Question

The equilibrium constant for the isomerisation of butane at ${25}^{o}C$ is $K_c=8$

Butane $\to$ Isobutane

If 5.8 g butane is introduced into a 12.5 L vessel at ${25}^{o}C$. What mass of isobutane will be present at equilibrium?

• A. 5.15 g
• B. 2.56 g
• C. 7.49 g
• D. 5.80 g

Answer 1

Answer: (A)

5.15 g

Initial mole of butane $=\frac{5.8}{58}=0.1$

$\underset{0.1-x}{Butane}\rightleftharpoons \underset{x}{lsobutane}$

$K_c=8=\frac{x}{0.1-x}orx=0.088$

m(isobutane)$=0.088\times 58=5.15g$