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Question

The equilibrium constant for the following reaction is 64. H2O(l)+CO(g)H2(g)+CO2(g). If the rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

A
0.4
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B
2.5
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C
6.2
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D
10.24
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Solution

The correct option is B 2.5
Given that the equilibrium constant for the reaction, H2O(l)+CO(g)H2(g)+CO2(g) is 64.

The expression for the equilibrium constant, K=kfkb
GIven, K=64, kf=160
Substituting values in the above expression, we get
64=160kb kb=2.5

Hence, the rate constant for the backward reaction is 2.5.

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