# What would happen to the rate of a reaction with rate law rate = k[NO]^2[H_2] if the concentration of NO were doubled?

$\text{Rate 1}$ $=$ $k {\left[N O\right]}^{2} \left[{H}_{2}\right]$
$\text{Rate 2}$ $=$ $k {\left[2 N O\right]}^{2} \left[{H}_{2}\right]$ $=$ $\text{Rate 2}$ $=$ $4 k {\left[N O\right]}^{2} \left[{H}_{2}\right]$
$\text{Rate 2"/"Rate 1}$ $=$ $\frac{4 k {\left[N O\right]}^{2} \left[{H}_{2}\right]}{k {\left[N O\right]}^{2} \left[{H}_{2}\right]}$ $=$ $4$