Question 15.3.4: Consider the equilibrium system involving two isomers of but......
Consider the equilibrium system involving two isomers of butane.
\begin{matrix} \hspace{135 pt} \enspace \enspace \enspace \underset{|}{CH_{3}} \\ H_{3}C – CH_{2}\underset{Butane}{-} CH_{2}-CH_{3} \rightleftarrows H_{3}C-\underset{\underset{\text{Isobutane}}{H}}{C} -CH_{3} \end{matrix} K = \frac{\left[isobutane\right] }{\left[butane\right] } = 2.5 at 20 °C
A flask originally contains 0.200 M butane. Calculate the equilibrium concentrations of butane and isobutane.
You are asked to calculate equilibrium concentrations of reactants and products in a reaction.
You are given a chemical equation and equilibrium constant and the initial concentration of a reactant.
Step 1. Use the ICE method to determine equilibrium concentrations. Use the unknown quantity x to define the equilibrium concentrations of butane and isobutane.
\qquad [butane] \rightleftarrows [isobutane]
Initial (M) 0.200 0
Change (M) -x +x
Equilibrium (M) 0.200 – x x
Step 2. Substitute the equilibrium concentrations into the equilibrium constant expression and determine the value of x, the change in butane and isobutane concentrations as the system approaches equilibrium.
K = 2.5 = \frac{x}{0.200-x}
x = 0.143
Step 3. Use the value of x to calculate the equilibrium concentrations of butane and isobutane.
[butane]_{equilibrium} = 0.200 − x = 0.057 M [isobutane]_{equilibrium} = x = 0.143 M
Is your answer reasonable? As a final check of your answer, confirm that these are equilibrium concentrations by substituting them into the equilibrium constant expression and calculating K.
K = \frac{\left[isobutane\right] }{\left[butane\right] } = \frac{0.143}{0.057} = 2.5