Exams › JEE Advanced › Chemistry
A first-order reaction reaches 50% completion in 20 minutes at 27 deg C and in 5 minutes at 47 deg C. Determine the activation energy of the reaction.
- 100 kJ/mol
- 55.14 kJ/mol
- 11.97 kJ/mol
- 6.65 kJ/mol
Correct answer: 55.14 kJ/mol
Solution
Since t(1/2) is inversely proportional to k, k2/k1 = 20/5 = 4. Using log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2) with T1 = 300 K, T2 = 320 K, Ea = 55.14 kJ/mol.
Related JEE Advanced Chemistry questions
- The given equation implies that:
- Given that C₁ represents the starting concentration of A, and C₁, C₂, and C₃ denote the concentrations of A, B, and C respectively at a specific time t, which equation can be used to calculate the values of k₁ and k₂?
- In the reaction M → N, the rate at which M is consumed becomes 8 times faster when the concentration of M is doubled. What is the reaction order with respect to M?
- For the reaction A(g) + B(g) ⇌ AB(g), the activation energy for the reverse reaction is greater than that for the forward reaction by 2RT (in J mol⁻¹). If the forward reaction's pre-exponential factor is four times that of the backward reaction, what is the magnitude of ΔG⁰ (in J mol⁻¹) at 300 K?
- A gaseous compound X decomposes into gaseous products Y and Z according to first-order kinetics (X -> Y + Z). Initially only X is present. Ten minutes after the reaction begins, the partial pressure of X is 200 mmHg and the total pressure of the mixture is 300 mmHg. What is the rate constant for this reaction?
- A catalyst is added to a first-order reaction at 500 K, causing its rate to become 2.718 times the original rate. The activation energy with the catalyst present is 4.15 kJ/mol. What was the activation energy before the catalyst was introduced? (Given: R = 8.3 J/K/mol, ln(2.718) = 1)
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →