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:

When two completion percentages of the SAME first-order reaction are compared, you do not need to calculate k at all -- the times are directly proportional to the log-ratios, so $t_2 = t_1 \times \frac{\log R_2}{\log R_1}$ gets you the answer in one step.

Answer 1

Step 1: The integrated rate law for a first-order reaction is presented as: \[ \log \left( \frac{[R]_0}{[R]} \right) = \frac{k \cdot t}{2.303} \] Definitions: - \( [R]_0 \) denotes the initial concentration. - \( [R] \) denotes the concentration at time \( t \). - \( k \) represents the rate constant. - \( t \) signifies the time elapsed. Given that the reaction is 25% complete after 40 minutes, 75% of the reactant remains. This leads to the calculation: \[ \frac{[R]_0}{[R]} = \frac{1}{0.75} = 1.33 \] Taking the logarithm of this value yields: \[ \log 1.33 = 0.125 \] Substituting these values into the rate law: \[ 0.125 = \frac{k \cdot 40}{2.303} \] Solving for the rate constant \( k \): \[ k = \frac{0.125 \times 2.303}{40} = 0.0069 \, \text{min}^{-1} \] Step 2: To determine the time for 80% completion, where \( \frac{[R]_0}{[R]} = \frac{1}{0.20} = 5 \), we first find the logarithm: \[ \log 5 = 0.69 \] Then, we substitute this into the rate law: \[ 0.69 = \frac{k \cdot t}{2.303} \] Using the previously calculated value for \( k \): \[ 0.69 = \frac{0.0069 \cdot t}{2.303} \] Solving for \( t \): \[ t = \frac{0.69 \times 2.303}{0.0069} = 230.3 \, \text{min} \] Consequently, the time required for the reaction to reach 80% completion is 230.3 minutes.