1.) Nitric oxide, NO reacts with H 2 to give N 2 O and H 2 O according to 2NO(g)

Question

1.) Nitric oxide, NO reacts with H2 to give N2O and H2O according to

2NO(g) + H2(g) N2O + H2O(g)

In a series of experiments, the following initial rates of disappearance of NO were obtained:

What is the rate law for the reaction?

2.) The rate of a particular reaction quadruples when the temperature is raised from 35° to 45° C. What is the activation energy of this reaction?

Initial Concentrations
[NO] (mole L–1)
[H2] Initial rate (mole L-1 sec-1) 6.4 x 10-3 4.4 x 10-3 5.2 x 10-5 12.8 x 10-3 4.4 x 10-3 2.0 x 10-4 6.4 x 10-3 9.0 x 10-3 1.04 x 10-4

Explanation / Answer

1.) Nitric oxide, NO reacts with H2 to give N2O and H2O according to

2NO(g) + H2(g) N2O + H2O(g)

In a series of experiments, the following initial rates of disappearance of NO were obtained:

Initial Concentrations
[NO]

(mole L–1)
[H2]

Initial rate (mole L-1 sec-1)

6.4 x 10-3

4.4 x 10-3

5.2 x 10-5

12.8 x 10-3

4.4 x 10-3

2.0 x 10-4

6.4 x 10-3

9.0 x 10-3

1.04 x 10-4

What is the rate law for the reaction?

A. k [NO]2 [H2]3

B. k [NO]2 [H2]

C. k [NO] [H2]2

D. k [NO]2 [H2]2

E. Cannot be determined with information provided

Nitric oxide, NO reacts with H2 to give N2O and H2O according to

2NO(g) + H2(g) N2O + H2O(g)

Here in experiments number 1 and 2 where the concentration of H2 is constant but the concentration of [NO] is doubled then the rate increase by factor 4 means the rate with respect NO of second order.

Here in experiments number 1 and 3 where the concentration of NO is constant but the concentration of [H2] iS doubled then the rate increase by factor 2 means the rate with respect H2 is first order.

Thus the rate is B. k [NO]2 [H2]

2.) The rate of a particular reaction quadruples when the temperature is raised from 35° to 45° C. What is the activation energy of this reaction?

A. 1.07 kJ mole–1

B. 60.6 kJ mole–1

C. 52.8 kJ mole–1

D. 112.9 kJ mole–1

E. 105.6 kJ mole–1

T1= 35 = 308

T2 = 45=318

Ea = R ln(k/k) / (1/T - 1/T)
= 8.314472J/molK ln(4) / (1/308K - 1/318K)

= 8.314472J/molK 1.38 / (3.25*10^-3K -3.14*10^-3K)

= 104309.18J/mol

=104.31 kJ/mol

E. 105.6 kJ mole–1

Initial Concentrations
[NO]

(mole L–1)
[H2]

Initial rate (mole L-1 sec-1)

6.4 x 10-3

4.4 x 10-3

5.2 x 10-5

12.8 x 10-3

4.4 x 10-3

2.0 x 10-4

6.4 x 10-3

9.0 x 10-3

1.04 x 10-4

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