For the gaseous reactions (I) and (II), the equilibrium constants are $X$ and $Y$ respectively. I.…
For the gaseous reactions (I) and (II), the equilibrium constants are $X$ and $Y$ respectively.
I. $\frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$
II. $2 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)$
Using the above reactions the equilibrium constant $Z$ for the reaction (III) given below is
III. $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons \mathrm{N}_2(g)+2 \mathrm{O}_2(g)$
$Z=X Y$
$Z=\frac{Y^2}{X}$
$Z=\frac{1}{X Y^2}$
$Z=\frac{1}{X^2 Y}$
Solution
For reaction
$\begin{aligned}
\frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g) & \rightleftharpoons \mathrm{NO}_2(g) \\
X & =K_{C_1}=\frac{\left[\mathrm{NO}_2\right]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}
\end{aligned}$
$\begin{aligned} & \text { For } 2 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g) \\ & Y=K_{C_2}=\frac{\left[\mathrm{N}_2 \mathrm{O}_4\right]}{\left[\mathrm{NO}_2\right]^2}\end{aligned}$
For reaction
$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons \mathrm{N}_2(g)+2 \mathrm{O}_2$
Substituting values from equation (i) and (ii) in equation (iii) we get.
$\begin{aligned}
& Z=\left(\frac{\left[\mathrm{NO}_2\right]}{X}\right)^2 / Y\left[\mathrm{NO}_2\right]^2 \\
& Z=\frac{\left[\mathrm{NO}_2\right]^2}{X^2 Y\left[\mathrm{NO}_2\right]^2}=\frac{1}{X^2 Y}
\end{aligned}$
$\begin{aligned} & \text { For } 2 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g) \\ & Y=K_{C_2}=\frac{\left[\mathrm{N}_2 \mathrm{O}_4\right]}{\left[\mathrm{NO}_2\right]^2}\end{aligned}$
For reaction
$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons \mathrm{N}_2(g)+2 \mathrm{O}_2$
Substituting values from equation (i) and (ii) in equation (iii) we get.
$\begin{aligned}
& Z=\left(\frac{\left[\mathrm{NO}_2\right]}{X}\right)^2 / Y\left[\mathrm{NO}_2\right]^2 \\
& Z=\frac{\left[\mathrm{NO}_2\right]^2}{X^2 Y\left[\mathrm{NO}_2\right]^2}=\frac{1}{X^2 Y}
\end{aligned}$