84. Calculate the pressures of NO, Cl2, and NOCI in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl2. (Hint: Kp is small; assume the reverse reaction goes to completion then comes back to equilibrium.) 2NO(8) +Cl2(8) 2NOCI(g) Kp = 2.5 × 103 Solution 2NO(g) + Cl2(g) --------------- 2 NOCl(g) 4.0atm      2.0atm                            0 -2x              -x                            +2x 4.0-2x       2.0-x                         +2x Kp= P^2NOCl/P^2NOxPCl2 2.5x10^3 = (2x)^2/(4.0-2x)^2 (2.0-x) for solving the equation x= 1.89 at equilibrium Pressure of NO= 4.0-2x=4.0-2(1.89)=0.22 atm Pressure of Cl2= 2.0-x=2.0-1.89= 0.11 atm Pressure of NOCl = 2x = 2x1.89= 3.78 atm .

1. 84. Calculate the pressures of NO, Cl2, and NOCI in an equilibrium mixture produced by the
reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl2. (Hint: Kp is small; assume the
reverse reaction goes to completion then comes back to equilibrium.) 2NO(8) +Cl2(8) 2NOCI(g)
Kp = 2.5 × 103
Solution
2NO(g) + Cl2(g) --------------- 2 NOCl(g)
4.0atm 2.0atm 0
-2x -x +2x
4.0-2x 2.0-x +2x
Kp= P^2NOCl/P^2NOxPCl2
2.5x10^3 = (2x)^2/(4.0-2x)^2 (2.0-x)
for solving the equation
x= 1.89
at equilibrium
Pressure of NO= 4.0-2x=4.0-2(1.89)=0.22 atm
Pressure of Cl2= 2.0-x=2.0-1.89= 0.11 atm
Pressure of NOCl = 2x = 2x1.89= 3.78 atm