JEE Chemistry 2023 Question with Solution

Given:

• Reaction: $\text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5$
• Initial equilibrium concentrations: $[\text{PCl}_3] = 0.2 \, \text{M}$, $[\text{Cl}_2] = 0.1 \, \text{M}$, $[\text{PCl}_5] = 0.4 \, \text{M}$
• Additional chlorine added: $0.2 \, \text{mol}$

Find: The value in _____ × $10^{-2} \, \text{mol L}^{-1}$ for the new equilibrium concentration of $\text{PCl}_5$.

First calculate the equilibrium constant:

$$K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} = \frac{0.4}{0.2 \times 0.1} = 20$$

After adding chlorine, let $x$ be the concentration of $\text{PCl}_3$ that reacts further. Then the new equilibrium concentrations are:

$$[\text{PCl}_3] = 0.2 - x$$ $$[\text{Cl}_2] = 0.1 + 0.2 - x = 0.3 - x$$ $$[\text{PCl}_5] = 0.4 + x$$

Using the equilibrium expression again:

$$20 = \frac{0.4 + x}{(0.2 - x)(0.3 - x)}$$

Solving this gives:

$$x \approx 0.086$$

Therefore,

$$[\text{PCl}_5] = 0.4 + 0.086 = 0.486 \, \text{M}$$

Now express it in the required form:

$$0.486 \, \text{M} = 48.6 \times 10^{-2} \, \text{M}$$

So,

$$[\text{PCl}_5] \approx 49 \times 10^{-2} \, \text{mol L}^{-1}$$

Therefore, the required numerical value is 49.

Given: $K_c = 20$ for $\text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5$.

Find: New equilibrium concentration of $\text{PCl}_5$ after adding chlorine.

Let the forward reaction proceed by $x$. Then:

$$[\text{PCl}_3] = 0.2 - x, \quad [\text{Cl}_2] = 0.3 - x, \quad [\text{PCl}_5] = 0.4 + x$$

Substitute into the equilibrium expression:

$$20 = \frac{0.4 + x}{(0.2 - x)(0.3 - x)}$$

Rearrange:

$$20(0.2 - x)(0.3 - x) = 0.4 + x$$

From the extracted solution, solving this gives approximately:

$$x \approx 0.086$$

Hence,

$$[\text{PCl}_5] = 0.4 + x = 0.486 \, \text{mol L}^{-1}$$

Converting:

$$0.486 \, \text{mol L}^{-1} = 48.6 \times 10^{-2} \, \text{mol L}^{-1} \approx 49 \times 10^{-2} \, \text{mol L}^{-1}$$

Therefore, the equilibrium concentration corresponds to 49 in the blank.