Question

# For $$A\longleftrightarrow B$$ 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 $${k}_{{p}_{1}}=2{k}_{{p}_{2}}$$ is:

A
1/2
B
1/3
C
1/4
D
2

Solution

## The correct option is B 1/2For the first equation                         $$A\rightarrow B$$At t=0                  1     0t=equilibrium  $$1-\alpha \quad 2\alpha$$Total moles=$$1+\alpha$$$$P_B=X_BP_T$$Where $$P_B$$is the pressure of B $$P_T$$is the total pressure $$X_B$$is the mole fraction of BPressure is $$p_1$$$$X_B=\dfrac{2\alpha}{1+\alpha}$$similarly$$X_A=\dfrac{1-\alpha}{1+\alpha}$$$$K_P1=\dfrac{(\dfrac{2\alpha}{1+\alpha})^2p_1^2}{\dfrac{1-\alpha}{1+\alpha }p_1}$$$$K_P1=\dfrac{4\alpha^2p_1}{1-\alpha^2}$$For the second equationgiven dissociation A and C is same so                        $$C\rightarrow D+E$$At t=0                1         0   0t=equilibrium$$1-\alpha\quad \alpha \quad \alpha$$Similarly total moles$$=1-\alpha+\alpha+\alpha=1-\alpha$$finding for moles of D$$X_D=\dfrac{\alpha}{1+\alpha}$$$$X_E=\dfrac{\alpha}{1+\alpha}$$$$X_C=\dfrac{1-\alpha}{1+\alpha}$$Total pressure is $$p_2$$$$P_k2=\dfrac{\dfrac{\alpha^2}{(1+\alpha)^2}p_2^2}{\dfrac{1-\alpha}{1+\alpha}p_2}$$$$=\dfrac{\alpha^2p_2}{1-\alpha^2}$$given $$k_p1=2k_p2$$$$\dfrac{K_p1}{K_p2}=2=\dfrac{4\alpha^2p_1}{1-\alpha^2}\times\dfrac{1-\alpha^2}{\alpha^2p_2}$$$$2=\dfrac{4p_1}{p_2}$$$$\dfrac{p_1}{p_2}=\dfrac{1}{2}$$Chemistry

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