Question

At 600 K, the rate constant for the first-order decomposition of nitroethane

is 1.9 × 10^-4s^-1 .

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NO}}_2\to {\mathrm{C}}_2{\mathrm{H}}_4+{\mathrm{HNO}}_2$

A sample of CH₃CH₂NO₂ is heated to 600 K, at which point its initial partial pressure is measured to be 0.078 atm. Calculate its partial pressure after 3.0 hours.

Short answer

The partial pressure is 0.010 atm after 3 hours.

Step-by-step solution

Calculate time for given reaction

Rate law of the reaction when partial pressure is given by the following expression

$2.303\log \frac{{\mathrm{P}}_{\mathrm{t}}}{{\mathrm{P}}_0}=-\mathrm{kt}.......\left(1\right)$

Here P₀ is the initial pressure, P_t is the final pressure, k is the rate constant and t is the time taken.

Convert time in hours to seconds

localid="1660231583743" role="math" $$\begin{gathered} \mathrm{t}=3.0hours \\ \mathrm{t}=3.0\mathrm{hours}\times \frac{60\mathrm{minutes}}{1\mathrm{hour}}\times \frac{60\mathrm{seaconds}}{1\min } \\ \mathrm{t}\hspace{0.17em}=10800s \end{gathered}$$

The total time is t =10800s

Calculation of final pressure Pt

Substituting below the values in equation 1,

$$\begin{gathered} \mathrm{T}\hspace{0.17em}=600\mathrm{K} \\ {\mathrm{P}}_{\mathrm{i}}=\hspace{0.17em}0.078\mathrm{atm} \\ \mathrm{t}=10800\mathrm{s} \end{gathered}$$

The following equation is obtained

$$\begin{gathered} 2.303\log \frac{\mathrm{Pt}}{0.078}\hspace{0.17em}=-\left(1.9\times {10}^{-4}\right)\left(10800\mathrm{s}\right) \\ {\mathrm{logp}}_t-\log \left(0.0078\right)=-\frac{\left(1.9\times {10}^{-4}\right)\left(10800\mathrm{s}\right)}{2.303} \\ {\mathrm{logp}}_t+1.11\hspace{0.17em}=-0.891 \\ {\mathrm{logp}}_t\hspace{0.17em}=-2.00 \\ {\mathrm{p}}_{\mathrm{t}}={10}^{-2.00} \\ {\mathrm{p}}_{\mathrm{t}}=0.010 \end{gathered}$$

Therefore the partial pressure after 3.0 hours will be 0.010 atm