chemistry pressure

Question

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Kp = 1.45 × 10⁻⁴ at 160 °C. A 1.00 L vessel at 160 °C is filled with PCl₅(g) at an initial pressure of 3.75 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Cl₂(g) at equilibrium?

Answer 1

Set up an ICE table as follows:

PCl₅(g) <==> PCl₃(g) + Cl₂(g) ... Kp = 1.45x10^-4

3.75....................0................0...........Initial

-x......................+x...............+x..........Change

3.75-x.................x.................x..........Equilibrium

Use the Kp expression to solve for x:

Kp = 1.45x10^-4 = [PCl₃][Cl₂] / [PCl₅]

1.45x10^-4 = (x)(x) / 3.75-x (assume x is small relative to 3.75 and ignore it in the denominator)

1.45x10^-4 = x² / 3.75

x² = 5.4375x10^-4

x = 0.0233 atm (note: this is only ~ 0.6% of 3.75 so above assumption was valid)

Equilibrium partial pressure of Cl₂ = x (see ICE table)

Equilibrium partial pressure of Cl₂ = 0.0233 atm