Chemical and Ionic Equilibrium - Result Question 9
####9. The values of $\frac{K _p}{K _C}$ for the following reactions at $300 K$ are, respectively (At $300 K, R T=24.62 dm^{3} atm mol^{-1}$ )
$$ \begin{aligned} & N _2(g)+O _2(g) \rightleftharpoons 2 NO(g) \ & N _2 O _4(g) \rightleftharpoons 2 NO _2(g) \ & N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) \end{aligned} $$
(2019 Main, 10 Jan I)
(a) $1,24.62 dm^{3} atm mol^{-1}, 606.0 dm^{6} atm^{2} mol^{-2}$
(b) $1,24.62 dm^{3} atm mol^{-1}, 1.65 \times 10^{-3} dm^{-6} atm^{-2} mol^{2}$
(c) $24.62 dm^{3} atm mol^{-1}, 606.0 dm^{6} atm^{-2} mol^{2}$, $1.65 \times 10^{-3} dm^{-6} atm^{-2} mol^{2}$
(d) $1,4.1 \times 10^{-2} dm^{-3} atm^{-1} mol, 606 dm^{6} atm^{2} mol^{-2}$
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) $\quad N _2(g)+O _2(g) \rightleftharpoons 2 NO(g)$
$$ \Rightarrow \quad(R T)^{2-(1+1)}=(R T)^{0}=1 $$
(ii) $\quad N _2 O _4(g) \rightleftharpoons 2 NO _2(g)$
$$ \Rightarrow \quad(R T)^{2-1}=R T=24.62 dm^{3} atmmol^{-1} $$
(iii) $\quad N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g)$
$\Rightarrow(R T)^{2-(3+1)}=(R T)^{-2}$
$$ \begin{aligned} & =\frac{1}{\left(24.62 dm^{3} atm mol^{-1}\right)^{2}} \ & =1.649 \times 10^{-3} dm^{-6} atm^{-2} mol^{2} \end{aligned} $$
