide shown in Reaction 16.1 is feasible.
Suggest why Reaction 16.1 does not take place at 25 °C despite being feasible.
[1]
© OCR 2025. You may photocopy this page. 4 of 65 Created in ExamBuilder
