Does the stoichiometry effect the value of the equilibrium constant of an equilibrium reaction? Will be there any change in the $K_c$ values for same reaction, if we write stoichiometry as a fraction or as a integer? For example, assuming that we calculate $K_c$ at same temperature:

$$\begin{align}
&amp;(1)  &amp;&amp;\ce{1/2A + 3/2B -&gt; C}\\
&amp;(2)  &amp;&amp;\ce{A + 3B -&gt; 2C}\\
\end{align}$$

If we calculate $K_c$ for above two equilibrium reactions, do we get different answers?
$$K_c = \mathrm{products \over reactants}$$ So this is done by multiplying the concentrations of the products and dividing by the concentrations of the reactants. The concentrations are put to the power of their coefficient. So in your problems at equilibrium $K_c$ would be:
\begin{align}
  K_c &amp;= \frac{[\ce{C}]^1}{[\ce{A}]^{\frac{1}{2}}[\ce{B}]^{\frac{3}{2}}}\tag1\\
  K_c &amp;= \frac{[\ce{C}]^2}{[\ce{A}]^1[\ce{B}]^3}\tag2\\
\end{align}

Note that $[\ce{A}]$ means concentration of $\ce{A}$.