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# What is the rate constant k for this reaction ?

For the following reaction the below data were measured.

2NO(g) + 2H2 (g) -----> N2 (g) + 2H2Og

Initial Concentrations (mol/L)

[NO] [H2]

0.10 0.10

0.10 0.20

0.20 0.10

Initital Rate of Formation of H2O (mol/Ls)

0.00123

0.00246

0.00492

What is the rate constant k for this reaction ?

### 3 Answers

- 1 decade agoFavorite Answer
Rate = k * [NO]^m * [H2]^n

0.00123 = k * 0.1^m * 0.1^n -------------------(1)

0.00246 = k * 0.1^m * 0.2^n -------------------(2)

0.00492 = k * 0.2^m * 0.1^n -------------------(3)

Divide (1) by (2)

0.00123/0.00246 = k/k * (0.1^m)/(0.1^m) * (0.1^n)/(0.2^n)

1/2 = (1/2)^n

Therefore n = 1

Divide (1) by (3)

0.00123/0.00492 = k/k * (0.1^m)/(0.2^m) * (0.1^n)/(0.1^n)

1/4 = (1/2)^m

4 = 2^m

Therefore m = 2

Now Substitute the values for m and n to the first (1) equation

0.00123 = k * 0.1^m * 0.1^n

0.00123 = k * 0.1^2 * 0.1^1

0.00123 = k * 0.001

k = 0.00123/0.001

k = 1.23

Units for k = (mol L-1)/{(mol2 L-2) * (mol L-1)}

= mol-2 L2 (litre squared per mol squared)

Thus, k = 1.23 mol-2 L2

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- 1 decade ago
When you double the [H2], while keeping [NO2] the same, the rate doubles; so the reaction is directly proportional to [H2]. When you double [NO2] while keeping [H2] the same, the rate becomes 4 times. So the reaction is second order in [NO2].

The rate law is Rate = k [NO2]^2 [H2].

Use one of the sets of values in this equation to get k:

0.00123 = k (0.1)^2*(0.1)

k = 0.00123/(0.1*0.1*0.1) = 1.23 L^2/mol^2.s

- docrider28Lv 41 decade ago
Rate equation is r= k[0.1]^2 [0.1]

Note that as you double the first reactant in trial #3 that the rate of the reaction quadruples (=4x) indicating that it is second order with respect to the first reactant.

k = r/([0.1]^2 [0.1]) = 1.23

Source(s): gen chem