Question

(i) The initial concentration of N\(_2\)O\(_5\) in the first-order reaction:
\[ \text{N}_2\text{O}_5 (\text{g}) \rightarrow 2 \text{NO}_2 (\text{g}) + \frac{1}{2} \text{O}_2 (\text{g}) \] was \( 1.2 \times 10^{-2} \) mol L\(^{-1}\). The concentration of N\(_2\)O\(_5\) after 60 minutes was \( 0.2 \times 10^{-2} \) mol L\(^{-1}\). Calculate the rate constant of the reaction at 318 K.
\([ \log 6 = 0.778 ]\)

Hint:

For first-order reactions, the rate constant \(k\) can be calculated using the integrated rate law, where the natural logarithm of the ratio of concentrations is directly proportional to time.

Answer 1

For a first-order reaction, the integrated rate law is given by:
\[ \ln \left( \frac{[A]_0}{[A]_t} \right) = k t \] Where:
\([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), and \(k\) is the rate constant.
Given:
\([A]_0 = 1.2 \times 10^{-2}\) mol L\(^{-1}\), \([A]_t = 0.2 \times 10^{-2}\) mol L\(^{-1}\), \(t = 60\) minutes, and \(\log 6 = 0.778\).
Substitute the values into the equation:
\[ \ln \left( \frac{1.2 \times 10^{-2}}{0.2 \times 10^{-2}} \right) = k \times 60 \] \[ \ln 6 = k \times 60 \] \[ 0.778 = k \times 60 \] Solving for \(k\):
\[ k = \frac{0.778}{60} = 0.01297 \, \text{min}^{-1} \]