Answer to Question #106849 in Physical Chemistry for mya 25

Question

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates.

The chemical equation for this reaction is
PCl3(g)+Cl2(g) −⇀ PCl5(g)

Calculate the new partial pressures, P, after equilibrium is reestablished.

Answer 1

Before adding:

K = P (PCl5)/(P(PCl3)*P(Cl2))

K = 217 / 13.2*13.2

K = 1.245 = const

After adding:

K = y / x²

y - P (PCl5)

x - P (Cl2) = P (PCl3)

1.245 = y / x²

x + x + y = 263 - total pressure!

Solving the system:

x = 13.75 Torr - Cl2 and PCl3

y = 235.5 Torr - PCl5