Question

The equilibrium constant, K, for the following reaction is 3.40×10-2 at 527 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 10.5 L container at 527 K contains 0.271 M PCl5, 9.60×10-2 M PCl3 and 9.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 19.5 L? [PCl5] = M [PCl3] = M [Cl2] = M

Answer 1

Answer

[PCl5] = 0.1309M

[PCl3] = 0.06671M

[Cl2] = 0.06671M

Explanation

PCl5(g) <-------> PCl3(g) + Cl2(g)

K = [PCl3][Cl2]/[PCl5] = 3.40 × 10^-2

After increasing volume of the container

Initial concentration

[PCl₅] = 0.271M/( 19.5L/10.5L) = 0.1459M

[PCl3] = 0.0960M/(19.5L /10.5L) = 0.05169M

[Cl2] = 0.0960M/(19.5L /10.5L) = 0.05169M

change in concentration

[PCl5] = -x

[PCl3] = +x

[Cl2] = +x

equilibrium concentration

[PCl5] = 0.1459 - x

[PCl3] = 0.05169 + x

[Cl2] = 0.05169 + x

so,

(0.05169 + x)²/(0.1459 - x) = 3.40 ×10^-2

solving for x

x = 0.01502

Therefore, at equilibrium

[PCl5] = 0.1459 - 0.01502 = 0.1309M

[PCl3] = 0.05169 + 0.01502 = 0.06671M

[Cl2] = 0.05169 + 0.01502 = 0.06671M