Question

The decomposition of NO₂(g) occurs by the following bimolecular elementary reaction:

$\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)$

The rate constant at 273 K is 2.3 × 10^-12 L mol^-1 s^-1, and the activation energy is 111 kJ/mol. How long will it take for the concentration of NO₂(g) to decrease from an initial partial pressure of 2.5atm to 1.5atm at 500. K? Assume ideal gas behaviour.

Short answer

It will take 113.4 x 10⁹ s.

Step-by-step solution

Firstly, we have to find k2 which is the rate of the reaction.

At 500K

$$\begin{gathered} \ln \frac{k_2}{k_1}=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ \ln \frac{k_2}{2.3\times {10}^{-12}}=\frac{111}{8.314}\left[\frac{1}{273}-\frac{1}{500}\right] \\ \frac{k_2}{2.3\times {10}^{-12}}=1.0224 \\ {k}_2=2.3515\times {10}^{-12}\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$mol-s$}\right. \end{gathered}$$

At 500K for the decrease of the partial pressure from 2.5atm to 1.5atm the time taken will be:

This is a second-order reaction, therefore using the formula 

$$\begin{gathered} k=\frac{1}{t}\left[\frac{1}{P_{N{O}^2}}-\frac{1}{{P^{+}}_{N{O}^2}}\right] \\ k=\frac{1}{t}\left[\frac{1}{1.5}-\frac{1}{2.5}\right] \\ 2.3515\times {10}^{-12}=\frac{1}{t}\left[\frac{50-30}{75}\right] \\ t=\frac{1}{2.3515\times {10}^{-12}}\left[0.2667\right] \\ t=0.1134\times {10}^{12} \\ t=113.4\times {10}^9\sec \end{gathered}$$