Answer:
Order w.r.t N2 = 3
Order w.r.t O3 = 1
Overall order = 4
Initial rate = 424 M/s
k = [tex]3.4*10^{3} M^{-3} s^{-1}[/tex]
Explanation:
The rate law is given as:
[tex]rate = K[N_{2} ]^{3} [O_{3} ][/tex]-----(1)
1) Order of a reaction can be defined as the power dependence of the concentration of the reactants on the rate of the reaction
Here, the reaction order in [tex]N_{2} = 3[/tex]
2) The reaction order in [tex]O_{3} = 1[/tex]
3) The overall reaction order is the sum of all powers, here it is:
[tex]3+1=4[/tex]
4) Initial rate, Rate1 = 53.0 M/s
i.e. from equation (1):
[tex]53.0 = K[N_{2} ]^{3} [O_{3} ][/tex]------(2)
When the concentration of N2 is doubled:
[tex]Rate2 = K[2N_{2} ]^{3} [O_{3} ][/tex]-----(3)
Dividing equation (3) by (2) gives:
[tex]Rate2=53.0M/s*8\\ = 424 M/s[/tex]
5) It is given that:
rate = 5.0*10^3 M/s
[N2]=1.1 M
[O2]=1.1 M
Based on equation(1), the rate constant k is given as:
[tex]k = \frac{Rate}{[N_{2}]^{3}[O_{3}] } \\= \frac{5.0*10^{3} M/s}{[1.1M]^{3} *[[1.1M]} =3.4*10^{3} M^{-3} s^{-1}[/tex]
