Exams › JEE Advanced › Chemistry
For a first-order reaction, the time required for 99% completion is 10 minutes. What is the time required for 99.9% completion of the same reaction? (Answer in minutes)
- 15
- 20
- 12
- 18
Correct answer: 15
Solution
For first-order kinetics, t = (2.303/k) * log([A]0/[A]). For 99% completion: [A] = 1% of [A]0, so t1 = (2.303/k) * log(100) = (2.303/k) * 2. For 99.9% completion: [A] = 0.1% of [A]0, so t2 = (2.303/k) * log(1000) = (2.303/k) * 3. Ratio: t2/t1 = 3/2. Therefore t2 = (3/2) * 10 = 15 minutes.
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 →