In: Chemistry
The following set of data was obtained by the method of initial rates for the reaction: 2NO(g) + O2 (g) ----> 2NO2(g)
Experiment # | [NO], M | [O2], M | Initial Rate, M/s |
1 | 0.0126 | 0.0125 | 1.41 x 10^-2 |
2 | 0.0252 | 0.0250 | 1.13 x 10^-1 |
3 | 0.0252 | 0.0125 | 5.64 x 10^-2 |
a) What is the rate law for the reaction?
b) If you triple the concentration of both reactants, what will happen to the rate?
c) What is the value of the rate constant?
2NO(g) + O2 (g) ----> 2NO2(g)
Rate = K[NO]^m[O2]^n
1.41*10^-2 = K(0.0126)^m(0.0125)^n --------------------> 1
1.13*10^-1 = K(0.0252)^m(0.025)^n ----------------------> 2
5.64*10^-2 = K(0.0252)^m (0.0125)^n ------------------> 3
equation 1 divide by equation 3
1.41*10^-2/5.64*10^-2 = K(0.0126)^m(0.0125)^n/K(0.0252)^m (0.0125)^n
0.25 = (0.0126/0.0252)^m
0.25 = (0.5)^m
(0.5)^2 = (0.5)^m
m = 2
equation 2 divide by equation 3
1.13*10^-1/5.64*10^-2 = K(0.0252)^m(0.025)^n/K(0.0252)^m (0.0125)^n
2 = (0.025/0.0125)^n
2^1 = (2)^n
n = 1
Rate law
Rate = K[NO]^2[O2]^1
b. Rate = K(3NO)^2(3O2)^1
Rate = 27K[NO]^2[O2]^1
Rate of reaction increases 27 times
c.
Rate = K[NO]^2[O2]^1
1.41*10^-2 = K(0.0126)^2(0.0125)^1
K = 1.41*10^-2/(0.0126)^2(0.0125)^1
= 7105 M^-2 sec^-1