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Effect of Catalyst

Equilibrium c...

Question

Equilibrium constant for the reaction is 4.0, then what will be the equilibrium constant for the reverse reaction

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Solution

$A + B \leftrightarrow C + D F o r t h e f o r w a r d d i r e c t i o n i f t h e e q u i l i b r i u m c o n s \tan t i s K, t h e n f o r t h e r e v e r s e d i r e c t i o n t h e e q u i l i b r i u m c o n s \tan t = \frac{1}{K} S o h e r e i t s = \frac{1}{4} = 0.25$

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