An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of
ID: 873852 • Letter: A
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 appropriate chemical equation is
An equilibrium mixture of PCls(g), PCls(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 Calculate the new partial pressures after equilibrium is reestablished. Number Torr PCi, Number Torr CI Number Torr PClExplanation / Answer
the reaction is
PCl3 + Cl2 ---> PCl5
Kp = PPCl5 / ( PPCl3 * PCl2)
kp = 217 / 13.2 * 13.2
kp = 1.2454
Given
total pressure at the moment of mixing = 263 torr
so
total initial pressure = 263 torr
total initial pressure = PPCl5 + PPCl3 + PCl2
given
PPCl5 = 217
PPCl3 = 13.2
PCl2 = ?
so
263 = 217 + 13.2 + PCl2
PCl2 = 32.8 torr
now
the system reaches equilirbium
PCl3 + Cl2 ---> PCl5
using ICE table
inital pressures of PCl3 , Cl2 , PCl5 are 13.2 , 32.8 , 217
change in pressures of PCl3 , Cl2 , PCl5 are -x , -x , +x
final pressures of PCl3 , Cl2 , PCl5 are 13.2 - x , 32.8 -x , 217 + x
now
Kp = PPCl5 / ( PPCl3 * PCl2)
1.2454 = (217 + x ) / ( 13.2-x) (32.8-x)
solving this we get
x= 6.404
so
partial pressures after equilirbium are
PPCl3 = 13.2 - x = 13.2 - 6.404 = 6.796 torr
PCl2 = 32.8 - 6.404 = 26.396 torr
PPCl5 = 217 + x = 217 + 6.404 = 223.404 torr