# The following data were obtained during the first order thermal decomposition of

Question:

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$

Calculate the rate of the reaction when total pressure is 0.65 atm.

Solution:

The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

After time, $t$, total pressure, $\mathrm{P}_{t}=\left(\mathrm{P}_{0}-p\right)+p+p$

$\Rightarrow \mathrm{P}_{t}=\mathrm{P}_{0}+p$

$\Rightarrow p=\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{0}$

Therefore, $\mathrm{P}_{\mathrm{o}}-p=\mathrm{P}_{\mathrm{o}}-\left(\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{\mathrm{o}}\right)$

= 2 P0 − Pt

For a first order reaction,

$k=\frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-p}$

$=\frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{t}}$

When $t=100 \mathrm{~s}, k=\frac{2.303}{100 \mathrm{~s}} \log \frac{0.5}{2 \times 0.5-0.6}$

= 2.231 × 10−3 s−1

When Pt = 0.65 atm,

P0 + p = 0.65

⇒ p = 0.65 − P0

= 0.65 − 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is

$p_{\mathrm{SOCl}_{2}}=\mathrm{P}_{0}-\mathrm{p}$

= 0.5 − 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate $=k\left(p_{\mathrm{SOCl}_{2}}\right)$

= (2.23 × 10−3 s−1) (0.35 atm)

= 7.8 × 10−4 atm s−1