3
O2
| Kinetics : Sample Rate Calculations |
Rate Laws
For each of the following, determine the rate law of the reaction and find the value of the rate constant (k) with proper units.
1) 2 O3 3
O2
|
Experiment # |
Initial [O3] (M) |
Initial Rate (M/sec) |
|
1 |
2.00 |
0.500 |
|
2 |
4.00 |
1.000 |
|
3 |
6.00 |
1.500 |
First, set up a generic rate law:
rate = k[A]X
Next, create a ratio with the rate laws for 2 experiments:
Simplify and solve for x:
The reaction is first order with respect to [O3] because the exponent was calculated to be 1.
Now we can write the rate law and solve for k by plugging in values for the rate and the [O3] from a single experiment:
2) N2O4 2
NO2
|
Experiment. # |
Initial [N2O4 ] (M) |
Initial Rate (M/sec) |
|
1 |
0.50 |
0.050 |
|
2 |
1.00 |
0.200 |
|
3 |
1.50 |
0.450 |
3) Xe + 3 F2 XeF6
Experiment. # |
Initial [Xe] (M) |
Initial [F2] (M) |
Initial Rate (M/sec) |
1 |
0.50 |
0.25 |
0.00156 |
2 |
1.50 |
1.00 |
0.05625 |
3 |
0.75 |
0.25 |
0.0032 |
4 |
1.50 |
0.25 |
0.01406 |
5 |
0.50 |
1.00 |
0.00625 |
The rate law includes the concentrations of all the reactants:
rate = k[Xe]X[F2]Y
Therefore, we must solve for both X and Y separately.
First choose two experiments in which the initial [F2] are the
same and set up a ratio with these experiments:
The [F2] cancel each other out ,and the equation simplifies to find that X = 2. The reaction is second order with respect to [Xe].
Next, we do the same to solve for Y. Choose two experiments in which the initial [Xe] are the same and set up a ratio.
The [Xe] cancel each other out, and the equation simplifies to find that Y = 1. The reaction is first order with respect to [F2].
Now we can write the rate law:
* The reaction is first order with respect to F2 and second order with respect to Xe.
Now we have to solve for k:
Next: "Collision Molecular Theory"