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  • <img alt="\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})" src="/latex-image/?%5Cmathrm%7BPCl%7D_%7B5%7D%28%5Cmathrm%7B%7Eg%7D%29%20%5Crightarrow%20%5Cmathrm%7BPCl%7D_%7B3%7D%28%5Cmathrm%7B%7Eg%7D%29&am

\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) In the above first order reaction, the concentration of \mathrm{PCl}_{5} reduces from initial concentration 50 \mathrm{~mol} \mathrm{} \mathrm{L}^{-1} \: to\: 10 \mathrm{~mol} \mathrm{} \mathrm{L}^{-1} in 120 minutes at 300 \mathrm{~K}. The rate constant for the reaction at 300 \mathrm{~K}$ is $x \times 10^{-2} \min ^{-1}. The value of x is _______ (nearest integer) [Given \log 5=0.6989 ]
 

Answers (1)

best_answer

The given reaction follows first order kinetics

\mathrm{P Cl_{5} \longrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}^{}}

C_{0}=50, C_{t}=10, t=120 \mathrm{~min}

As we know,

for a first order reaction.

\begin{aligned} &\ln \left(\frac{C_{0}}{C_{t}}\right)=k t \\ &\Rightarrow \ln \left(\frac{50}{10}\right)=k(120) \\ &\Rightarrow 120 k=2.303 \log 5 \end{aligned}

\begin{aligned} \Rightarrow k &=\frac{2.303 \times 0.6989}{120} \\ &=0.013 \\ &=1.3 \times 10^{-2} \mathrm{~min} \end{aligned}

Hence, the correct answer is (1).

Posted by

sudhir.kumar

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