Question

For the equilibrium CH3CH2CH2CH3 (g)《===》ISO BUTANE (g) if the value of kc is 3 the percentage by mass of iso butane in the equilibrium mixture would be

Answer 1

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

Explanation: These two compounds are the structural isomers and are always present in equilibrium.

$Butane\rightleftharpoons Isobutane$

$k_c$ (equilibrium constant) for this reaction is written as:

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

It is given that, $k_c=3.0$

We can write,

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

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

For a given reaction,

[Butane] + [Isobutane] = 1

[Butane] + 3[Butane] = 1

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

[Butane] = 0.25

[Isobutane] = 0.75

Mass percentage can be calculated as:

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

$\text{Mass }\%\text{ of isobutane}=\frac{0.75}{1}\times 100$

Mass percentage of Isobutane = 75%

Answer 2

Answer:

Explanation:

Let x and 1-x be the concentration of product and reactant at eqlbm

Then, Kc=x/1-x

Or

3(1-x)=x

We get x=0.75

x×100=75%