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Consider the following gaseous equilibria with equilibrium constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively.
$\begin{aligned} \mathrm{SO}_{2}(\mathrm{~g})+& \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \\ 2 \mathrm{SO}_{3}(g) & \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}$
The equilibrium constants are related as
Options:
$\begin{aligned} \mathrm{SO}_{2}(\mathrm{~g})+& \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \\ 2 \mathrm{SO}_{3}(g) & \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}$
The equilibrium constants are related as
Solution:
1426 Upvotes
Verified Answer
The correct answer is:
$\mathrm{K}_{1}^{2}=\frac{1}{\mathrm{~K}_{2}}$
For the reaction,
$$
\begin{aligned}
&\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \\
&\text { Equilibrium constant, } \mathrm{K}_{1}=\frac{\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{SO}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}} \quad \ldots \text { (i) }
\end{aligned}
$$
For the reaction,
$$
\begin{aligned}
&2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\
&\text { equilibrium constant, } \mathrm{K}_{2}=\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}} \quad \ldots \text { (ii) }
\end{aligned}
$$
On squaring both sides in Eq (i), we get
$$
\mathrm{K}_{1}^{2}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} \quad \ldots \text { (iii) }
$$
Eqs. (ii) $\times$ Eq (iii), we get
$$
\mathrm{K}_{1}^{2} \times \mathrm{K}_{2}=1
$$
or
$$
\begin{aligned}
\mathrm{K}_{2} &=\frac{1}{\mathrm{~K}_{1}^{2}} \\
\mathrm{~K}_{1}^{2} &=\frac{1}{\mathrm{~K}_{2}}
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \\
&\text { Equilibrium constant, } \mathrm{K}_{1}=\frac{\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{SO}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}} \quad \ldots \text { (i) }
\end{aligned}
$$
For the reaction,
$$
\begin{aligned}
&2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\
&\text { equilibrium constant, } \mathrm{K}_{2}=\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}} \quad \ldots \text { (ii) }
\end{aligned}
$$
On squaring both sides in Eq (i), we get
$$
\mathrm{K}_{1}^{2}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} \quad \ldots \text { (iii) }
$$
Eqs. (ii) $\times$ Eq (iii), we get
$$
\mathrm{K}_{1}^{2} \times \mathrm{K}_{2}=1
$$
or
$$
\begin{aligned}
\mathrm{K}_{2} &=\frac{1}{\mathrm{~K}_{1}^{2}} \\
\mathrm{~K}_{1}^{2} &=\frac{1}{\mathrm{~K}_{2}}
\end{aligned}
$$
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