Answer
1.1121×$10^{-5}$
Work Step by Step
Here $K_{c}$=$\frac{[NO]^{2}}{[N_{2}][O_{2}]}$
Given [NO]=$1.1×10^{-5} mol/L $
$[N_{2}]=6.4×10^{-3} mol/L$
And $[O_{2}]=1.7×10^{-3} mol/L$
Then $ K_{c}=\frac{[1.1×10^{-5}]^{2}}{[6.4×10^{-3}] [1.7×10^{-3}]}$
=1.1121×$10^{-5}$
