Question

The equilibrium composition for the reaction PCl₃ + Cl₂ ⇌ PCI5, at 298 K is given below. 
[PCl₃]_eq = 0.2 mol L^-
[Cl₂]eq = 0.1 mol L^-1,
[PCl₅]_eq = 0.40 mol L^-1 
If 0.2 mol of Cl₂ is added at the same temperature, the equilibrium concentrations of PCl5 is ___ × 10^-2 mol L^-1
Given: K_c for the reaction at 298 K is 20

Answer 1

Correct Answer: 49

Equilibrium Constant Calculation Problem 

Given:

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

Reaction:

\( \text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5 \)

Initial Concentrations:

\( [\text{PCl}_3] = 0.2 \text{ M}, [\text{Cl}_2] = 0.1 \text{ M}, [\text{PCl}_5] = 0.4 \text{ M} \)

At Equilibrium, assuming x M of PCl3 reacts:

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

Equilibrium Constant Expression:

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

Given: \( K_c = 20 \)

Substituting values into the equation:

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

Solving for x (details omitted, assumed to be solved):

\( x \approx 0.086 \)

Concentration of PCl5:

\( [\text{PCl}_5] = 0.4 + x = 0.4 + 0.086 = 0.486 \text{ M} \)

\( [\text{PCl}_5] = 0.486 \text{ M} = 48.6 \times 10^{-2} \text{ M} \)

\( [\text{PCl}_5] \approx 49 \times 10^{-2} \text{ M} \)

Final Answer:

Ans. 49

Answer 2

We are given the equilibrium constant Kc for the reaction at 298 K as 20.
\(K_c =\) \(\frac{[PCl_5]}{[PCl_3][Cl_2]} =\) \(\frac{0.40}{0.20 \times 0.10}\) \(= 20\)
After adding 0.2 mol of \(Cl_2\), the new concentrations become: 
\(PCl_3 = 0.2 - x,\)

\(Cl_2 = 0.2 + x, \)

\(PCl_5 = 0.4 + x.\)
Solving for x, we find that x = 0.084. Thus, the equilibrium concentration of \(PCl_5\) is 0.484 \(mol L^{-1}\).