1. Rate Law (Honors)
Mathematical expressions relating the rate of reaction to the concentration of reactants.
Rate of a reaction can only be determined from experimental data, not from a balanced equation.
Crash Course: Kinetics and Rate Law

2. Rate = k [A]x [B]y Rate Law Equation
Must be determined experimentally from lab data to determine what “order” to use
Rate = k [A]x [B]y
Specific rate constant at that temperature for this reaction
[ ] = shorthand for “the concentration of”

3. “Order” of a Reaction
Refers to the power to which the conc. of a reactant is raised to express the observed relationship between conc. and rate.
First Order: X 1
if concentration of reactant doubles, rate doubles
Second Order: X 2
if concentration of reactant doubles, rate quadruples
Zero Order: X 0
if concentration of reactant doubles, it has no effect on overall rate
Overall Order Of Reaction:
the sum of all the reactants “orders”

4. HONORS PACKET PRACTICE
Pages 8, 9, 10, 11

5. Reaction Mechanism Problems
Show series of steps from reactants to products.
You need to be able to:
Write the net equation
Identify any catalyst or reaction intermediates present.
If rate determining step is indicated, how may you speed it up?

6. Reaction Mechanism Problems
Reaction Intermediates: are formed first as
a temporary product then become reactant
in later step
Catalysts: put in as a reactant first and later
come back out as a product (not altered by
reaction)
Both DO NOT appear in net equation

7. Reaction Mechanism Problems
Step 1: A + B → AB
Step 2: AB + A → A2B
Net: 2A + B → A2B
Intermediate?: __________
Catalyst?: ____________

8. Step 1: H2O2 + I-1 → H2O + IO-1
Step 2: H2O2 + IO-1 → H2O + O2 + I-1
Net: 2H2O2 → 2H2O + O2
Intermediate?: _________
Catalyst?: __________

9. Step 1: Cl2 + AlCl3 → AlCl4-1 + Cl+1
Step 2: Cl C6H6 → C6H5Cl + H+1
Step 3: H+1 + AlCl4-1 → AlCl3 + HCl
Net: Cl2 + C6H6 → C6H5Cl + HCl
Intermediate?: ____________
Catalyst?: ___________

10. Step 1: NO2 + NO2 → NO3 + NO (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
Net: NO2 + CO → CO2 + NO
Intermediate?: _________
Catalyst?: _____________
How could I speed up overall rxn?
______________________

11. Stoichiometry and ∆H (Honors)
Remember the quantity of energy indicated in a reaction are for the number of moles indicated.
If given grams or a different number of moles use stoich to determine the energy.
Ex: How much energy is released when
11.5g C2H5OH is burned?
C2H5OH + 3O2 → 2CO2 + 3H2O
∆H = kJ

12. Honors
If 30 grams of octane (C8H18) is burned how much energy is released?

13. ∆H of Formation (Honors)
∆Hf represents the heat energy absorbed or released when I mole of a compound is formed from its elements. (at 25 °C, 1atm)
Stability of a compound is linked to ∆Hf
Lower in energy = more stable
- ∆Hf indicates the compound is lower in energy than the separate elements and more stable
+ ∆Hf indicates higher energy products and more unstable compounds

14. Thermodynamics Table Shows Heat of Formation ∆Hf to form
1 mole of a compound from its elements
NOTE:
Pure Elements in their standard
states are not do not need to be
“formed” so have a ∆Hf value = zero

15. Using ∆Hf Tables (Honors)
Which compound is most stable?
NaCl(s) ∆Hf = kJ/mole
C2H4(g) ∆Hf = kJ/mole
BaCO3(s) ∆Hf = kJ/mole

16. Calculating ∆Hrxn from ∆Hf (Honors)
You can calculate the overall change in enthalpy (∆H) in a reaction by using ∆Hf values of the reactants and products:
We will be basically be assuming that a rxn takes place by breaking down reactants into their elements and then reassembling them as products.
∆Hrxn = Σ (∆Hf Products) - Σ (∆Hf Reactants)
Σ means “the sum of”

17. Enthalpy Calculations
H°rxn = Hf(products) - Hf(reactants)
Elements in their standard states are not
included in the ΔHrxn calculations because
ΔHf° for an element in its standard state is zero.

18. 2C2H6 (g) + 7O2(g) → 4CO2 (g) + 6H2O(l)
Find ∆H for this reaction using ∆Hf values from the tables
2C2H6 (g) + 7O2(g) → 4CO2 (g) + 6H2O(l)
Note:
Pure Elements have ∆Hf = 0 and can be disregarded
Remember ∆Hf values are for one mole only
Multiply ∆Hf values by coefficient in balanced equation
Answer:
∆Hrxn = (4(-393.5) + 6( )) - (2(-84.68))
= kJ

19. Find ∆H for this reaction using ∆Hf values from the tables
4NH3 (g) + 5O2 → 6H2O (g) + 4NO
Answer:
∆Hrxn = (6( ) + 4(90.37)) - (4(-46.19))
= kJ

20. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:
Calculate H° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
Hf° (kJ/mol)
Na(s) 0
H2O(l) –286
NaOH(aq) –470
H2(g) 0 H° = –368 kJ
[2(–470) + 0] – [0 + 2(–286)] = –368 kJ
ΔH = –368 kJ
Crash Course: Enthalpy & Hess’s Law