12.3 Rate Law
Rate Laws and Reaction Order
A rate law is the mathematical relationship between reactant concentrations and the rate of reaction.
r = k[reactant(s)]order
r = rate
lower case k is the rate constant while uppercase Keq is the equillibrium constant covered in chapter ??.
order = the reactants’ concentration’s influence over reaction speed. This can be any integer, fraction, or even zero.
[reactants] = The reactants concentration, and yes you can have several reactants together, such as [X][Y][Z].
We can graph reaction rate against concentration as as a line. Because order is a power, this means that the line plotting rate versus concentration often curves and indicates a nonlinear relationship. If the order of a reaction is one (1), the rate versus concentration graph resembles y = mx + b, and k is the line’s slope.
We need an illustration for this yesterday.
For example, the rate of the gas-phase decomposition of dinitrogen pentoxide is directly proportional to N2O5’s concentration. That means the order = 1.
2N2O5 → 4NO2 + O2
rate = k [N2O5]1
Even though N2O5’s stoichiometric coefficient is 2, the equation’s order is 1, and k has nothing to do with the stoichiometric equation. We must learn both k and order by experimentation.
For a typical equation such as:
nA + mB → products
the rate law will be
r = k[A]x[B]y….
order = x + y
Note, the order of this equation is x + y, the sum of the powers, which symbolize each reactant’s concentration’s influence on reaction rate. Also x and y, the powers that add to form the order have no relation to the stoichiometric coefficients, n and m. Like k, we determine reaction order experimentally.
Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law.
Therefore another way to write the rate law equestion is:
k = concentration(1-order)/time
Reaction Order
As shown above, order in the rate law is the sum of the exponents in its concentration terms. This is the reactants’ concentration’s influence. In the N2O5 decomposition equation the order and only exponent is 1. This makes N2O5 decomposition a first order reaction.
When rate laws grow more complicated, we have order for each component as well as order for the entire system. As an example, consider a reaction. Let’s examine such a rate law. The equation is.
A + 3B + 2C → P
The experimentally determined rate law is:
r = k[A]1 [B]2
Remember the 1 is often implied and left out.
The overall order for this equation is the sum of its powers or 3.
The order for A is first order (1).
The order for B is second order or 2.
As for C, any number to zero-eth power is 1. When we make the 0 explicit we have…
r = k[A]1[B]2[C]0
Our sample equation is also zero order for C. A zero order reactant means that the rate of reaction occurs indepently of that reactant’s concentration, once there is a sufficient amount for the reaction to occur.
This is where I’m going to stop for 1/27/23. I’m sorry I did not get further.
Of course one might ask, how do we determine the effect each reactant’s concentration has on reaction speed? The answer is by experimentation. We can run the reaction with all reactant concentrations except one hled constant. We can learn the reaction rate, and then compare rates at different reaction concentrations.
For the reaction:
NH4+ + NO2– –> N2 + 2H20
We run three experiments. In the figure below, experiments 1–3 show how [NH4+] affects rate, while experiments 4–6 show how [NO2−] affects rate. Doubling NH4+‘s concentration doubles the rate, and doubling NO2-‘s concentration also doubles the rate.
Table 12:3 Rate vs. Conentration for NH4+ + NO2– –> N2 + 2H20
Therefore: r = k [NH4+] [NO2−]
The order of this reaction is 1 for [NH4+] and also 1 for [NO2–]. The sum of both orders for reactants, the sum for the reaction as a whole is 2, so it is a second order reaction. Remember the lowercase k in this formula is the rate constant. It is temperature dependent and determined by experiment.
Practice Problem Example 2
6H+ + BrO3– + 5Br– → 3 Br2 + 3 H2O
The rate for this equation is:
rate = k[Br–]1[BrO3–]1[H+]2
What happens to this rate if, in separate experiments:
We double [BrO3–]?
We increase pH is increased by one unit? (Remember pH is based on powers of ten and increasing decreases the H+ ion.
Dilute the solution to twice its volume(half its concentration), with the pH kept constant by use of a buffer?
Solutions:
(a) Doubling [BrO3–]’s concentration doubles the reaction rate, because the reaction has an order of 1 for [BrO3–].
(b) Increasing the pH by one unit decreases the [H+] by a factor of 10. Because this reaction is second order for [H+], decreasing [H+] ten times decreases the rate by a factor of ten squared or 100.
(c) Reducing the concentrations of both Br2 and BrO3– by half their original cuts the rate in half, so reducing both reactants’ concentration cuts the reaction rate by half, and by half again. Think 1/2 * 1/2 = 1/4.
