Question

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr,...

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 appropriate chemical equation is

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

Calculate the new partial pressures after equilibrium is reestablished.

P PCl3 = ? torr

P PCl2 = ? torr

P PCl5 = ? torr

Solutions

Expert Solution

[pPCl3] = 6.80 torr

[pCl2] = 26.4 torr

[pPCl5] = 223.4 torr

Explanation

pressure equilibrium constant, Kp = [pPCl5]eq / [pPCl3]eq[pCl2]eq

Kp = (217.0 torr) / (13.2 torr) * (13.2 torr)

Kp = 1.2454

Total pressure = [pPCl5]eq + [pPCl3]eq + [pCl2]eq

Total pressure = (217.0 torr) + (13.2 torr) + (13.2 torr)

Total pressure = 243.4 torr

pressure Cl2 added = (total pressure after adding Cl2) - (total pressure before adding Cl2)

pressure Cl2 added = (263.0 torr) - (243.4 torr)

pressure Cl2 added = 19.6 torr

total pressure Cl2 = (13.2 torr) + (19.6 torr)

total pressure Cl2 = 32.8 torr

ICE table PCl3 Cl2 PCl5
Initial 13.2 torr 32.8 torr 217.0 torr
Change -x -x +x
Equilibrium 13.2 torr - x 32.8 torr - x 217.0 torr + x

Kp = [pPCl5]eq / [pPCl3]eq[pCl2]eq

1.2454 = (217.0 torr + x) / [(13.2 torr - x) * (32.8 torr - x)]

solving for x, x = 6.4 torr

[pPCl3]eq = 13.2 torr - x

[pPCl3]eq = 13.2 torr - 6.4 torr

[pPCl3]eq = 6.80 torr

[pCl2]eq = 32.8 torr - x

[pCl2]eq = 32.8 torr - 6.4 torr

[pCl2]eq = 26.4 torr

[pPCl5]eq = 217.0 torr + x

[pPCl5]eq = 217.0 torr + 6.4 torr

[pPCl5]eq = 223.4 torr

queen_honey_blossom answered 2 years ago
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