Question

Consider the following reaction: 2 NO(g) + 2H2(g) → N2(g) + 2 H2O(g) The rate law for this reaction is first order in H2 and second order in NO. What would happen to the rate if the initial concentration of NO tripled while all other factors stayed the same? The rate will increase by a factor of 9. The rate will decrease by a factor of 3. The rate will double. The rate will triple. The rate will remain constant.

Answer 1

Answer: The rate will increase by a factor of 9.

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)`

Given: Order with respect to
`NO` = 2

Order with respect to
`H_2` = 1

Thus rate law is:

`Rate=k[NO]^2[H_2]^1`

k= rate constant

It is given that the initial concentration of NO is tripled while all other factors stayed the same

`Rate'=k[3* NO]^2[H_2]^1`

`Rate'=k[3]^2[NO]^2[H_2]^1`

`Rate'=k* 9[NO]^2[H_2]^1`

`Rate'=9* Rate`

Thus the rate will increase by a factor of 9.