An equilibrium mixture of PCl5g PCl3g and Cl2g has partial p

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

Solution

P = 217 PCl5

P = 13.2 PCl3

P = 13.2 CL2>

Calculate equilibrium Kp

Kp = PCl5 / (PCl3*PCl2) = 217/(13.2^2) =1.245

PCl3 + Cl2 --> PCl5

PT = 263 instantanoeus

PT = 217 +13.2 +13.2 x = 262

x = 18.6 torr of Cl2 was added

In equilibrium

PPCl5 = 217 + x

PPCl3 = 13.2 - x

PCL2 = 31.8 -x

substitute in Kp

Kp = PCl5 / (PCl3*PCl2)

1.245 = (217 + x)/(13.2 - x)/(31.8 -x)

solve for x

(13.2 - x)/(31.8 -x) = (217+x)/1.245

419.8 -45x + x^2 = 174.3 + 0.803x

x^2 -45.803x + 245.5 = 0

x = 6.2

Subtitute in all concnetations

PPCl5 = 217 +6.2 = 223.2

PPCl3 = 13.2 - 6.2 = 7

PCL2 = 31.8 -6.2 = 25.6