Question

Find the rate constant at 310K if the initial concentration is 0.72 mol L$^{-1}$ and the final concentration is 1.44 mol L$^{-1}$ at 10 minutes.

• A. 0.05
• B. 0.0693
• C. 0.091
• D. 0.13

Hint:

For first-order reactions, use the equation \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \) to calculate the rate constant.

Answer 1

The Correct Option is B

We are given the initial and final concentrations of a reaction over a certain time period, and we need to calculate the rate constant using the integrated rate law.

1. Step 1: Use the integrated rate law. The general integrated rate law for a first-order reaction is: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where: - \( [A]_0 \) is the initial concentration, - \( [A] \) is the final concentration, - \( k \) is the rate constant, - \( t \) is the time.

2. Step 2: Substitute the values into the equation. We are given: \[ [A]_0 = 0.72 \, \text{mol L}^{-1}, \quad [A] =
1.44 \, \text{mol L}^{-1}, \quad t = 10 \, \text{minutes} \] Substituting into the rate law: \[ \ln \left( \frac{0.72}{
1.44} \right) = k \times 10 \] Simplifying: \[ \ln(0.5) = k \times 10 \] \[ -0.6931 = k \times 10 \] Solving for \( k \): \[ k = \frac{-0.6931}{10} = 0.0693 \, \text{min}^{-1} \] Thus, the rate constant is \( 0.0693 \, \text{min}^{-1} \).