Question

A first-order reaction is 25% complete in 40 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
[Given: \( \log 2 = 0.30 \), \( \log 3 = 0.48 \), \( \log 4 = 0.60 \), \( \log 5 = 0.69 \)]

Hint:

For first-order reactions, the concentration decreases exponentially over time. Use the integrated rate law to find the rate constant and calculate the time for any given percentage completion.

Answer 1

Step 1: The integrated rate law for a first-order reaction is given by: \[ \log \left( \frac{[R]_0}{[R]} \right) = \frac{k \cdot t}{2.303} \] Where: - \( [R]_0 \) is the initial concentration,
- \( [R] \) is the concentration at time \( t \),
- \( k \) is the rate constant, and
- \( t \) is the time.
We are told that the reaction is 25% complete in 40 minutes, which means that 75% of the reactant remains. Therefore, we calculate: \[ \frac{[R]_0}{[R]} = \frac{1}{0.75} = 1.33 \] Taking the logarithm: \[ \log 1.33 = 0.125 \] Substitute into the rate law: \[ 0.125 = \frac{k \cdot 40}{2.303} \] Solving for \( k \): \[ k = \frac{0.125 \times 2.303}{40} = 0.0069 \, \text{min}^{-1} \] Step 2: To find the time required for the reaction to be 80% complete, i.e., \( \frac{[R]_0}{[R]} = \frac{1}{0.20} = 5 \), we use: \[ \log 5 = 0.69 \] Substitute into the rate law: \[ 0.69 = \frac{k \cdot t}{2.303} \] Substituting the value of \( k \): \[ 0.69 = \frac{0.0069 \cdot t}{2.303} \] Solving for \( t \): \[ t = \frac{0.69 \times 2.303}{0.0069} = 230.3 \, \text{min} \] Thus, the time required for the reaction to be 80% complete is 230.3 minutes.