5. The decomposition of ozone is important to many atmospheric reactions: TRENO

ID: 1037593 • Letter: 5

Question

5. The decomposition of ozone is important to many atmospheric reactions: TRENO LINC O(g)02(g)+O(g) A study of the kinetics of the reaction resulted in the following data: h? Temperature (K) Rate Constant (Mos) Temperature (K) Rate Constant (Mes") 600 3.37 x 101300 7.83 x 10 1.45 x 108 2.46 x 10 3.93 x 10 5.93 x 10 8.55 x 10 1.19 x 10 700 4.85 x 104 1400 1500 1600 1700 1800 1900 8.SH 8003.58 x 105 900 1000 1100 1200 1.70 x 10 5.90 x 10° 1.63 x 107 3.81 x 10 ?a Ea--(m , R) R, Determine by graphical analysis the Ea for the reaction (using EXCEL) by plotting In k vs. 1/T. Be sure to attach a copy of the EXCEL graph with your assignment. Determine the rate constant, k at 298 K (Show workl) a) E (1 R T y-m x+ b Given b)

Explanation / Answer

We shall need the following table as shown below.

Temperature (K)

Rate Constant (M-1.s-1)

1/(Temperature) (K-1)

ln (Rate Constant)

600

3.37*103

0.001667

8.1227

700

4.85*104

0.001428

10.7893

800

3.58*105

0.001250

12.8610

900

1.70*106

0.001111

14.3461

1000

5.90*106

0.001000

15.5905

1100

1.63*107

0.000909

16.6067

1200

3.81*107

0.000833

17.4557

1300

7.83*107

0.000769

18.1760

1400

1.45*108

0.000714

18.7922

1500

2.46*108

0.000667

19.3208

1600

3.93*108

0.000625

19.7893

1700

5.93*108

0.000588

20.2007

1800

8.55*108

0.000555

20.5666

1900

1.19*109

0.000526

20.8972

Plot ln (Rate Constant) vs 1/(Temperature) as below.

The linear regression equation is

y = -11182x + 26.777.

The integrated equation is

ln k = -Ea/R*(1/T) + ln A

Comparing the two equations,

-11182 = -Ea/R

=====> Ea = 11182*R = 11182*(8.314 J/mol.K) = 92967.148 J/mol.K = (92967.148 J/mol.K)*(1 kJ/1000 J) = 92.967148 kJ/mol.K ? 92.967 kJ/mol.K (ans).

Again, we have x = 1/T; therefore, when T = 298 K, we have,

x = 1/(298 K) = 0.003356 K-1.

Put x = 0.003356 K-1 in the regression equation and obtain

y = -11182*(0.003356) + 26.777 = -37.527 + 26.777 = -10.75

We have

ln (Rate Constant) = -10.75

====> Rate constant = exp (-10.75) = 2.14*10-5

Therefore, k = 2.14*10-5 M-1.s-1 (ans).