Question

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed rate of disappearance of NO is 9.3×10−5M/s.
(a) What is the rate of disappearance of O2 at this moment?
(b) What is the value of the rate constant?
(c) What are the units of the rate constant?
(d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8?

Answer 1

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(c) The units of rate constant is:  $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in  $O_{2}$.

Then we can say that, rate = k$[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$.

    Rate of disappearance of  $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*($9.3*10^{-5}$) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*($9.3*10^{-5}$)

    rate = $4.65*10^{-5}$ M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

    k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

(c) The units of the rate constant can be obtained from (2).

    k = $\frac{rate}{[NO]^{2}[O_{2}]}$

    Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     $rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24