For the equilibrium,
If the value of Kc is 3, the percentage by mass of iso-butanein the equilibriummixture would be:-


a) 75%
b) 90%
c) 30%
d) 60%
Kindly answer sir/mam.

1. K_c for the above reaction is represented as :
     
               $$\begin{gathered} K_c = \frac{\left[isobu\tan e\right]}{\left[bu\tan e\right]} \\ 3=\frac{\left[isobu\tan e\right]}{\left[bu\tan e\right]} \\ \left[isobu\tan e\right] = 3\left[bu\tan e\right] \end{gathered}$$

2. So concentration of isobutane is thrice of butane in a 100% solution of the two.
       Thus isobutane = 75%
                 butane = 25%

Correct answer is (a).