Question

For the equilibrium, $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{3\left(g\right)}\rightleftharpoons C{H}_3-\underset{C{H}_3(iso-butane)}{\underset{|}{CH}}-C{H}_3\left(g\right)$
If the value of $K_C$ is $3.0$, the percentage by mass of iso-butane in the equilibrium mixture would be :

• A. $75\%$
• B. $90\%$
• C. $30\%$
• D. $60\%$

Answer 1

The correct option is A $75\%$

For the equilibrium :-

Now, $K_C=\frac{\left[isobutane\right]}{\left[butane\right]}$

$\Rightarrow 3=\frac{C^{{}^{\prime }}}{C-C^{{}^{\prime }}}$

Now, $C$ or $C^{{}^{\prime }}$ $\alpha$ Mass of the species (compound)

$\Rightarrow 3=\frac{M^{{}^{\prime }}}{M-M^{{}^{\prime }}}$

$\Rightarrow \frac{M-M^{{}^{\prime }}}{M^{{}^{\prime }}}=\frac{1}{3}\Rightarrow \frac{M}{M^{{}^{\prime }}}-1=\frac{1}{3}$

$\Rightarrow \frac{M^{{}^{\prime }}}{M}=\frac{3}{4}$

where, $M^{{}^{\prime }}$ and $M$ are mass of isobutane & butane respectively.

So, mass % of iso-butane $=\frac{M^{{}^{\prime }}}{M}\times 100$% $=\frac{3}{4}\times 100$% $=75$%