JEE Chemistry 2023 Question with Solution

Given:

• $X\text{(g)} \rightleftharpoons Y\text{(g)} + Z\text{(g)}$ with $K_{p1} = 3$
• $A\text{(g)} \rightleftharpoons 2B\text{(g)}$ with $K_{p2} = 1$
• The degree of dissociation $\alpha$ is the same in both cases.

Find: The value of $x$ in $\frac{P_1}{P_2} = x:1$.

For reaction 1,

$$K_{p1} = \frac{\left(\frac{\alpha}{1+\alpha} \times P_1\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)P_1}$$

So,

$$3 = \frac{\alpha^2 P_1}{1-\alpha^2}$$

For reaction 2,

$$K_{p2} = \frac{\left(\frac{2\alpha}{1+\alpha} \times P_2\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)P_2}$$

Thus,

$$1 = \frac{4\alpha^2 P_2}{1-\alpha^2}$$

Now divide the two expressions:

$$\frac{K_{p1}}{K_{p2}} = \frac{P_1}{4P_2}$$

Substituting the given values,

$$\frac{3}{1} = \frac{P_1}{4P_2}$$

Hence,

$$P_1 : P_2 = 12 : 1$$

Therefore, $x = 12$.

Given:

• $X\text{(g)} \rightleftharpoons Y\text{(g)} + Z\text{(g)}$ with $K_{p1} = 3$
• $A\text{(g)} \rightleftharpoons 2B\text{(g)}$ with $K_{p2} = 1$
• Same initial amount and same degree of dissociation $\alpha$.

Find: The ratio $\frac{P_1}{P_2}$.

For $X\text{(g)} \rightleftharpoons Y\text{(g)} + Z\text{(g)}$, let initial moles be $n$. At equilibrium, moles are:

• $X = n(1-\alpha)$
• $Y = n\alpha$
• $Z = n\alpha$

Total moles at equilibrium = $n(1+\alpha)$. Therefore, the partial pressures are proportional to mole fractions, giving

$$p_X = \frac{1-\alpha}{1+\alpha} P_1, \quad p_Y = \frac{\alpha}{1+\alpha} P_1, \quad p_Z = \frac{\alpha}{1+\alpha} P_1$$

Hence,

$$K_{p1} = \frac{p_Y p_Z}{p_X} = \frac{\left(\frac{\alpha}{1+\alpha}P_1\right)^2}{\left(\frac{1-\alpha}{1+\alpha}P_1\right)} = \frac{\alpha^2 P_1}{1-\alpha^2}$$

So,

$$3 = \frac{\alpha^2 P_1}{1-\alpha^2}$$

For $A\text{(g)} \rightleftharpoons 2B\text{(g)}$, again take initial moles as $n$. At equilibrium, moles are:

• $A = n(1-\alpha)$
• $B = 2n\alpha$

Total moles at equilibrium = $n(1+\alpha)$. Thus,

$$p_A = \frac{1-\alpha}{1+\alpha} P_2, \quad p_B = \frac{2\alpha}{1+\alpha} P_2$$

Therefore,

$$K_{p2} = \frac{p_B^2}{p_A} = \frac{\left(\frac{2\alpha}{1+\alpha}P_2\right)^2}{\left(\frac{1-\alpha}{1+\alpha}P_2\right)} = \frac{4\alpha^2 P_2}{1-\alpha^2}$$

So,

$$1 = \frac{4\alpha^2 P_2}{1-\alpha^2}$$

Taking ratio,

$$\frac{3}{1} = \frac{\alpha^2 P_1}{1-\alpha^2} \div \frac{4\alpha^2 P_2}{1-\alpha^2} = \frac{P_1}{4P_2}$$

Thus,

$$P_1 = 12P_2$$

Therefore, the required value is $12$.