Obtain the relation between equilibrium constant $K$ and $K'$ for forward and reverse reactions.
(N/A) Consider the synthesis of $HI$ as the forward reaction:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)} \quad (Eq.-i)$
According to the law of chemical equilibrium,the equilibrium constant $K$ is:
$K = \frac{[HI]^2}{[H_2][I_2]} = X \quad (Eq.-ii)$
For the reverse reaction (decomposition of $HI$):
$2 HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \quad (Eq.-iii)$
The equilibrium constant $K'$ for this reverse reaction is:
$K' = \frac{[H_2][I_2]}{[HI]^2} = \frac{1}{X} \quad (Eq.-iv)$
Comparing $(Eq.-ii)$ and $(Eq.-iv)$,we get:
$K = \frac{1}{K'} \quad \text{or} \quad K \times K' = 1$
Thus,the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)} \quad (Eq.-i)$
According to the law of chemical equilibrium,the equilibrium constant $K$ is:
$K = \frac{[HI]^2}{[H_2][I_2]} = X \quad (Eq.-ii)$
For the reverse reaction (decomposition of $HI$):
$2 HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \quad (Eq.-iii)$
The equilibrium constant $K'$ for this reverse reaction is:
$K' = \frac{[H_2][I_2]}{[HI]^2} = \frac{1}{X} \quad (Eq.-iv)$
Comparing $(Eq.-ii)$ and $(Eq.-iv)$,we get:
$K = \frac{1}{K'} \quad \text{or} \quad K \times K' = 1$
Thus,the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.
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