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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
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B
1/3
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C
1/4
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D
2
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Solution

The correct option is B 1/2
For the first equation
                         $$A\rightarrow B$$
At t=0                  1     0
t=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 B
Pressure 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 equation
given dissociation A and C is same so
                        $$C\rightarrow D+E$$
At t=0                1         0   0
t=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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