Partial Pressure

Question

Knowing Kp = Kc(RT)Δng (R=0.0821 L atm/mol K)

The reaction

2HCl (g) ↔H2 (g) + Cl2 (g)

has Kc= 3.2 ×10−34 at 25 Celsius. If a reaction vessel contains initially 0.45 atm of HCl which then reacts to reach equilibrium, what will be the partial pressure of H2 and Cl2

Answer 1

2HCl(g) <==> H2(g) + Cl2(g)

0.45 atm.........0...........0........Initial

-2x...............+x...........+x.......Change

0.45-2x..........x.............x........Equilibrium

Kp = (H₂)(Cl₂) / (HCl)²

To find Kp, we use Kp = Kc(RT)∆n where ∆n is change in # moles of gas = 0 since there are 2 moles of gas on each side of the balanced equation.

Kp = Kc(RT)º = Kc(RT) = 3.2x10^-34 x 0.0821 x 298

Kp = 7.8x10^-33

7.8x10^-33 = (x)(x) / (0.45-2x)²

Solve for x and that will be the partial pressures of both H₂ and Cl₂ (in atm)