Equilibrium at Certain Temperatures

Question

The K_eq for the reaction 2 HI_(g)  H_2(g) + I_2(g) has a value of 1.85 x 10^-2 at 425°C. If 0.18 mol of HI is placed in a 2.0 L flask and allowed to come to equilibrium at this temperature, what will be the equilibrium [I₂]?



Answer 1

These types of problems are usually best approached using an ICE table.

2HI(g) <==> H₂(g) + I₂(g) ... balanced equation

0.18 mol......0............0............Initial

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

0.18-2x........x...........x.............Equilibrium

Next, write the equilibrium expression, plug in the equilibrium values, and solve for x...

Keq = 1.85x10^-2 = [H₂][I₂] / [HI]²

1.85x10^-2 = (x)(x) / (0.18-2x)²

sq root of both sides leads to

0.136 = x / 0.18 - 2x

x = 0.0193 moles

Plug this value into the Equilibrium line of the ICE table to find moles of each species @ equilibrium

moles HI = 0.18 - 2x = 0.18 - 0.0386 = 0.141 moles HI

moles H₂ = x = 0.0193

moles I₂ = x = 0.0193

Finally, since the reaction takes place in a 2.0 L flask, convert these values to M (moles/liter)....

Equilibrium concentration are...

[HI] = 0.141 mols HI / 2.0 L = 0.071 M (2 sig. figs.)

[H₂] = 0.0193 mols H2 / 2.0 L = 9.7x10-3 M (2 sig. figs.)

[I₂] = 0.0193 mols I2 / 2.0 L = 9.7x10-3 M (2 sig. figs.)