Question

r =k[A] for a reaction, 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes.

Answer 1

Correct Answer: 399

To solve this problem, we need to determine the time required for 90% decomposition of A when the given rate expression is \( r = k[A] \). This represents a first-order reaction. The integrated rate law for a first-order reaction is:

\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]

Given that 50% of A is decomposed in 120 minutes, the remaining concentration of A is 50% of the initial concentration \([A]_0\). Thus, \([A] = 0.5[A]_0\) and the equation becomes:

\[ \ln(2) = k \times 120 \]

\( k \) can be derived as:

\[ k = \frac{\ln(2)}{120} \]

Next, for 90% decomposition, the remaining concentration is 10% of the initial, so \([A] = 0.1[A]_0\). Using the rate law again:

\[ \ln\left(\frac{[A]_0}{0.1[A]_0}\right) = kt_{\text{90\%}} \]

\[ \ln(10) = \frac{\ln(2)}{120} \times t_{\text{90\%}} \]

Solving for \( t_{\text{90\%}} \) gives:

\[ t_{\text{90\%}} = \frac{120 \times \ln(10)}{\ln(2)} \]

Calculating this expression:

\[ t_{\text{90\%}} \approx \frac{120 \times 2.302}{0.693} \approx 398.63 \text{ minutes} \]

The computed time for 90% decomposition is approximately 399 minutes, which fits within the given range of 399 to 399 minutes.

Answer 2

For a first-order reaction:

\[ t_{1/2} = 120 \, \text{min} \]

For 90% completion:

\[ t = \frac{2.303}{k} \log \left( \frac{a}{a - x} \right) \] \[ t = \frac{2.303 \times 120}{0.693} \log \left( \frac{100}{10} \right) \] \[ t = 399 \, \text{min}. \]