Question

At 600 K, the rate constant for the first-order decomposition of nitroethane ${\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{CH}}}_{\mathbf{2}}{\mathbf{\text{NO}}}_{\mathbf{2}}\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{4}}\mathbf{}\mathbf{+}{\mathbf{\text{HNO}}}_{\mathbf{2}}$is 1.9 × 10^-4s^-1 . A sample of ${\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{CH}}}_{\mathbf{2}}{\mathbf{\text{NO}}}_{\mathbf{2}}$

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

Short answer

The final pressure is around 100 atm.

Step-by-step solution

Step-1: Time of the reaction:

Rate law or first order reaction when initial and final pressures are given

$2.303\log \left(\frac{P_0}{P_t}\right)=kt.......\left(1\right)$

Where $P_0\text{and}{P}_t$ are initial and final pressure k is the rate constant of the reaction.

Conversion of hours to seconds

$\begin{array}{c}t=3.0hour\\ t=3.0hour\times \frac{60\min }{1hour}\times \frac{60s}{1\min }\\ t=10800s\end{array}$

Step-2: Final pressure after time t.

Substituting all the values in equation 1

$\begin{array}{rcl}og\left(0.078\right)-\log p_t& =& -\frac{\left(1.9\times {10}^{-4}{s}^{-1}\right)\left(10800s\right)}{2.303}\\ \log \left(0.078\right)-\log p_t-& =& -\frac{\left(1.9\times {10}^{-4}{s}^{-1}\right)\left(10800s\right)}{2.303}\\ 1.11-\log p_t+& =& -0.891\\ -\log p_t& =& -0.891-1.11\end{array}$

solve further as,

$\begin{array}{rcl}& & \begin{array}{c}-\log p_t=-2.00\end{array}\\ & & \begin{array}{l}-\log p_t=-2.00\\ \begin{array}{c}\log p_t=2.00\\ P_t={10}^{2.00}\\ P_t=100\end{array}\end{array}\end{array}$

Hence the final pressure is 100 atm.