Exams › JEE Advanced › Chemistry
A substance A decomposes simultaneously via three parallel first-order pathways: (i) A —k1—> B, with k1 = 5 * 10⁻² s⁻¹ (ii) 2A —k2—> C, with k2 = 3 * 10⁻² s⁻¹ (iii) 3A —k3—> D, with k3 = 5 * 10⁻³ s⁻¹ All rate constants are pseudo-first-order in A. Calculate the average lifetime of substance A (in seconds).
- 200/17 s (approximately 11.76 s)
- 20 s
- 1/0.05 s (20 s)
- 100/8.5 s (approximately 11.76 s)
Correct answer: 200/17 s (approximately 11.76 s)
Solution
The effective first-order rate constant for parallel decay is k_eff = k1 + k2 + k3 = 0.05 + 0.03 + 0.005 = 0.085 s⁻¹. The average (mean) lifetime is tau = 1/k_eff = 1/0.085 = 200/17 approximately 11.76 s.
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 →