1.1 What is Chemistry?

http://www.ck12.org Chapter 18. Kinetics

Notice that the starting concentrations of NO and H 2 were varied in a specific way. In order to compare the rates of
reaction and determine the exponent associated with each reactant, the initial concentration of each reactant must be
changed while the other is held constant.

Comparingexperiments 1 and2: the concentration of NO was doubled, while the concentration of H 2 was held
constant. Doubling the concentration of NO quadrupled the initial reaction rate (5.00× 10 −^5 /1.25× 10 −^5 = 4).
Therefore, the exponent associated with the concentration of NO must be 2. In other words, Rate∝[NO]^2. Because
22 = 4, doubling the value of [NO] increases the rate by a factor of four.

Comparingexperiments 2 and3: the concentration of H 2 was doubled while the concentration of NO was held
constant. Doubling the concentration of H 2 doubled the initial rate of the reaction (1.00× 10 −^4 /5.00× 10 −^5 = 2).
Therefore, the exponent associated with the concentration of H 2 is 1 (Rate∝[H 2 ]^1 ). Because 2^1 = 2, doubling the
value of [H 2 ] also doubles the reaction rate.

The overall rate law incorporates both of these results into a single equation:

Rate=k[NO]^2 [H 2 ]

Note that exponents of 1 are generally not included, since [A]^1 = [A]. Based on the exponents associated with each
concentration, we would say that this reaction issecond-orderwith respect to NO andfirst-orderwith respect to
H 2. The overall reaction order is the sum of the individual reaction orders, or the sum of all exponents in the rate
law. You can see that this would be a third-order reaction overall.

Additionally, now that we know the form of the rate law, we can plug in data from any of the experiments to
determine the value of k for this reaction under the conditions (temperature, pressure, etc.) at which the data was
measured. Using the data from experiment 1:

Rate=k[NO]^2 [H 2 ]

1. 25 × 10 −^5

M

s

=k[ 0 .0050 M]^2 [ 0 .0020 M]

k= 250

1

M^2 ·s

Let’s consider another hypothetical reaction of the form A + B→C + D. Under a certain set of conditions, the
following data were collected:

TABLE18.2:

Experiment [A] (M) [B] (M) Initial Rate (M/s)

1 0.015 0.025 0.112

2 0.030 0.025 0.224

3 0.030 0.050 0.224

Comparing experiments 1 and 2, we see that doubling the concentration of A while holding [B] constant doubles the
rate. Therefore, this reaction is first-order with respect to A (Rate∝[A]^1 ). Plugging this into our generic rate law,
we get the following:

Rate=k[A]x[B]y

Rate=k[A]^1 [B]y

Rate=k[A][B]y