Honors Only Material

1. Honors Only Material

2. 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 = -1367 kJ

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

4. ∆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

5. Using ∆Hf Tables (Honors) Which compound is most stable? NaCl(s) ∆Hf = -410.9 kJ/mole C2H4(g)∆Hf = 52.30 kJ/mole BaCO3(s)∆Hf = -1216.3 kJ/mole

6. Calculating ∆Hrxn from ∆Hf (Honors) You can calculate the overall change in enthalpy (∆H) by using ∆Hf values 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 (∆HfProducts) - Σ (∆HfReactants) Σ means “the sum of”

7. 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 ∆Hfvalues by coefficient in balanced equation Answer: ∆Hrxn = (4(-393.5) + 6(-285.85)) - (2(-84.68)) = -3119.74 kJ

8. Find ∆H for this reaction using ∆Hf values from the tables 4NH3 (g) + 5O2 → 6H2O (g) + 4NO Answer: ∆Hrxn = (6(-136.10) + 4(90.37)) - (4(-46.19)) = -270.36 kJ

9. 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?

10. 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

11. Reaction Mechanism Problems Step 1: A + B → AB Step 2: AB + A → A2B Net: 2A + B → A2B Intermediate?: AB Catalyst?: None

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

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

14. Step 1: NO2 + NO2→ NO3 + NO (slow) Step 2: NO3 + CO → NO2 + CO2 (fast) Net: NO2 + CO→ CO2 + NO Intermediate?: NO3 Catalyst?: How could I speed up overall rxn? Increase concentration of NO2

15. 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.

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

17. “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”

18. Crash Course http://www.youtube.com/watch?v=7qOFtL3VEBc&safe=active