# What is the value for the reverse reaction?

My question deals with the reaction $\ce{2NO(g) + O2(g) -> 2NO2(g)}$ with $K_c = 1752$.

This is what I have so far; $K_c$= products over reactants, so is the reverse reaction reactants over products to find the answer Such as
$$1752 = \frac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$ So then does the answer become $$1752 = \frac{[x]^2}{[x]^2[x]}$$

${K_c}$ of reverse reaction should just be the reciprocal of forward ${K_c}$. So, ${K_c}$ of reverse reaction is $\frac{1}{1752}$.
Now, I am not sure, why you used 2nd equation. Suppose you have 1 unit conc of $\ce{NO}$ and $\ce{O_2}$, then after time $t$, if the conversion is $x$, then your equation should be
$1752 = \frac{[2x]^2}{([1-x])^3}$.