tying up the loose ends in chemical kinetics. reaction mechanisms here is a sample reaction...
TRANSCRIPT
Mechanisms, CatalystsIntermediates and k
Tying up the loose ends in chemical kinetics
Reaction mechanismsHere is a sample reaction mechanism
Step 1 ClO- + H2O HOCl + OH-
Step 2 Br- + HOCl HOBr + Cl-
Step 3 OH- + HOBr H2O + BrO-
There are several questions they will ask you about these reaction mechanisms1.What is the overall reaction2. Identify “intermediates” and “catalysts”
Reaction mechanismsThe overall reaction
Step 1 ClO- + H2O HOCl + OH-
Step 2 Br- + HOCl HOBr + Cl-
Step 3 OH- + HOBr H2O + BrO-
This is easy, just add up the reactions like you did when you were working on Hess’s law. Cross out all the chemicals that appear on both sides and then add up.
Reaction mechanismsThe overall reaction
Step 1 ClO- + H2O HOCl + OH-
Step 2 Br- + HOCl HOBr + Cl-
Step 3 OH- + HOBr H2O + BrO-
ClO- + Br- Cl- + BrO-This is the overall reaction. Not so bad. You don’t even need to flip any equations, or multiply the coefficients. All you do is cancel and add.
Reaction mechanisms“Intermediates” and “catalysts”
Step 1 ClO- + H2O HOCl + OH-
Step 2 Br- + HOCl HOBr + Cl-
Step 3 OH- + HOBr H2O + BrO-
ClO- + Br- Cl- + BrO-The “intermediates” are crossed out in black.
They are all the chemicals that cancel out that are not present in on the reactant side of the first step or the product side of the last step.
Reaction mechanisms“Intermediates” and “catalysts”
Step 1 ClO- + H2O HOCl + OH-
Step 2 Br- + HOCl HOBr + Cl-
Step 3 OH- + HOBr H2O + BrO-
ClO- + Br- Cl- + BrO-In the reaction above H2O is a “catalyst”. Water
is present on the reactant side of the first step and on the product side of the last step. This makes H2O a catalyst.
Reaction mechanisms“Intermediates” and “catalysts”
Step 1 H2O2 + I- OI- +H2O
Step 2 H2O2 + OI- I- + H2O + O2
1. Write the overall reaction2. Indentify both the catalyst and the
intermediate.
Free energy diagram
As we discussed earlier in thermodynamics, not all spontaneous reaction occur at observable rates.
A reaction that occurs slowly is likely to have a high activation energy (Ea).
What is this called again?
Is it positive or negative?
Free energy diagram
What is the effect of a catalyst?
A catalyst lowers the activation energy!
In this class (i.e. for the AP exam) there are two ways to increase the rate of a reaction.
1. Add a catalyst2. Increase the
temperature.
How to make it fasterIf the temperature of a non-reversible
reaction is increased then the reaction rate will increase.The higher temperature increases the average kinetic energy of the molecules.
Therefore more molecules have sufficient energy to overcome the activation energy (transition state energy).
Catalyst
A catalyst lowers the activation energy!
If the activation energy is lowered then less energy is required to go from reactants to products.
Therefore the reaction will proceed more quickly.
What you will be askedThey will try to trick you!!! (the AP people
that is)Here are the tricks1) The will get you to say ΔG is negative and
that the reaction is spontaneous. They will then ask you to comment on the speed of the reaction.
ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected by any thermodynamic data you could possibly present me.
What you will be askedHere’s trick number 22) You will be told that concentration,
pressure, or volume has been change or that an inert extra chemical has been added to your system.
ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected by anything else.
What you will be askedHere’s trick number 33) You will be told that anything other than a
temperature change or the addition of a catalyst has occured.
ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected. If temperature goes ↑ then k goes ↑, therefore rate ↑. If temperature goes down then k goes down, therefore rate decrease.If a catalyst is added then activation energy goes down, therefore the rate of reaction increases
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (slow)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (fast)
First, write the overall reaction
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (slow)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (fast)
Next, draw a line under the slow step
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (slow)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (fast)
Next, draw a line under the slow step
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (slow)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (fast)
Last, add up all the reactants up to the slow step only
H2O2 + I-
This makes: Rate = k [H2O2][I-]
Order of reaction from mechanismWhat if the location of the slow step changes
Mechanism
Step 1 H2O2 + I- HOI + OH- (fast)
Step 2 HOI + I- I2 + OH- (slow)
Step 3 2OH- + H3O+ 4H2O (fast)
First, write the overall reaction.
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (fast)
Step 2 HOI + I- I2 + OH- (slow)
Step 3 2OH- + H3O+ 4H2O (fast)
Next, draw a line under the slow step
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (fast)
Step 2 HOI + I- I2 + OH- (slow)
Step 3 2OH- + H3O+ 4H2O (fast)
Next, draw a line under the slow step
H2O2 + 2I- + 2H3O+ I2 + 4H2O
Order of reaction from mechanismThis is really not hard at all. I’ll show you.
Mechanism
Step 1 H2O2 + I- HOI + OH- (slow)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (fast)
Last, add up all the reactants up to the slow step only
H2O2 + 2I-
This makes: Rate = k [H2O2][I-]2
The coefficients become the exponents
You work itWhat if the location of the slow step changes
Mechanism
Step 1 H2O2 + I- HOI + OH- (fast)
Step 2 HOI + I- I2 + OH- (fast)
Step 3 2OH- + H3O+ 4H2O (slow)
Limitless practice, just click the circle below