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

Jan 12, 2016

If we assume that $\Delta \left[N O\right] = 0$, then the rate is halved. Try making up numbers and performing the calculation.

Let ["NO"] = "0.2 M" and ["H"_2] = "0.5 M", and let $k = \text{2}$ $\frac{1}{\text{M"^2*"s}}$.

$\textcolor{g r e e n}{r \left(t\right)} = 2 \cdot {0.2}^{2} \cdot 0.5$

$\textcolor{g r e e n}{\text{= 0.04 M/s}}$

Okay, now do the same thing when $\left[{H}_{2}\right] = \text{0.25 M}$. Can you predict what you get without writing out any actual math?

What does that tell you about the order of ${H}_{2}$ in the reaction? What, then, is the order of $N O$ in the reaction? How do you know?