The following data were collected for the reaction between hydrogen and nitric o

Question

The following data were collected for the reaction between hydrogen and nitric oxide at 700 C:

2H2(g)+2NO(g) -> 2H2O(g)+N2(g)

a) What is the rate law for the reaction?

b) Calculate the rate constant for the reaction

c) More careful studies if the reaction show that the rate law over a wide range of concentrations of reactants should be:

rate= (k1[NO]2[H2])/(1+k2[H2])

What happens to the rate law at very high and very low hydrogen concentrations?

Experiment [H2]/M [NO]/M initial rate/M*s-1 1 0.010 0.025 2.4x10-6 2 0.0050 0.025 1.2x10-6 3 0.010 0.0125 0.60x10-6

Explanation / Answer

What is the rate law for the reaction?

When compared experiments 1 and 3:

Concentration of Hydrogen kept constant

Concentration of Nitric oxide halved.

This affected the rate of reaction to decrease by 4 fold.

This can only happen when rate order of reaction is second order w.r.t. Nitric oxide.

Similarly

When compared experiments 1 and 2:

Concentration of Hydrogen halved.

Concentration of Nitric oxide kept constant

This affected the rate of reaction to decrease by 2 fold.

This can only happen when rate order of reaction is first order w.r.t. hydrogen.

Rate law of the reaction = k[H2][NO]2

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b) Calculate the rate constant for the reaction:

Rate constant, k = Rate of the reaction / [H2][NO]2

From experiments 1:

Rate of the reaction = 2.4 x 10-6 M.s-1

[H2] = 0.01 M

[NO] = 0.025 M

rate constant, k = 2.4 x 10-6 / 0.01 x (0.025)2

= 2.4 x 10-6 / 0.01 x 0.000625

= 2.4 x 10-6 / 6.25 x 10-6

= 0.384

The rate constant for the reaction = = 0.384

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c) More careful studies if the reaction show that the rate law over a wide range of concentrations of reactants should be:

rate= (k1[NO]2[H2])/(1+k2[H2])

What happens to the rate law at very high and very low hydrogen concentrations?

A suggested mechanism for this reaction is as follows:

2 H2(g) + 2 NO(g) N2(g) + 2 H2O(g)

H2 + NO =(reversible) fast step

X + NO = Y + H2O slow step

Y + H2 = N2 + H2O fast step

Since hydrogen is involved in a second step during reduction of NO and that step is also a fast step not a rate limiting step. Hence very high and very low hydrogen concentrations do not have appreciable effect on rate law of reaction.

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