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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, K_p = [pPCl₅]_eq / [pPCl₃]_eq[pCl₂]_eq

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

K_p = 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

K_p = [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 1 year ago

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