Question

The reaction $\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(\mathit{g}\right)\mathbf{\to }\mathbf{}\mathit{S}{\mathit{O}}_{\mathbf{2}}\left(\mathit{g}\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(\mathit{g}\right)$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

Step-1:  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

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

Where $P_{0}and{P}_t$are initial and final pressure k is the rate constant of the reaction.

STEP-2:   Time of the reaction:

Substituting all the given values in equation 1

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

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