Question

FOR THE EQUILIBRIUM BUTANE TO ISOBUTANE IF THE VALUE OF KC IS 3.0 THE PERCENTAGE BY MASS OF ISOBUTANE IN THE EQUILIBRIUM MIXTURE WOULD BE

Answer 1

Answer: Isobutane present will be 75% in the equilibrium mixture.

Explanation: Butane and Isobutane are the structural isomers and are present in equilibrium.

Reaction follows:

                               $Butane\rightleftharpoons Isobutane$

The equilibrium constant, $k_c$ of the reaction is written as:

                             $k_c=\frac{[Isobutane]}{[Butane]}$

In the question, it is given that $k_c=3.0$

We can write

$3=\frac{[Isobutane]}{[Butane]}$

$Isobutane=3\times [Butane]$

For a reaction,

[Butane] + [Isobutane] = 1

[Butane] + 3[Butane] = 1

$[Butane]=\frac{1}{4}$ = 0.25

[Isobutane] = 0.75

$\text{Mass percentage}= \frac{\text{Mass of the component}}{\text{Total mass of the solution}}\times 100$

$\text{Mass Percentage of Isobutane}=\frac{0.75}{1}\times 100$

Mass percentage of Isobutane = 75%