Question

Question: Butane exists as two isomers, n−butane and isobutane.

\({K_P} = 2.5\;at\;2{5^o}C\)

What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?

Short answer

The pressure of isobutane is 0.87 atm.

Step-by-step solution

Given Data

Given information:

\({\text{n - butane}} \rightleftharpoons {\text{isobutane}}\)

• The total pressure at equilibrium is 1.22 atm
• The equilibrium constant is K_p=2.5

We have to find the pressure of isobutane (x) in a container.

\(\begin{array}{*{20}{c}}{{P_{n - {\rm{\;butane\;}}}} + {P_{{\rm{isobutane\;}}}} = 1.22{\rm{atm}}}\\{{P_{{\rm{isobutane\;}}}} = x}\\{{P_{n - {\rm{\;butane\;}}}} = 1.22{\rm{amt}} - x}\end{array}\)

Determine the value of x:

Now, we will find the value of x

\begin{array}{*{20}{c}}{{K_p} = \frac{{{P_{{\rm{isobutane\;}}}}}}{{{P_{n - {\rm{\;butane\;}}}}}}}\\{2.5 = \frac{x}{{1.22 - x}}}\\{x = 3.05 - 2.5x}\\{3.5x = 3.05}\\{x = 0.87{\rm{atm}}}\end{array}

Therefore, the pressure of isobutane is

\({P_{{\rm{isobutane\;}}}} = x = 0.87{\rm{\;atm\;}}\)