For the reaction (1) and (2) : A B + C ....(1) D 2E ....(2) Given… - Vidyadip

MCQ

For the reaction $(1)$ and $(2)$ :
$A \rightleftharpoons B + C ....(1)$
$D \rightleftharpoons 2E ....(2)$
Given $K_{P_1} : K_{P_2} : : 9 : 1$
If the degree dissociation of $A$ and $D$ be same then the total pressure at equilibria $(1)$ and $(2)$ are in the ratio (Assume reaction are started with equal number of moles of $A$ and $D$).

A
$3:1$
✓
$36:1$
C
$1:1$
D
$0.5:1$

✓

Answer

Correct option: B.

$36:1$

b
$\mathop {\mathop A\limits_1 }\limits_{1 - \alpha } \rightleftharpoons \mathop {\mathop B\limits_0 }\limits_\alpha + \mathop {\mathop C\limits_0 }\limits_{\alpha \,\, = (1 + \alpha )} \,\,\,\,\,\,\,\,\,\,\,$

${K_{{p_1}}} = \frac{{{\alpha ^2}}}{{1 - {\alpha ^2}}},{p_1}$

$\mathop {\mathop D\limits_1 }\limits_{1 - \alpha } \rightleftharpoons \mathop {\mathop {2E}\limits_0 }\limits_{2\alpha \,\,\, = (1 + \alpha )} $

${\mathrm{K}_{p_{2}}=\frac{40^{2}}{1-\alpha^{2}} \cdot \mathrm{P}_{2}}$

$\frac{\mathrm{K}_{\mathrm{p}_{\mathrm{r}}}}{\mathrm{K}_{\mathrm{p}_{2}}}=\frac{\mathrm{p}_{1}}{4 \mathrm{p}_{2}}$

so $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=4 \cdot \frac{\mathrm{Kp}_{1}}{\mathrm{Kp}_{2}}=4 \times 9=36: 1$

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