Suppose that before new equilibrium had been reached, a rate of forward reactions exceeded a rate of reverse ones by $N_{A} x$. Then
\[
\begin{gathered}
\Delta \nu_{1}=-2 x, \quad \Delta \nu_{2}=-x, \quad \Delta \nu_{3}=2 x \\
\Delta \nu=\Delta \nu_{1}+\Delta \nu_{2}+\Delta \nu_{3}=-x
\end{gathered}
\] Let us write the ideal gas law for the equilibrium states:
$p V=\left(\nu_{1}+\nu_{2}+\nu_{3}\right) R T, \quad(p+\Delta p)(V+\Delta V)=\left(\nu_{1}+\nu_{2}+\nu_{3}-x\right) R T$.
For small increments the solution of this equations is
\[
\frac{\Delta p}{p}+\frac{\Delta V}{V}=-\frac{x}{\nu_{1}+\nu_{2}+\nu_{3}}\tag{2}
\] Now let us derive an equation to determine $\Delta V / V$ and $x$. The quantity
\[
\frac{\left(\nu_{1} / V\right)^{2}\left(\nu_{2} / V\right)}{\left(\nu_{3} / V\right)^{2}}
\] is proportional to the reaction rates and it depends only on temperature, hence, it is the same for the two equilibrium states:
\[
\begin{gathered}
\frac{\nu_{1}^{2} \nu_{2}}{\nu_{3}^{2} V}=\frac{\left(\nu_{1}+\Delta \nu_{1}\right)^{2}\left(\nu_{2}+\Delta \nu_{2}\right)}{\left(\nu_{3}+\Delta \nu_{3}\right)^{2}(V+\Delta V)} \\
\text { whence } \quad \frac{\Delta V}{V}=2 \frac{\Delta \nu_{1}}{\nu_{1}}+\frac{\Delta \nu_{2}}{\nu_{2}}-2 \frac{\Delta \nu_{3}}{\nu_{3}}
\end{gathered}
\] Using (1), one obtains
\[
\frac{\Delta V}{V}=-x\left(\frac{4}{\nu_{1}}+\frac{1}{\nu_{2}}+\frac{4}{\nu_{3}}\right)\tag{3}
\]
Solving together (2) and (3), one finds:
\[
\begin{gathered}
x=\frac{\Delta p}{p}\left(\frac{4}{\nu_{1}}+\frac{1}{\nu_{2}}+\frac{4}{\nu_{3}}-\frac{1}{\nu_{1}+\nu_{2}+\nu_{3}}\right)^{-1} \\
\frac{\Delta V}{V}=-\frac{\Delta p}{p} \frac{\frac{4}{\nu_{1}}+\frac{1}{\nu_{2}}+\frac{4}{\nu_{3}}}{\frac{1}{\nu_{1}}+\frac{1}{\nu_{2}}+\frac{4}{\nu_{3}}-\frac{1}{\nu_{1}+\nu_{2}+\nu_{3}}}
\end{gathered}
\] Since $x$ is known, $\Delta \nu_{1}, \Delta \nu_{2}$, and $\Delta \nu_{3}$ follow from (1).