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The rate constant for the forward reaction  A(g)\rightleftharpoons 2B(g)  is 1.5\times10^{-3}\, s^{-1}\, \, at \: 100K. If 10^{-5} moles of A and 100 moles of B are present in a 10 liter vessel at equilibrium then rate constant for the backward reaction at this temp. is

Option: 1

\mathrm{1.5\times10^{4}\: L\, mol^{-1}\, s^{-1}}


Option: 2

\mathrm{1.5\times10^{11}\: L\, mol^{-1}\, s^{-1}}


Option: 3

\mathrm{1.5\times10^{10}\: L\, mol^{-1}\, s^{-1}}


Option: 4

\mathrm{1.5\times10^{-11}\: L\, mol^{-1}\, s^{-1}}


Answers (1)

best_answer

First, calculate the concentration of A & B.

[A]=\frac{10^{-5}}{10}= 10^{-6}mol/lit

[B]=\frac{100}{10}= 10\, mol/lit

Calculate equilibrium constant  K_{eq}= \frac{[B]^{2}}{[A]}= 10^{8}\, mol/lit

K_{eq}= \frac{K_{p}}{K_{b}}

\therefore 10^{8}=\frac{1.5\times10^{-3}s^{-1}}{K_{b}}\, \, \,

\therefore K_{b}=1.5\times10^{-11} L\, mol^{-1}\, s^{-1}

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Sayak

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