PLEASE ANSWER PARTS C and D! The following data were collected for the reaction

Question

PLEASE ANSWER PARTS C and D!

The following data were collected for the reaction between hydrogen and nitric oxide at 700 °C 2H2(g) + 2NO (g) 2H20(g) + N2(g) INOJ (M) 0.025 0.025 0.0125 Initial rate (M s) (H2l (M) 0.010 0.0050 0.010 Experiment 2.4 × 10-6 1.2 x 10-6 0.60 × 10-6 (a) What is the rate law for the reaction? (b) Calculate the rate constant for the reaction. (c) Suggest a plausible mechanism that is consistent with the rate law. Assume that an oxygen atom is the intermediate. (d) More careful studies of the reaction show that the rate law over a wide range of concentration of the reactions is as shown below. What happens to the rate law at very high and very low concentration of hydrogen? k[NO1 1H21

Explanation / Answer

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

choose point 1 and 2 so NO cnacels

(2.4*10^-6)/(1.2*10^-6) = (0.01/0.005)^a

a = ln((2.4*10^-6)/(1.2*10^-6)) / ln((0.01/0.005))

a = 1

1st order with respect to H2

now, choose points 1 and 3

(2.4*10^-6)/(0.6*10^-6) = (0.025/0.0125)^b

b = ln((2.4*10^-6)/(0.6*10^-6)) / ln((0.025/0.0125))

b = 2

2nd order with respect to NO

then

Rate = k*[H2][NO2]^2

b)

rate constant, choose any point

2.4*10^-6 = k*(0.01)(0.025^2)

k = (2.4*10^-6) / ((0.01)(0.025^2)) = 0.384

c)

NO + NO = N2O2; (fast)

N2O2+ H2 = N2O + H2O (slow)

N2O + H2 = N2 + H2O (fast)

d)

if [H2]

too high -->

1 + k2[H2] --> K2[H2]

Rate = k[NO2]^2 * [H2 ] / k2*[H2] = k [ NO2]^2

the rate depends mainly on NO2

if [H2] = too low

the rate depends mainly on H2

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