UPSC Civil Services (Mains), Chemistry optional · 2024 · Paper I
Q1(e). Temperature at which the uncatalysed reaction matches catalase at 20 °C
Question
The activation energy for the decomposition of hydrogen peroxide, 2H₂O₂(aq) → 2H₂O(l) + O₂(g), is 42 kJ mol⁻¹, whereas when the reaction is catalyzed by the enzyme catalase it is 7.0 kJ mol⁻¹. Calculate the temperature that would cause the uncatalyzed reaction to proceed as rapidly as the enzyme-catalyzed decomposition at 20 °C. Assume the frequency factor A to be the same in both cases.
Answer in brief
T = 293.15 K × 42/7.0 = 1759 K (about 1486 °C).
Key points
- 'As rapidly' at the same concentrations means equal rate constants; since A is the same, the Arrhenius exponents must be equal.
- A e−Ea/RT = A e−E′a/RT₀, so Ea/T = E′a/T₀ and T = T₀ Ea/E′a = 293.15 × 42/7.0 = 1759 K.
- The activation energy is six times larger, so the absolute temperature must be six times higher.
- At 20 °C catalase speeds the reaction up by e14.36 = 1.7 × 10⁶.
- Catalase is a haem enzyme that opens a lower-energy path through an Fe(IV)O intermediate; ΔG and K stay the same.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversArrhenius equationactivation energycatalaseenzyme catalysishydrogen peroxide decomposition
All parts of Question 1
- 1(a)(i) Most probable distance of the 1s electron5 marks
- 1(a)(ii) Lattice energies of NaBr and NaI above the calculated values5 marks
- 1(b)(i) Separation factor for ²³⁵UF₆ and ²³⁸UF₆ in one effusion stage5 marks
- 1(b)(ii) Unit cell and the cubic Bravais lattices5 marks
- 1(c) Normal boiling point of mercury10 marks
- 1(d)(i) Will CCl₂F₂ condense at 18 atm and 20 °C?5 marks
- 1(d)(ii) Overvoltage and its applications5 marks
- 1(e) Temperature at which the uncatalysed reaction matches catalase at 20 °C10 marks