Answer:
p'PCl3 = 6.8 torr
p'Cl2 =26.4 torr
p'PCl5 =223.4 torr
Step-by-step explanation:
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 after equilibrium is reestablished. [in torr]
pPCl3
pCl2
pPCl5
Step 1: Data given
Partial pressure before adding chlorine gas:
Partial pressure of PCl5 = 217.0 torr
Partial pressureof PCl3 = 13.2 torr
Partial pressureof Cl2 = 13.2 torr
A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing
Step 2: The equation
PCl3(g)+Cl2(g) ⇔ PCl5(g)
Step 3: The expression of an equilibrium constant before adding chlorine gas
Kp = pPCl5 / (pPCl3 * pCl2)
Kp = 217.0 / (13.2 * 13.2)
Kp = 1.245
Step 4: The expression of an equilibrium constant after adding chlorine gas
Partial pressure of PCl5 = 217.0 torr
Partial pressure of PCl3 = 13.2
Partial pressure of Cl2 = TO BE DETERMINED
Step 5: The total pressure of the system
Ptotal = pPCl5 + pPCl3 + pCl2
263.0 torr = 217.0 torr + 13.2 torr + pCl2
pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr
Step 6: The initial pressure
The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)
pPCl3 = 13.2 torr
pCl2 = 32.8 torr
pPCl5 = 217.0 torr
Step 7: The pressure at the equilibrium
p'PCl3 = (13.2 -x) torr
p'Cl2 = (32.8 - x) torr
p'PCl5 = (217.0 + x) torr
Step 8: The equilibrium constant
'Kp = p'PCl5 / (p'PCl3 * p'Cl2)
1.245 = (217.0+x) / ((13.2-x)(32.8-x)
x = 6.40 torr
p'PCl3 = 13.2 -6.40 = 6.8 torr
p'Cl2 = 32.8 - 6.40 =26.4 torr
p'PCl5 = 217.0 + x) 6.4 = 223.4 torr