#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?

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Answer 1 · 2017-08-14T08:14:01

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

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.