H 2(g) + Cl 2(g) ↔ 2HCl (g) K = 4.4 x 10 -2 at 0 ̊C

Question

Initially, 1.5 mol of hydrogen and 1.5 mol of chlorine are injected into a 0.75 L refrigerated reaction vessel cooled to 0 ̊C.

a) Find the concentrations of H 2 , Cl 2 and HCl at equilibrium.

b) Calculate the amount (in moles) of each entity at equilibrium.

c) Calculate the reaction extent as a percent reaction

Answer 1

1.5 mol / 0.75 L = 2 M final concentration of each gas.

H₂(g) + Cl₂(g) <==> 2HCl(g) ... K = 4.4x10^-2

a) Use an ICE table to help find the equilibrium concentrations:

H₂(g) + Cl₂(g) <==> 2HCl(g)

2 M.......2 M..............0 M.........Initial

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

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

K= 4.4x10^-2 = [HCl]² / [H₂][Cl₂]

4.4x10^-2 = (2x)² / (2-x)(2-x)

Solve for x:

x = 0.19 M (be sure to check my math)

Equilibrium concentration are:

[H₂] = 2 - 0.19 = 1.81 M

[Cl₂] = 2 - 0.19 = 1.81 M

[HCl] = 2 x 0.19 = 0.38 M

b) moles at equilibrium are:

H₂ = 1.81 mol/L x 0.75 L = 1.36 moles

Cl₂ = 1.81 mol/L x 0.75 L = 1.36 moles

HCl = 0.38 mol/L x 0.75 L = 0.285 moles

c) reaction extent:

1.50 mols H₂ originally present

1.36 mols H₂ at equilibrium

0.14 moles H₂ converted during the reaction

0.14 mols / 1.50 mols (x100%) = 9.3%