Medium
If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.
(N/A) The equilibrium of $HI$ synthesis at a definite temperature is as follows:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ $\quad (Eq.-I)$
The equilibrium constant $K_c$ is:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = X$ $\quad (Eq.-II)$
On multiplying the $HI$ synthesis equation by a factor $n$,we get:
$nH_{2(g)} + nI_{2(g)} \rightleftharpoons 2nHI_{(g)}$ $\quad (Eq.-III)$
For this reaction,the new equilibrium constant $K'_c$ is:
$K'_c = \frac{[HI]^{2n}}{[H_2]^n[I_2]^n} = \left( \frac{[HI]^2}{[H_2][I_2]} \right)^n = X^n$ $\quad (Eq.-IV)$
Thus,$K'_c = (K_c)^n$.
It should be noted that because the equilibrium constants $K_c$ and $K'_c$ have different numerical values,it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ $\quad (Eq.-I)$
The equilibrium constant $K_c$ is:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = X$ $\quad (Eq.-II)$
On multiplying the $HI$ synthesis equation by a factor $n$,we get:
$nH_{2(g)} + nI_{2(g)} \rightleftharpoons 2nHI_{(g)}$ $\quad (Eq.-III)$
For this reaction,the new equilibrium constant $K'_c$ is:
$K'_c = \frac{[HI]^{2n}}{[H_2]^n[I_2]^n} = \left( \frac{[HI]^2}{[H_2][I_2]} \right)^n = X^n$ $\quad (Eq.-IV)$
Thus,$K'_c = (K_c)^n$.
It should be noted that because the equilibrium constants $K_c$ and $K'_c$ have different numerical values,it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.
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