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Question

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps:

CO(g)+Cl2(g)COCl(g)+Cl(g)\mathrm { CO } ( g ) + \mathrm { Cl } _ { 2 } ( g ) \rightleftharpoons \operatorname { COCl } ( g ) + \mathrm { Cl } ( g )

At

25C25 ^ { \circ } \mathrm { C }

the rate constants for the forward and reverse reactions are

1.4×1028M1s11.4 \times 10 ^ { - 28 } M ^ { - 1 } \mathrm { s } ^ { - 1 }

and

9.3×1010M1s19.3 \times 10 ^ { 10 } \mathrm { M } ^ { - 1 } \mathrm { s } ^ { - 1 }

respectively. What is the value for the equilibrium constant at

25C?25 ^ { \circ } \mathrm { C } ?

Solution

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Answered 2 years ago
Answered 2 years ago
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CO(g)+Cl2(g)COCl(g)+Cl(g)\mathrm{ CO(g) + Cl_2(g) \rightleftharpoons COCl(g) + Cl(g) }

The rate constant for forward reaction is kf=1.4×1028 M1s1\mathrm{k_f = 1.4 \times 10^{-28}\ M^{-1}s^{-1} }

The rate constant for reverse reaction is kr=9.3×1010 M1s1\mathrm{k_r = 9.3 \times 10^{10}\ M^{-1}s^{-1} }

The value of equilibrium constant is

Keq=kfkr=1.4×1028 M1s19.3×1010 M1s1=1.5×1039\begin{align*} \mathrm{ K_{eq} } &= \mathrm{ \frac { k_f } { k_r } }\\ &= \mathrm{ \frac{ 1.4 \times 10^{-28}\ M^{-1}s^{-1} } { 9.3 \times 10^{10}\ M^{-1}s^{-1} } }\\ &= 1.5 \times 10^{-39} \end{align*}

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