Question

The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel. (a) (b) (c) The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 × 10−6 L2/mol2/s.

Answer 1

`1.6* 10^(-6) M/s` is the rate of formation of nitrogen dioxide.

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+O_2\rightarrow 2NO_2`

Given: Order with respect to
`O_2` = 1

Order with respect to
`NO` = 2

Thus rate law is:

`R=k[NO]^2[O_2]^1`

k= rate constant

`R=k[NO]^2[O_2]^1`

`=5.8* 10^(-6) s/L^2/mol ^2* (0.75 M)^2* ( 0.50 M)^1`

`R = 1.6* 10^(-6) M/s`

`1.6* 10^(-6) M/s` is the rate of formation of nitrogen dioxide.