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

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”

“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”

HONORS PACKET PRACTICE Pages 8, 9, 10, 11

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?

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

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

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

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?: ____________ Catalyst?: ___________

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? ______________________

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

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

∆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

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 www.sciencegeek.net/APchemistry/Powerpoints/6_Hess'sLaw.ppsx

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

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”

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.

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(-285.85)) - (2(-84.68)) = -3119.74 kJ

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

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 https://www.youtube.com/watch?v=SV7U4yAXL5I