Reactions Rates
From MyMCAT
Contents |
Introduction
Why do some reactions proceed rapidly, such as in the case of chemical explosions, while others proceed too slowly to observe? There are several ways of developing equations to show how the rates of chemical reactions change and the equations are called rate expressions.
The Differential Rate Law
One way in identifying reaction rates is by measuring the rate of the reaction initially at known concentrations of reactants. This process however can only be done at the immediate start of the reaction because after any given length of time the concentrations of the reactants will drop (since they are being used up) and so there will no longer be a valid concentration to mathematically work with.
The rate law states that a reaction will be proportional to the product of the reactant concentrations to some power. Thus, for a reaction with reactants A and B, the rate is proportional to the product of ([A] to some power m) and ([B] to some power n). To create an equation instead of a proportion, a rate constant k is multiplied in.
![r\; =\; k\left[ A \right]^{m}\left[ B \right]^{n}](/w/images/math/5/e/d/5ed5ddd49d8bff8f2e371bd23202c434.png)
Example
Consider the hypothetical reaction with reactants A, B and C where following initial rate data was collected:
| Experiment | [A] mol/L | [B] mol/L | [C] mol/L | Initial rate, r (molL-1s-1) |
|---|---|---|---|---|
| 1 | 0.400 | 1.600 | 0.0600 | 4.86 |
| 2 | 0.800 | 1.600 | 0.0600 | 9.72 |
| 3 | 0.400 | 0.800 | 0.0600 | 4.86 |
| 4 | 0.800 | 1.600 | 0.1800 | 87.5 |
The rate law for this experiment is then
The idea is to compare experiments where all the concentrations except one are held constant so that we can determine how the rate changes when one concentration changes. Comparisons where two concentrations or more change don't work because you would never know how each of the concentration changes were effecting the rate!
Finding the power of A
To isolate the A term, we need the concentrations of B and C to remain constant. Looking at the table we can see that in trials 1 and 2, [A] changes, but [B] and [C] do not, so this is perfect.
One can then determine the power of A either mathematically or simply through examination as most values on the MCAT will be trivial to calculate.
2Rate(1) = Rate(2)
2*4.86 = 9.72 2k[A]x[B]y[C]z = k[A]x[B]y[C]z
The [B] and [C] terms can be removed since they do not change, the rate constant k can be divided out leaving:
2(0.400)x = (0.800)x
Gather like terms to solve for x:
2 = (0.800)x / (0.400)x
2 = (0.800/0.400)x
2 = 2x
(Recall that m1 = m )
2 = 21 so x = 1
Alternatively, one could have simply looked at the rates and decided, between trial 1 and 2, the concentration of A doubles, and the rate also doubles (2 * 4.86 = 9.72!) thus the power must have just been 1.
Finding the power of B
To isolate the B term, we need the concentrations of A and C to remain constant. Looking at the table we can see that in trials 1 and 3, [B] goes down, but [A] and [C] do not change, so this is perfect.
One can then determine the power of B again either mathematically or simply through examination. (Mathematically the steps will be the same as above.) Looking at the rates, between trial 1 and 3, the concentration of B halves but the rate does not change at all. For the rate to be independent of B, [B] must be removed from the rate equation. To do this, the power of B must simply be 0, that way the B term just becomes 1 and in no way effects the rate. (Anything to the power of 0 is always 1).
Finding the power of C
Compare Experiments 2 and 4:
[A] and [B] are constant, [C] is tripled, and the rate is increased by a factor of nine. 9Rate(2) = Rate(4) 9k[A]x[B]y[C]z = k[A]x[B]y[C]z 9(0.0600)z = (0.1800)z 9 = 3z z = 2
The rate expression is now Rate = k[A]1[B]0[C]2 Which simplifies to Rate = k[A][C]2
Often the value of k must be solved for. This can be done trivially by simply using one of the trials and plugging in all the concentrations and the rate into the rate equation and solving for k.
