Question

For $A⟷B$ equilibrium constant at total pressure $p_1$ is $k_{p_1}$ and for $C\rightleftharpoons D+E$ equilibrium constant at total pressure $p_2$ is $k_{p_2}$. If both reaction are gaseous homogeneous reations and degree of dissociation of $A$ and $C$ are same, then the ratio of $p_1/p_2$ if $k_{p_1}=2k_{p_2}$ is:

• A. 1/2
• B. 1/3
• C. 1/4
• D. 2

Answer 1

Answer: (A)

1/2

For the first equation

$A\to B$

At t=0 1 0

t=equilibrium $1-\alpha 2\alpha$

Total moles=$1+\alpha$

$P_B=X_B{P}_T$

Where $P_B$is the pressure of B

$P_T$is the total pressure

$X_B$is the mole fraction of B

Pressure is $p_1$

$X_B=\frac{2\alpha }{1+\alpha }$

similarly

$X_A=\frac{1-\alpha }{1+\alpha }$

$K_{P}1=\frac{(\frac{2\alpha }{1+\alpha }{)}^2{p}_1^2}{\frac{1-\alpha }{1+\alpha }{p}_1}$

$K_{P}1=\frac{4{\alpha }^2{p}_1}{1-{\alpha }^2}$

For the second equation

given dissociation A and C is same so

$C\to D+E$

At t=0 1 0 0

t=equilibrium$1-\alpha \alpha \alpha$

Similarly

total moles$=1-\alpha +\alpha +\alpha =1-\alpha$

finding for moles of D

$X_D=\frac{\alpha }{1+\alpha }$

$X_E=\frac{\alpha }{1+\alpha }$

$X_C=\frac{1-\alpha }{1+\alpha }$

Total pressure is $p_2$

$P_{k}2=\frac{\frac{{\alpha }^2}{(1+\alpha {)}^2}{p}_2^2}{\frac{1-\alpha }{1+\alpha }{p}_2}$

$=\frac{{\alpha }^2{p}_2}{1-{\alpha }^2}$

given $k_{p}1=2k_{p}2$

$\frac{K_{p}1}{K_{p}2}=2=\frac{4{\alpha }^2{p}_1}{1-{\alpha }^2}\times \frac{1-{\alpha }^2}{{\alpha }^2{p}_2}$

$2=\frac{4p_1}{p_2}$

$\frac{p_1}{p_2}=\frac{1}{2}$