1022 PM sessionmasteringchemistrycom C Welcome Stude Maze Ru

10:22 PM session.masteringchemistry.com C Welcome Stude.. Maze Runner: T..Elementary Dif.. Watch The Wal...

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

Kp = P_PCl5/(P_PCl3 × P_Cl2) = 0.120

Initial partial pressure

P_PCl5 = 0.500atm

P_PCl3 = 0.300atm

P_Cl2 = 0.500atm

Change in Partial pressure

P_PCl5 = +x

P_PC3 = - x

P_Cl2 = - x

equillibrium partial presssure

P_PCl3 = 0.300 - x

P_Cl2 = 0.500 - x

P_PCl5 = 0.500 + x

therefore,

0.500 + x/((0.300 - x) (0.500 - x)) = 0.120

solving for x

x = - 0.4204

Therefore, at equillubrium,

Partial pressure of PCl5 = 0.500+(-0.4204) = 0.0796atm

Partial pressure of PCl3 = 0.300 - (-0.4204) = 0.7204atm

Partial pressure of Cl2 = 0.500 - ( - 0.4204) = 0.9204atm