Question

Initial concentration of N\(_2\)O\(_5\) in a first-order reaction was \(1.24 \times 10^{-2}\) mol L\(^{-1}\) at 310 K, which remained \(0.20 \times 10^{-2}\) mol L\(^{-1}\) after 30 minutes. Calculate velocity constant at 310 K. \((\log_{10} 6.2 = 0.7924)\)

Hint:

Always use \(\log_{10}\) in first-order kinetics formula: \(k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\).

Answer 1

Step 1: First-order rate equation.
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] Step 2: Substitution.
\[ k = \frac{2.303}{30} \log \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} = \frac{2.303}{30} \log (6.2) \] Step 3: Simplification.
\[ k = \frac{2.303}{30} \times 0.7924 \approx 0.044 \, min^{-1} \] Step 4: Conclusion.
The velocity constant is \(\approx 0.044 \, min^{-1}\).