H_2(g) + 2NO(g) -> H_20(g) + N_2O(g). At high temperatures, doubling the concentration of H_2 doubles the rate of reaction, while doubling the concentration of NO increases the rate fourfold. How do you write a rate law for this reaction?

Aug 14, 2017

We gots ${H}_{2} \left(g\right) + 2 N O \left(g\right) \rightarrow {H}_{2} O \left(g\right) + {N}_{2} O \left(g\right)$

Explanation:

Now $\text{rate}$ $= k {\left[{H}_{2}\right]}^{x} {\left[N O\right]}^{y}$, where $x$ and $y$ are exponents, whose magnitude we have to assess by experiment.

So ${\text{rate}}_{1} = k {\left[{H}_{2}\right]}^{x} {\left[N O\right]}^{y}$

So ${\text{rate"_2=k[2H_2]^x[NO]^y=2xx"rate}}_{1}$, clearly $x = 1$

But ${\text{rate"_3=k[H_2][2NO]^y=4xx"rate}}_{1}$, clearly $y = 2$.

And we write the overall rate law.....$\text{rate} = k \left[{H}_{2}\right] {\left[N {O}_{2}\right]}^{2}$

Note that this conveniently follows the stoichiometric equation given; it did not NECESSARILY have to follow this rate law.