Question

`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?

Answer 1

We gots `H_2(g) +2NO(g) rarrH_2O(g) + N_2O(g)`

Explanation:

Now `"rate"` `=k[H_2]^x[NO]^y`, where `x` and `y` are exponents, whose magnitude we have to assess by experiment.

So `"rate"_1=k[H_2]^x[NO]^y`

So `"rate"_2=k[2H_2]^x[NO]^y=2xx"rate"_1`, clearly `x=1`

But `"rate"_3=k[H_2][2NO]^y=4xx"rate"_1`, clearly `y=2`.

And we write the overall rate law.....`"rate"=k[H_2][NO_2]^2`

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