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4.21 The following data were obtained during the first order thermal decomposition of $SO_2 Cl_2$ at a constant volume.
$( SO_2Cl_2 g) \rightarrow SO_2 (g) + Cl_2 (g)$

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

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The thermal decomposition of $SO_{2}Cl_{2}$ is shown here;

After t time, the total pressure $p_{t}$ = $p_{0}-p+(p+p) = p_{0}+p$

So, $p = p_{t}-p_{0}$

thus, $p_{0}-p = 2p_{0}-p_{t}$

for first order reaction,

$k= \frac{2.303}{t}\log\frac{p_{0}}{p_{0}-p}$
$= \frac{2.303}{360}\log\frac{p_{0}}{2p_{0}-p_{t}}$
now putting the values of pressures, when t = 100s
$k = \frac{2.303}{100}\log\frac{0.5}{2*0.5-0.6}$
$= 2.231 \times 10^{-3}\ s^{-1}$

when $p_{t} = 0.65\ atm$

$p = p_{t}-p_{0}$
= 0.65 - 0.5
= 0.15 atm

So, $p(_{SO_{2}Cl_{2}}) = p_{0}-p$
= 0.5 - 0.15
= 0.35 atm

Thus, rate of reaction, when the total pressure is 0.65 atm
rate = k( $p(_{SO_{2}Cl_{2}})$ )
= $2.31\times 10^{-3}\times 0.35$
= 7.8 $\times 10^{-4}\ atm\ s^{-1}$

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manish

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