Answer :
Answer:
(a) The rate of disappearance of [tex]O_{2}[/tex] is: [tex]4.65*10^{-5}[/tex] M/s
(b) The value of rate constant is: 0.83036 [tex]M^{-2}s^{-1}[/tex]
(c) The units of rate constant is: [tex]M^{-2}s^{-1}[/tex]
(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:
[tex]2NO(g)+O_{2}->2NO_{2}[/tex]
rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = [tex]\frac{1}{2}\frac{d}{dt}[NO_{2}][/tex] -----(1)
According to the question, the reaction is second order in NO and first order in [tex]O_{2}[/tex].
Then we can say that, rate = k[tex][NO]^{2}[O_{2}][/tex] -----(2)
where k is the rate constant.
The rate of disappearance of NO is given:
[tex]-\frac{d}{dt}[NO][/tex] = [tex]9.3*10^{-5}[/tex] M/s.
(a) From (1), we can get the rate of disappearance of [tex]O_{2}[/tex].
Rate of disappearance of [tex]O_{2}[/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex]) M/s = [tex]4.65*10^{-5}[/tex] M/s.
(b) The rate of the reaction can be obtained from (1).
rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex])
rate = [tex]4.65*10^{-5}[/tex] M/s
The value of rate constant can be obtained by using (2).
rate constant = k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]
k = [tex]\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}[/tex] = 0.83036 [tex]M^{-2}s^{-1}[/tex]
(c) The units of the rate constant can be obtained from (2).
k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]
Substituting the units of rate as M/s and concentrations as M, we get:
[tex]\frac{Ms^{-1} }{M^{3}}[/tex] = [tex]M^{-2}s^{-1}[/tex]
(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.
[tex]rate\alpha [NO]^{2}[/tex]
If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of [tex](1.8)^{2}[/tex] = 3.24