Question
At a certain temperature, only $$50\% \,HI$$ is dissociated into $${H_2}$$ and $${I_2}$$ at equilibrium. The equilibrium constant is :
A.
1.0
B.
3.0
C.
0.5
D.
0.25
Answer :
1.0
Solution :
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HI \rightleftharpoons \,\,\,\,\,\,\,\frac{1}{2}{H_2} + \frac{1}{2}{I_2} \cr
& {\text{at}}\,t = 0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr
& {\text{at eq}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\left( {1 - 0.5} \right)\,\,\,\,\,\left( {0.5} \right)\,\,\,\,\,\,\left( {0.5} \right) \cr
& \left( {\because \,\,\alpha = \frac{{50}}{{100}} = 0.5} \right) \cr
& {\text{Now}}\,\,{K_C} = \frac{{{{\left( {0.5} \right)}^{\frac{1}{2}}}{{\left( {0.5} \right)}^{\frac{1}{2}}}}}{{0.5}} = 1 \cr} $$