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 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.

Answer 2

Concept:

• For the same first-order reaction, the rate constant $k$ is common to every stage, so the ratio of times for two different extents of completion equals the ratio of their log terms -- $k$ cancels out entirely.

Step 1: Set up the ratio relation.
For a first-order reaction, $t = \frac{2.303}{k}\log\left(\frac{[R]_0}{[R]}\right)$. Since $k$ is the same constant throughout the same reaction, comparing two stages gives:
$\frac{t_2}{t_1} = \frac{\log R_2}{\log R_1}$, where $R_1, R_2$ are the respective $\frac{[R]_0}{[R]}$ ratios -- the $\frac{2.303}{k}$ factor cancels completely.

Step 2: Find the log ratios at each stage.
At 25% completion, 75% remains, so $R_1 = \frac{100}{75} = \frac{4}{3}$, giving $\log R_1 = \log 4 - \log 3 = 0.60-0.48 = 0.12$, at $t_1 = 40$ minutes.
At 80% completion, 20% remains, so $R_2 = \frac{100}{20} = 5$, giving $\log R_2 = \log 5 = 0.69$.

Step 3: Apply the ratio directly to find $t_2$.
$t_2 = t_1 \times \frac{\log R_2}{\log R_1} = 40 \times \frac{0.69}{0.12} = 40 \times 5.75 = 230$ minutes

Step 4: Find $k$ from the first data point if needed separately.
$k = \frac{2.303}{40}\log\left(\frac{4}{3}\right) = \frac{2.303 \times 0.12}{40} \approx 0.0069\ \text{min}^{-1}$

Final Answer: $k \approx 0.0069\ \text{min}^{-1}$, and the reaction is 80% complete in about 230 minutes.