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  • The following data were obtained during the first order thermal decomposition of SO_2 Cl_2 at a constant volume. ( SO_2 Cl_2 g) gives SO_2 (g) plus Cl_2 (g)

The following data were obtained during the first-order thermal decomposition of \mathrm{SO}_2 \mathrm{Cl}_2 at a constant volume.
\left(\mathrm{SO}_2 \mathrm{Cl}_2 \mathrm{~g}\right) \rightarrow \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})

       

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

Answers (1)

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The thermal decomposition of \mathrm{SO}_2 \mathrm{Cl}_2 is shown here;

\begin{tabular}{lllcc} & $\mathrm{SO}_2 \mathrm{Cl}_{2(g)}$ & $\longrightarrow$ & $\mathrm{SO}_{2(g)}+\mathrm{Cl}_{2(g)}$ \\ At $t=0$ & $\mathrm{P}_0$ & & 0 & 0 \\ At $t=t$ & $\mathrm{P}_0-\mathrm{p}$ & & p & p \end{tabular}

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=2 p_0-p_t

for first order reaction,

\begin{aligned} k & =\frac{2.303}{t} \log \frac{p_0}{p_0-p} \\ & =\frac{2.303}{360} \log \frac{p_0}{2 p_0-p_t}\end{aligned}


now putting the values of pressures, when t = 100s


\begin{aligned} k & =\frac{2.303}{100} \log \frac{0.5}{2 * 0.5-0.6} \\ & =2.231 \times 10^{-3} \mathrm{~s}^{-1}\end{aligned}

when p_t=0.65 \mathrm{~atm}

p=p_t-p_0
    = 0.65 - 0.5
   = 0.15 atm

So, p\left(\mathrm{SO}_2 \mathrm{Cl}_2\right)=p_0-p
                            = 0.5 - 0.15
                            = 0.35 atm

Thus, the rate of reaction, when the total pressure is 0.65 atm


\begin{aligned} \text { rate } & =\mathrm{k}\left(p\left(\mathrm{SO}_2 \mathrm{Cl}_2\right)\right) \\ & =2.31 \times 10^{-3} \times 0.35 \\ & =7.8 \times 10^{-4} \text { atm s }\end{aligned}

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manish

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