HO2(g)+O3(g)-->OH(g)+2O2(g) The rate of this reaction was studied in the presenc

Question

HO2(g)+O3(g)-->OH(g)+2O2(g)

The rate of this reaction was studied in the presence of a large excess of ozone. Determine the psuedo-first-order rate constant and the second-order rate constant for the reaction from the following date:

Pseudo-first-order rate constant= ________________s^-1

Pseudo-second-order rate constant=_______________M^-1

Time [HO2] (M) [O3](M) 0 3.2 x 10^-6 1.0 x 10^-3 10 2.9 x 10^-6 1.0 x 10^-3 20 2.6 x 10^-6 1.0 x 10^-3 30 2.4 x 10^-6 1.0 x 10^-3 80 2.4 x 10^-6 1.0 x 10^-3

Explanation / Answer

(i)
in the presence of a large excess of ozone,
the psuedo-first-order rate law = k1avg*(HO2)

ln([A]o/[A]) = k*t
[A]o = 3.2 x 10^-6

k1 = ln((3.2 x 10^-6)/[A]) /t
t=10
k11 = ln((3.2 x 10^-6)/[2.9 x 10^-6])/10 = 0.0098 s^-1
t=20
k11 = ln((3.2 x 10^-6)/[2.6 x 10^-6])/20 = 0.0104 s^-1
t=30
k11 = ln((3.2 x 10^-6)/[2.4 x 10^-6])/30 = 0.0116 s^-1
t=80
k11 = ln((3.2 x 10^-6)/[2.4 x 10^-6])/80 = 0.00435 s^-1

k1avg =(k11+k12+k13+k14)/4= 0.0090375 s^-1

(ii)
the Pseudo-second-order rate law = k2*(HO2)^2

1/[A] - 1/[A]o = k*t; [A]o = 3.2 x 10^-6

1/[A] - 1/[3.2 x 10^-6] = k*t

k2 = -(1/[3.2 x 10^-6] - 1/[A]) /t

t=10
k21 = -(1/3.2 x 10^-6)-1/[2.9 x 10^-6])/10 = 3233 M^-1 s^-1
t=20
k22 = -(1/3.2 x 10^-6)-1/[2.6 x 10^-6])/20 = 3606 M^-1 s^-1
t=30
k23 = -(1/3.2 x 10^-6)-1/[2.4 x 10^-6])/30 = 3472 M^-1 s^-1
t=80
k24 = -(1/3.2 x 10^-6)-1/[2.4 x 10^-6])/80 = 1302 M^-1 s^-1

k2avg = (k21+k22+k23+k24)/4= 2903.25 M^-1 s^-1

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