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For the opposing \mathrm{P+Q \frac{k_f}{k_b} R+S}

The forward reaction has values \mathrm{E_a=100 \, \mathrm{kJmol}^{-1}} and \mathrm{A=10^{10} \, \mathrm{M}^{-1} \mathrm{~s}^{-1}.} The equilibrium concentration. of \mathrm{{P, Q, R}} and \mathrm{S} are \mathrm{1 M, 2 M, 5 M} and \mathrm{4 M} respectily, y, at \mathrm{700 K. } At this temperature, values of \mathrm{k_f} and \mathrm{k_b} respectively are?

Option: 1

\mathrm{20 \mathrm{~m}^{-1} \mathrm{~s}^{-1} \text { and } 2 \, \, \mathrm{M^{-1}s}^{-1}}


Option: 2

\mathrm{345 \mathrm{~M}^{-1} \mathrm{~s}^{-1} \text { and } 34.5 \mathrm{~M}^{-1} \mathrm{~s}^{-1}}


Option: 3

\mathrm{34.5 \, \mathrm{M}^{-1} \mathrm{~s}^{-1} \text { and } 3.45 \mathrm{~M}^{-1} \mathrm{~s}^{-1}}


Option: 4

\mathrm{200 \mathrm{~M}^{-1} \mathrm{~s}^{-1} \text { and } 20 \, \mathrm{M}^{-1} \mathrm{~s}^{-1}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \because k_f=A e^{-Ea / R T} \\\\ & k_f=10^{10} \exp \left(-\frac{100000}{8.314 \times 700}\right) \\\\ & k_f=10^{10} \times 3.45 \times 10^{-8} \\\\ & k_f=345 \mathrm{~m}^{-1} \mathrm{~s}^{-1} \\\\ & \text { Now, } k_{e q}=\frac{[R][s]}{[P][Q]}=\frac{5 \times 4}{1 \times 2}=10=\frac{k_f}{k_b} \\\\ & k_b=\frac{k_f}{10}=34.5 \mathrm{~m}^{-1} \mathrm{~s}^{-1} \\ & \end{aligned}}

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himanshu.meshram

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