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 :The percentage of isobutane in the mixture is 75%

Explanation :

The equilibrium reaction can be written as,

$Butane \rightleftharpoons Isobutane$

Let us assume that we have 100% butane initially

Let x be the % of butane that isomerizes and converts to isobutane.

Since the amount of reactant decreases, we have -x for the reactant side and +x for the product side.

Please refer to attached ICE table.

The equilibrium constant for the above reaction is written as,

$k_{c}=\frac{[Isobutane]}{[Butane]}$

Let us plug in the values from the ICE table.

$k_{c}=\frac{(x)}{(100-x)}$

Kc for the reaction is 3.0

$3.0=\frac{(x)}{(100-x)}$

on simplifying we get,

$3(100-x)= x$

$300-3x = x$

$300 = 4x$

x = 75.

The percentage of isobutane in the mixture is 75%.

Answer 2

Answer is in the attachment attached below ....