Question

The initial concentration of N$_2$O$_5$ in the following first order reaction $N_2O_5(g) \longrightarrow 2NO_2(g) + \tfrac{1}{2} O_2(g)$ was $1.24 \times 10^{-2}$ mol L$^{-1}$ at 318 K. The concentration of N$_2$O$_5$ after 60 minutes was $0.20 \times 10^{-2}$ mol L$^{-1}$. Calculate the rate constant of the reaction at 318 K.

Hint:

For first-order reactions, the half-life is independent of concentration: $t_{1/2} = \dfrac{0.693}{k}$.

Answer 1

Step 1: Formula for first-order rate constant.
\[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] where, \([R]_0 = 1.24 \times 10^{-2} \, mol L^{-1}\) (initial concentration) 
\([R] = 0.20 \times 10^{-2} \, mol L^{-1}\) (concentration after 60 min) 
\(t = 60 \, min = 60 \times 60 = 3600 \, s\) 

Step 2: Substitution.
\[ k = \frac{2.303}{3600} \log \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} \] \[ k = \frac{2.303}{3600} \log (6.2) \] \[ \log (6.2) \approx 0.792 \] \[ k = \frac{2.303 \times 0.792}{3600} \] \[ k \approx 5.06 \times 10^{-4} \, s^{-1} \] 

Conclusion:
The rate constant of the reaction at 318 K is: \[ \boxed{5.06 \times 10^{-4} \, s^{-1}} \]