Question

The reaction

${\mathrm{SO}}_2{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{SO}}_2\left(\mathrm{g}\right)+\mathrm{C}{\mathrm{l}}_2\left(g\right)$

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 reaction takes place at $3.2\times {10}^{2}s$

Step-by-step solution

Relationship between the pressure of gas and rate constant

The reaction in the gaseous state is the initial pressure and is the final pressure at time t and k is the rate constant at particular temparature is given by the equation

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

STEP-2 : Calculating  time required  for the decrease in pressure

Substitutingrate constant values,initial pressure, and final pressure values in equation 1

$$\begin{gathered} 2.303\log \left[\frac{P_0}{P_t}\right]=\mathrm{kt}............\left(1\right) \\ \mathrm{t}=\frac{2.303}{2.2\times {10}^{-5}}\log 2s \\ \mathrm{t}=1.0\times {10}^5\times 0.301s \\ \mathrm{t}=3.0\times {10}^4 \end{gathered}$$

Therefore the reaction takes place at 3.0 x 10⁴ when the partial pressure of the gas decreases from 0.5 atm to 1.0 atm.