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: Kc for the reaction at 298 K is 20
[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: Kc for the reaction at 298 K is 20
Correct Answer: 49
Approach Solution - 1
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
Approach Solution -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 \(mol L^{-1}\).
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