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  • <img alt="\mathrm{\mathrm{PCl}_5(\mathrm{~g}) \rightarrow \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BPCl%7

\mathrm{\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{\mathrm{PCl}_5} reduces from initial concentration \mathrm{60 \mathrm{molL}^{-1} \ to \ 10 mol L^{-1}} in 135 minutes at 300 K. The rate constant for the reaction at 300 K is \mathrm{ x \times 10^{-3} \mathrm{~min}^{-1}}. Calculate value of x {given log 6=0.77}

Option: 1

25


Option: 2

10


Option: 3

30


Option: 4

2


Answers (1)

best_answer

\mathrm{\mathrm{PCl}_ 5(\mathrm{~g}) \underset{30 \mathrm{OK}}{\stackrel{\text { Iorder }}{\longrightarrow}} \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})}

\mathrm{\begin{aligned} & t=0 ; \quad \mathrm{PCl}_5=60 \mathrm{M} \\ & t=180 \mathrm{~min} ; \quad \mathrm{PC}_5=10 \mathrm{M} \end{aligned}}

\mathrm{\begin{aligned} & k=\frac{2.303}{t} \log \frac{\left[A_0\right]}{\left[A_t\right]} \\ & k=\frac{2.303}{180} \log \frac{60}{10} \\ & k=\frac{2.303}{180} \times 0.77 \\ & k=9.8 \times 10^{-3}=10 \times 10^{-3} \end{aligned}}

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vishal kumar

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