Question

For given two equilibria attained in a container which are correct if degree of dissociation of A and A' are $\alpha$ and ${\alpha }^{\prime }$.
$A\left(s\right)\rightleftharpoons 2B\left(g\right)+C\left(g\right);K_{p_1}=8\times {10}^{-2}$
$A^{\prime }\left(s\right)\rightleftharpoons 2B\left(g\right)+D\left(g\right);K_{p_2}=2\times {10}^{-2}$

• A. $\frac{K_{p_1}}{K_{p_2}}={\left[\frac{(3{\alpha }^{\prime }+2\alpha )}{(3\alpha +2{\alpha }^{\prime })}\right]}^3\times \frac{\alpha }{{\alpha }^{\prime }}$
• B. $P_C^{\prime }/P_D^{\prime }=4$
• C. $P_B^{\prime }=2P_C^{\prime }+2P_D^{\prime }$
• D. $\alpha >{\alpha }^{\prime }$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $P_C^{\prime }/P_D^{\prime }=4$
B $\alpha >{\alpha }^{\prime }$
C $P_B^{\prime }=2P_C^{\prime }+2P_D^{\prime }$
D $\frac{K_{p_1}}{K_{p_2}}={\left[\frac{(3{\alpha }^{\prime }+2\alpha )}{(3\alpha +2{\alpha }^{\prime })}\right]}^3\times \frac{\alpha }{{\alpha }^{\prime }}$

$A\left(s\right)\rightleftharpoons \underset{2\alpha +{\alpha }^{\prime }}{2B\left(g\right)}+\underset{\alpha }{C\left(g\right)}$

$A^{\prime }\left(s\right)\rightleftharpoons \underset{2{\alpha }^{\prime }+{\alpha }^{\prime }}{2B}+\underset{{\alpha }^{\prime }}{D}$

$K_{p_1}=(2\alpha +2{\alpha }^{\prime }{)}^2\times \alpha \times {\left[\frac{P}{(3\alpha +2{\alpha }^{\prime })}\right]}^3$

$K_{p_2}=(2\alpha +2{\alpha }^{\prime }{)}^2\times {\alpha }^{\prime }\times {\left[\frac{P}{(3{\alpha }^{\prime }+2\alpha )}\right]}^3$

$\frac{K_{p_1}}{K_{p_2}}=\frac{[3{\alpha }^{\prime }+2\alpha {]}^3}{[3\alpha +2\alpha {]}^3}\times \frac{\alpha }{{\alpha }_1}$

Also, $K_{p_1}=(P_B^{\prime }{)}^2\times P_c^{\prime }$

$K_{p_2}=(P_B^{\prime }{)}^2\times P_p^{\prime }$

or $\frac{P_c^{\prime }}{P_D^{\prime }}=\frac{K_{p_1}}{K_{p_2}}=\frac{8\times {10}^{-2}}{2\times {10}^{-2}}=4$

Also, $P_B^{\prime }=2\times (P_c^{\prime }+P_D^{\prime })$

$K_{p_1}>K_{p_2}\therefore \alpha >{\alpha }_1$

Hence, all the options are correct.