Equilibrium - Result Question 11
####12. The values of $K p_1$ and $K p_2$ for the reactions $X \rightarrow Y+Z$
[2008]
and $A \rightarrow 2 B$
are in the ratio of $9: 1$. If degree of dissociation of $X$ and $A$ be equal, then total pressure at equilibrium for (a) and (b) are in the ratio :
(a) $3: 1$
(b) $1: 9$
(c) $36: 1$
(d) $1: 1$
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Solution:
- (c) Given reaction are
$X \rightarrow Y+Z$
and $A \rightarrow 2 B$
Let the total pressure for reaction (i) and (ii) be $P_1$ and $P_2$ respectively, then
$\frac{K _{P_1}}{K _{P_2}}=\frac{9}{1}$
(given)
After dissociation,
At equilibrium
$X \rightarrow \quad \rightarrow+Z$ $(1-\alpha) \underset{\alpha}{\rightarrow} \quad \mathrm{\alpha}$
[Let 1 mole of $X$ dissociate with $\alpha$ as degree of dissociation]
Total number of moles $=1-\alpha+\alpha+\alpha=(1+\alpha)$
Thus $P_X=(\frac{1-\alpha}{1+\alpha}) \cdot P_1 ; P_Y=(\frac{\alpha}{1+\alpha}) P_1$;
$ P_Z=(\frac{\alpha}{1+\alpha}) \cdot P_1 $
$\therefore K _{P_1}=(\frac{\alpha}{1+\alpha}) . P_1 \times \frac{\alpha}{(1+\alpha)}$.
$P_1 /(\frac{1-\alpha}{1+\alpha}) \cdot P_1$
Similarly for
$ \begin{equation*} A \rightarrow 2 B \tag{i} \end{equation*} $
At equilibrium $(1-\alpha) 2 \alpha$
We have,
$K _{P_2}=(\frac{2 \alpha P_2}{1+\alpha})^{2} /(\frac{1-\alpha}{1+\alpha}) P_2$
Dividing (i) by (ii), we get
$\frac{K _{P_1}}{K _{P_2}}=\frac{\alpha^{2} \cdot P_1}{4 \alpha^{2} \cdot P_2}$ or $\frac{K _{P_1}}{K _{P_2}}=\frac{1}{4} \cdot \frac{P_1}{P_2}$ or $9=\frac{1}{4} \cdot \frac{P_1}{P_2}$
or $\frac{P_1}{P_2}=\frac{36}{1}$ or $P_1: P_2=36: 1$
$ [\therefore \frac{K _{P_1}}{K _{P_2}}=\frac{9}{1}] $
