JEE Chemistry 2023 Question with Solution

Given:

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

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

First calculate the equilibrium constant:

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

Let $x \, \text{M}$ more of $\text{PCl}_3$ react after adding $\text{Cl}_2$. Then the new concentrations at equilibrium are:

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

Using the equilibrium constant expression:

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

Since $K_c = 20$,

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

Solving this gives:

$$x \approx 0.086$$

Therefore,

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

Now express it as $\times 10^{-2} \, \text{mol L}^{-1}$:

$$0.486 \, \text{M} = 48.6 \times 10^{-2} \, \text{M}$$ $$[\text{PCl}_5] \approx 49 \times 10^{-2} \, \text{mol L}^{-1}$$

Therefore, the required numerical answer is 49.

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

Find: The new equilibrium concentration of $\text{PCl}_5$ after adding $0.2$ mol of $\text{Cl}_2$.

From the given equilibrium data:

$$K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} = \frac{0.40}{0.20 \times 0.10} = 20$$

After adding $\text{Cl}_2$, take the new concentrations as:

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

Substitute into the equilibrium expression:

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

From the extracted working,

$$x \approx 0.086$$

So,

$$[\text{PCl}_5] = 0.4 + 0.086 = 0.486 \, \text{mol L}^{-1}$$

Hence,

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

Therefore, the answer is 49.