Question

The system is originally at equilibrium with [butane] = 1.0 M and [isobutane] = 2.5 M.If 0.50 mol/L of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of butane?

Answer 1

1.14 M

Step-by-step explanation:

Let's consider the following reaction.

butane ⇄ isobutane

We can use the concentrations at equilibrium to calculate the equilibrium constant.

Kc = [isobutane] / [butane]

Kc = 2.5 / 1.0

Kc = 2.5

If we add 0.50 M of isobutane, we get [isobutane] = 2.5 + 0.50 = 3.0 M.

This will be an initial concentration in an ICE chart.

butane ⇄ isobutane

I 1.0 3.0

C +x -x

E 1.0+x 3.0-x

The equilibrium constant is:

Kc = 2.5 = [isobutane] / [butane]

2.5 = (3.0-x) / (1.0+x)

2.5 + 2.5x = 3.0-x

x = 0.14

The equilibrium concentration of butane is:

[butane] = 1.0+x = 1.14 M