Table of Contents:
Rate Law Introduction
Rate Law is an equation that shows how the rate of the reaction is affected by the concentration of reactants.
X is the order with respect to reactant A.
Y is the order with respect to reactant B.
The overall order of the reaction is equal to the sum of powers of the reactants in the rate law.
Let's take a look at an example of a rate law.
The reaction is second order with respect to NO2 and first order with respect to H2. The overall order of reaction is third (2+1 = 3).
How to determine the rate law from experimental data
Problem: Determine the rate law for the following reaction:
F2(g)+2ClO2(g) -> 2FClO2(g)
Choose two experiments where the concentration of one of the reactants changes and the concentration of the other one stays the same.
Write out the rate law equation for each experiment plugging the values from the table in.
Divide one experiment over the other to find the power (order) of the reactant.
Repeat with other reactants until all orders are known.
The rate is first order with respect to F2 and first order with respect to ClO2. The overall rate is second order (1+1).
Alternatively, if the numbers in the chart are easy, we can determine the orders without plugging numbers into the rate law.
For example, looking at the experiment 1 and 3, we notice that as the concentration of F2 doubles, the rate double as well. This means F2 must be to the first power.
Similarly, looking at the experiment 1 and 2, as the concentration of ClO2 quadruples, the rate quadruples as well. ClO2 must be to the first power.
Notes:
If the concentration changes but the rate stays the same, the power/order for that reactant must be zero (it doesn't affect the rate).
If the concentration doubles, but the rate quadruples, the power/order for the reactant must be 2. The order may also be other numbers including fractions and could be negative.
To determine the rate constant, we choose any of the experiments present and plug in the values of initial concentrations and rate into the rate law that we determined to solve for k.
Note: Some professors ask students to find the rate constant for every experiment given and take their average.
How to determine the correct units for the rate constant k
Alternatively, you can plug in the units yourself and cancel to determine the units of the rate constant. The units for rate are M/s and the units for the concentrations of reactants are M.
Integrated Rate Laws
Problem: The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.80 10^-4/s at 45C.
If the initial concentration is 1.65 10-2 mol/L, what is the concentration after 825 s?
Solution: The integrated rate law for the first order is
[A]=?
[A]0=1.65 * 10-2 mol/L
k=4.80 *10^-4/s
t=825 s
Now, we plug and chug
ln([A]/1.65 10-2 mol/L) = -4.80 10^-4/s(825 s)
[A]=0.0111 mol/L
Collision Theory
Collision theory states that in order for products to form, reactants must collide with enough energy and proper orientation.
Activation energy is the minimum amount of energy needed for reactants to collide successfully and produce products
Arrhenius equation shows the dependence of rate constant on temperature
Elementary Reactions
Elementary reaction is a single step reaction whose reactant coefficients can be used as orders for the rate law.
Molecularity represents the number of reactant molecules shows in the equation.
A-> products : UNIMOLECULAR (one reactant molecule)
A+B --> products OR 2A --> products : BIMOLECULAR (two reactant molecules)
A+B+C --> products : TERMOLECULAR (three reactant molecules)
A reaction mechanism represents a series of elementary reactions (steps) that compromise an overall reaction.
Key Notes:
The slowest step in the reaction mechanism is the rate determining step and dictates the rate law.
Reaction intermediate is a product of one step and reactant of another. It cancels out and does not appear in the net equation
The overall equation is created by adding all the elementary steps and cancelling species seen on both sides
Catalyst speeds up the reaction by lowering the activation energy. Catalyst will be a reactant of one step and product of another. It will be cancelled and not seen in the net equation.
How to write the rate law based on elementary steps
Ozone reacts with nitrogen dioxide to produce oxygen and dinitrogen pentoxide.
O3(g) + 2NO2(g) --> O2(g) + N2O5(g)
The proposed mechanism is
O3+ NO2 -->NO3+O2 (slow)
NO3+ NO2 --> N2O5 (fast)
What is the rate law predicted by this mechanism?
The rate law is based on the SLOW step. Coefficients for the reactants in the slow step become orders in the Rate Law. Both rectants have a coefficient of 1 in the elementary step and will be to the first power in the rate law.
Rate= k[O3][NO2]
How to write the rate law based on Mechanisms with an Initial Fast Step
A proposed mechanism is
N2O2 and N2O are intermediates because they are a product of one step and reactant of the subsequent step and get cancelled out
Since N2O2 is an intermediate, it cannot be a part of the rate law and we need to use the fast equilibrium reaction above the slow step to get rid of the intermediate.
At equilibrium, the rate of forward and reverse reactions are equal.
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