Answer :

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.

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