Question

At 600 K, the rate constant for the first-order decomposition of nitroethane $C{H}_{3}C{H}_{2}N{O}_2\to C_2{H}_4+HN{O}_2$

is 1.9 × 10^-4s^-1 . A sample of $C{H}_{3}C{H}_{2}N{O}_2$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 0.010 atm

Step-by-step solution

Time of the reaction:

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

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

Where P₀ and P_t are initial and final pressure k is the rate constant of the reaction.

Conversion of hours to seaconds

$$\begin{gathered} \text{t}=3.0\text{hours} \\ \text{t}=3.0\text{hour}\times \frac{60min}{1\text{hour}}\times \frac{60s}{1min} \\ \text{t}=10800s \end{gathered}$$

Final pressure  after time t:

Substituting all the values in eqaution 1

$$\begin{gathered} 2.303\log \left(\frac{0.078}{P_t}\right)=kt.......\left(1\right) \\ \log 0.078-\log P_t=-\frac{\left(1.9\times {10}^{-4}\right)\left(10800s\right)}{2.303} \\ 1.11-\log {\text{P}}_t=-0.891 \\ -\log {\text{P}}_t=-0.891-1.11 \\ -\log {\text{P}}_t\text{=-2.00} \\ \text{Cancelling-signonbothsides} \\ \log {\text{P}}_t\text{=2.00} \\ {P}_t={10}^{2.00} \\ {P}_t=100\text{atm} \end{gathered}$$

Hence the final pressure is 100 atm