Question

The reaction ${\text{SO}}_2{\text{Cl}}_2\left(g\right)\to {\text{SO}}_2\left(g\right)+{\mathrm{Cl}}_2\left(g\right)$ is first order, with a rate constant of

2.2 × 10^-5s^-1 at 320°C. The partial pressure of SO₂Cl₂(g) in a sealed vessel at 320°C is 1.0 atm. How long will it take for the partial pressure of SO₂Cl₂(g) to fall to 0.50 atm?

Short answer

The time taken for the reaction is$3.2\times {10}^{4}s$

Step-by-step solution

Rate expression:

For the first order reaction which occurs in gaseous state,when initial and final pressures are given the rate expression is given as below

$2.303\log \left(\frac{P_0}{P_t}\right)\hspace{0.17em}=\text{kt}........\left(1\right)$

Where $P_0$ and P_t are initial and final pressure k is the rate constant of the reaction.

Time of the reaction:

Substituting all the given values in equation 1

$$\begin{gathered} 2.303\log \left(\frac{1.0}{0.5}\right)\hspace{0.17em}=2.2\times {10}^{-5}........\left(1\right) \\ t=\frac{2.303}{2.2\times {10}^{-5}}\log 2s \\ t=1.05\times {10}^{-5}\times 0.301 \\ t=3.2\times {10}^{4}s \end{gathered}$$

Hence the time taken for the reaction is$3.2\times {10}^{4}s$