Observing Reaction Orders
Rate-vs.-Concentration Proportionality
We can find the rate exponent (power) for a reactant we can plot instantanteous rate at different times against its concentration.
For single reactant decomposition:
A → products
The rate is the slope, ∆[A]/∆time.
This is a series of instantious rates we find by drawing tangents and measuring each ones slope.
Remember if doubling the concentration of A doubles the rate, then the reaction is first-order for [A]. If doubling [A] quadruples reaction rate, the reaction is second-order for [A].
Problem Example 3
The table below shows rate and concentration data for:
2 N2O5 → 4 NO2 + O2
Table 12.4
| Time (min) | p(N2O5) | [N2O5] mol L-1 | Rate mol L-1 min-1 |
| 0 | 301.6 | 0.0152 | |
| 10 | 224.8 | 0.0113 | 3.4 × 10–4 |
| 20 | 166.7 | 0.0084 | 2.5 |
| 30 | 123.2 | 0.0062 | 1.8 |
| 40 | 92.2 | 0.0046 | 1.3 |
| 69.1 | 69.1 | 0.0035 | 1.0 |
Solution: We use the ideal gas law to convert the partial pressures of N2O5 to molar concentrations. We plot these to over time, and take slopes of tangents to obtain instantaneous rates. We can then plot these rates against both [N2O5] and [N2O5]2. Because rates appear directly proportional to [N2O5]1, this reaction is first order with regard to [N2O5] and first order overall (Remember N2O5 is the only reactant!).
Initial Rate Method
Plotting rates against concetration becomes unwieldy with multi-reactant equations. Instead, we measure only the rate near the beginning of the reaction. We repeat the experiment with different starting concentrations of each reactant while keeping the other reactants’ concentrations the same.
There is a YouTube here.
| This graph shows five different runs for the decomposition of N205, with five different initial concentrations of this ingredient. We then compare the instantanteous raction rates (tangent slopes) at times t=0. |  |
| We then plot the five initial equation rates against N2O5‘s molar concentration. These rates are directly proportional to [N2O5], and the slope on this graph is k, the reaction constant. |  |
rate = (5.2 × 10–3) [N2O5] mol L–1 s–1
Problem Example 4
Gas-phase reduction of nitric oxide by hydrogen has the equation:
2 NO + 2 H2 → N2 + 2 H2O
The table below plots reactants’ partial pressures against reaction rate measured in torrs s-1.
| Experiment | P(NO) | P(H2) | Initial rate (torr s–1) |
| 1 | 359 | 300 | 1.50 |
| 2 | 300 | 300 | 1.03 |
| 3 | 152 | 300 | 0.25 |
| 4 | 300 | 289 | 1.00 |
| 5 | 300 | 205 | 0.71 |
| 6 | 200 | 147 | 0.51 |
These six experiments comprise two groups. The first varies NO’s pressure, while in the second group we vary H2‘s pressure.
This table shows reactant concentrations as pressure. With gaseuous reactants, their concentrations are directly proportional to their partial pressures at constant temperature and volume. Because we are comparing the ratios of pressures and rates, the units cancel out.
Solution:
In experiments 2 nad 3, reducing NO by a factor of two, reduces output by a factor of 4. Four is two squared, so the reaction is second order, with regard to [NO].
In experiment 4 and 6, we reduce hydrogen’s pressure by a factor of approximately 2, and H2‘s pressure drops by half. The reaction is first-order for [H2].
The rate law is:
rate = k[NO]2[H2]1.
The order of this equation over all is 3.
Dealing with Multiple Reactants: The Isolation Method
Multiple reactant experiments are not always practical for find individual reactants’ orders. In this case, we run experiments with excess concentrations of all reactants except one, to find that reactant’s order. For the reaction A + B + C → products, we want to find the rate law order with respect to [B].
We set [B]o to 0.020 M
We set [A]o to 2.00M
We set [C]0 to 2.00 M
Notice the order of magnitude difference in concentrations.
When we run the reaction, changes in [A] and [C] are minescule on a percentage basis. In fact they are often smaller than the experimental error so we can neglect them. Flooding the reaction with the other reactants effectively isolates the reactant of interest.
It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is
A + B + C → products
and we need to find the order with respect to [B] in the rate law. If we set
[B]o to 0.020 M and let [A]o = [C]o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By “flooding” the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested.
Video: Differential rate Method