Question

In: Chemistry

The following set of data was obtained by the method of initial rates for the reaction:...

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?

Solutions

Expert Solution

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


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