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: 75% of isobutane is present in the equilibrium mixture.

Explanation: For the reaction,

$Butane\rightleftharpoons Isobutane$

the $k_c$ (equilibrium constant) can be written as:

$k_c=\frac{[isobutane]}{[butane]}$

We are given that $k_c$ is 3.0,

$3=\frac{[isobutane]}{[butane]}$

$[Isobutane]=3\times [butane]$

This means that in a 100% of solution, 3 parts of isobutane is present and one part of butane is present

Butane present will be 25% and

Isobutane present will be 75% in equilibrium mixture.