Chemical and Ionic Equilibrium 1 Question 9
9. The values of $\frac{K_{p}}{K_{C}}$ for the following reactions at $300 \mathrm{~K}$ are, respectively (At $300 \mathrm{~K}, R T=24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$ )
$$ \begin{aligned} & \mathrm{N}{2}(g)+\mathrm{O}{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \ & \mathrm{N}{2} \mathrm{O}{4}(g) \rightleftharpoons 2 \mathrm{NO}{2}(g) \ & \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}{3}(g) \end{aligned} $$
(2019 Main, 10 Jan I)
(a) $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}, 606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$
(b) $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}, 1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$
(c) $24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}, 606.0 \mathrm{dm}^{6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$, $1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$
(d) $1,4.1 \times 10^{-2} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \mathrm{~mol}, 606 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$
10 Consider the following reversible chemical reactions,
$$ \begin{aligned} A_{2}(g)+B_{2}(g) & \stackrel{K_{1}}{\rightleftharpoons} 2 A B(g) \ 6 A B(g) & \stackrel{K_{2}}{\rightleftharpoons} 3 A_{2}(g)+3 B_{2}(g) \end{aligned} $$
The relation between $K_{1}$ and $K_{2}$ is (2019 Main, 9 Jan II)
(a) $K_{2}=K_{1}^{3}$
(b) $K_{1} K_{2}=3$
(c) $K_{2}=K_{1}^{-3}$
(d) $K_{1} K_{2}=\frac{1}{3}$
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Solution:
- We know that, the relationship between $K_{p}$ and $K_{C}$ of a chemical equilibrium state (reaction) is
$$ K_{p}=K_{C}(R T)^{\Delta n_{g}} \Rightarrow \frac{K_{p}}{K_{C}}=(R T)^{\Delta n_{g}} $$
where, $\quad \Delta n_{g}=\Sigma n_{\text {Products }}-\Sigma n_{\text {Reactants }}$
(i) $\quad \mathrm{N}{2}(g)+\mathrm{O}{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
$$ \Rightarrow \quad(R T)^{2-(1+1)}=(R T)^{0}=1 $$
(ii) $\quad \mathrm{N}{2} \mathrm{O}{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$
$$ \Rightarrow \quad(R T)^{2-1}=R T=24.62 \mathrm{dm}^{3} \mathrm{atmmol}^{-1} $$
(iii) $\quad \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
$\Rightarrow(R T)^{2-(3+1)}=(R T)^{-2}$
$$ \begin{aligned} & =\frac{1}{\left(24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}\right)^{2}} \ & =1.649 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2} \end{aligned} $$
