• question_answer The forward rate constant for the reversible gaseous reaction ${{C}_{2}}{{H}_{6}}2C{{H}_{3}}$ is$3.14\times {{10}^{+2}}{{s}^{-1}}$ at 200 K. What is the rate constant for the backward reaction at this temperature, if ${{10}^{-5}}$ moles of $C{{H}_{3}}$ and 100 mol of ${{C}_{2}}{{H}_{6}}$ are present in 10 L vessel at equilibrium? A) $1.57\times {{10}^{14}}Lmo{{l}^{-1}}{{s}^{-1}}$ B) $3.14\times {{10}^{15}}Lmo{{l}^{-1}}{{s}^{-1}}$ C) $1.57\times {{10}^{7}}Lmo{{l}^{-1}}{{s}^{-1}}$ D) $3.14\times {{10}^{10}}Lmo{{l}^{-1}}{{s}^{-1}}$

Correct Answer: B

Solution :

 ${{k}_{eq}}=\frac{{{k}_{1}}}{{{k}_{b}}}=\frac{{{[C{{H}_{3}}]}^{2}}}{[{{C}_{2}}{{H}_{6}}]}$ $[C{{H}_{3}}]=\frac{{{10}^{-5}}}{10}mol{{L}^{-1}}={{10}^{-6}}mol{{L}^{-1}}$ $[{{C}_{2}}{{H}_{6}}]=\frac{100}{10}=10mol{{L}^{-1}}=10M$ ${{k}_{eq}}=\frac{{{k}_{f}}}{{{k}_{b}}}$ $\frac{{{[{{10}^{-6}}]}^{2}}}{10}=\frac{3.14\times {{10}^{2}}{{s}^{-1}}}{{{k}_{b}}}$ ${{k}_{b}}=3.14\times {{10}^{3+12}}Lmo{{l}^{-1}}{{s}^{-1}}$ $=3.14\times {{10}^{15}}Lmo{{l}^{-1}}{{s}^{-1}}$

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