You cannot remove the exponents regardless.
The expression for Kc would be:
[tex]\sf\Large K_c= \frac{[CH_4]^2[Cl_2]^3}{[H_2]^3[CHCl_3]^2}[/tex]
You cannot reduce the exponents, as this would change the Kc values also. Also, I don't believe you would get a compound in both the reactant and product, so you would not be able to reduce it.
Also
2CHCl₃(g) + 3H₂(g) --> 2CH₄(g) + 3Cl₂(g) would have a different Kc value than
4CHCl₃(g) + 6H₂(g) --> 4CH₄(g) + 6Cl₂(g).
The Kc value would be squared because the coefficients were all doubled. Had all the coefficients been tripled, then the new Kc value would be to the power of 3 of the original equation.
I hope that answers any of the questions you may have had.
