Question

The decomposition of a compound A follows first-order kinetics. The concentration of A at time t = 0 is 1.0 mol L^-1. After 60 minutes, it reduces to 0.25 mol L^-1. What is the initial rate of the reaction at t = 0? (Take ln 2 = 0.693)

• A. 0.0115 mol L$^{-1}$ min$^{-1}$
• B. 0.0173 mol L$^{-1}$ min$^{-1}$
• C. 0.277 mol L$^{-1}$ min$^{-1}$
• D. 0.0364 mol L$^{-1}$ min$^{-1}$

Hint:

For first-order reactions, use the integrated rate law to calculate the rate constant and then determine the rate of the reaction.

Answer 1

The Correct Option is A

For a first-order reaction, the integrated rate law is given by: \[ \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \] Where: - \([A]_0 = 1.0 \, \text{mol L}^{-1}\) (initial concentration) - \([A]_t = 0.25 \, \text{mol L}^{-1}\) (concentration after 60 minutes) - \(k\) is the rate constant - \(t = 60 \, \text{minutes} = 1 \, \text{hour} = 60 \, \text{minutes}\) First, calculate the rate constant \(k\) using the integrated rate law: \[ \ln \left( \frac{1.0}{0.25} \right) = k \times 60 \] \[ \ln (4) = k \times 60 \] \[ 0.693 = k \times 60 \] \[ k = \frac{0.693}{60} = 0.01155 \, \text{mol L}^{-1} \, \text{min}^{-1} \] Thus, the initial rate of the reaction is \(0.0115 \, \text{mol L}^{-1} \, \text{min}^{-1}\).