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} \text{ 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} \text{ atm mol}^{-1}, 606.0$ $ dm^{6}$ $ atm^{2} $ $mol^{-2}$
(b) $1,24.62$ $ dm^{3}\text{ atm mol}^{-1}, 1.65 \times 10^{-3}$ $ dm^{-6} $ $atm^{-2} $ $mol^{2}$
(c) $24.62$ $ dm^{3}\text{ 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} $ $mol^{-2}$
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Answer:
Correct Answer: 9. (b)
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}$ $ atm $ $mol^{-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} \text {atm mol} ^{-1}\right)^{2}} \\ & =1.649 \times 10^{-3} dm^{-6} atm^{-2} mol^{2} \end{aligned} $
BETA
