ThermochemistryThermochemistryStudy of the transfer of Study of the transfer of

energy in chemical reactionsenergy in chemical reactions

Heat and TemperatureHeat and TemperatureWhat’s the difference??What’s the difference??

Temperature (C0): measures the average kinetic energy of particles.

Instrument: thermometer

Units: Celcius, Kelvin, F

HeatHeatHeat (Q): Q= m ∆t Cp

(Total Thermal Energy of a system)

Instrument: calorimeter p.519

Units: Joules, Cal

Heat of Reaction (∆H)Heat of Reaction (∆H) The net energy released or absorbed The net energy released or absorbed

in a chemical reaction. in a chemical reaction. Thermochemical equationThermochemical equation: :

includes the amount of energy includes the amount of energy released/absorbed.released/absorbed.Write the thermochemical equation of the Write the thermochemical equation of the formation of water if 483.6 kJ of energy are formation of water if 483.6 kJ of energy are produced from 2 moles of hydrogen.produced from 2 moles of hydrogen.

2H2H22 (g) + O (g) + O22 (g) (g) 2H 2H22O (l) + 486.3 kJ O (l) + 486.3 kJ (exo)(exo)

Always include states of matter!Always include states of matter!

Enthalpy (heat) Changes (Enthalpy (heat) Changes (ΔΔH)H)The amount of energy absorbed or lost by a The amount of energy absorbed or lost by a

system.system. How to calculate How to calculate ΔΔH:H: A) A) from a graphfrom a graph: : Heat Heat productsproducts – Heat – Heat

reactantsreactants

If If ΔΔH = negative: exothermic (Energy Lost)H = negative: exothermic (Energy Lost)

If If ΔΔH = positive: endothermic (Energy absorbed)H = positive: endothermic (Energy absorbed)

B) ∆H can be calculated using:Hess’s Law :The overall ΔH is equal to the sum of the ΔH for individual steps.

Molar heat of formation: ∆Hf is the net heat released or absorbed when one mole of a compound is formed by the combination of its elements. (synthesis)∆Hf values (Appendix A-14) are based on one mole of product. -∆Hf indicates a stable compound.

Molar heat of combustion: ∆Hc is the heat released by the complete combustion of one mole of a substance.∆Hc values (Appendix A-5) are based on one mole of reactant.

Calculate the Calculate the ΔΔHHrxnrxn for: for: 2SO2SO22 + O + O22 2SO 2SO33

Equation must be balanced.Equation must be balanced. Table A-14 (Table A-14 (ΔΔHHff))

ΔΔHHff SO SO2 2 = (-296.8)kJ/mol = (-296.8)kJ/mol

ΔΔHHff SO SO3 3 = (-395.7)kJ/mol = (-395.7)kJ/mol

∆∆HHrxnrxn = (∆H = (∆Hff products – (∆H products – (∆Hf f reactants)reactants) (-791.4) - (-593.6)(-791.4) - (-593.6)

Multiply both SO2 and SO3 by 2 because we have 2 moles of each.

-593.6 kJ

-791.4 kJ

Answer: -197.8 KJ

Sample problems(continued)Sample problems(continued)

2) Calculate the heat of reaction for:2) Calculate the heat of reaction for: 2 H2 H22OO22 2H 2H22O + OO + O22ꜛ (must be ꜛ (must be

balanced)balanced) Find ∆HFind ∆Hff table A-14 table A-14

∆∆HHff H H22OO22 (-187.8 kJ/mol) (2) = (-187.8 kJ/mol) (2) = -375.6 kJ-375.6 kJ

∆∆HHff O O22 (0.00 kJ/mol) (0.00 kJ/mol)

∆∆HHff H H22O (-285.8 kJ/mol) (2) = O (-285.8 kJ/mol) (2) = -571.6 kJ-571.6 kJ

∆∆HHrxnrxn = ∆H = ∆Hff products - ∆H products - ∆Hf f reactantsreactants (-571.6) - (-375.6) (-571.6) - (-375.6) ANSWER: - 196.0 kJ ANSWER: - 196.0 kJ

Calculating ∆H from Calculating ∆H from ∆H∆Hcc values.values. 1. Calculate the heat of formation for:1. Calculate the heat of formation for: CC(s)(s) + 2H + 2H2(g)2(g) CH CH4(g) 4(g)

Make sure the equation is balanced.Make sure the equation is balanced. Use Use ∆H∆Hc c values of C, Hvalues of C, H22 and methane (tbl. A-5) and methane (tbl. A-5)

NOTE:NOTE: ∆H∆Hc c values are based on one mole ofvalues are based on one mole of

REACTANTREACTANT. . Therefore:Therefore:

∆∆HHff = ∆H = ∆Hcc of of REACTANTSREACTANTS – (∆H – (∆Hcc PRODUCTS) PRODUCTS)

C = -393.5 kJ/mol CH4 = -890.8 kJ/mol

H = (-285.8 kJ) (2) = (-965.1 kJ) – (-890.8kJ) ANSWER: ∆Hf = -74.3 kJ

17.2 Two 17.2 Two Driving Forces Driving Forces of Reactions…Allow of Reactions…Allow one to predict if a reaction will occur one to predict if a reaction will occur spontaneously.spontaneously.

Enthalpy Enthalpy ((ΔΔH)H):: exothermic reactions exothermic reactions favored (-∆H) (products at favored (-∆H) (products at lowerlower energy) energy)

Entropy Entropy ((ΔΔS)S):: measures disorder of a measures disorder of a system. (+∆S)system. (+∆S)High disorderHigh disorder favored. favored.

What does high disorder look like? What does high disorder look like? Increase in randomness (disorder): Increase in randomness (disorder):

Solid to liquid to gasSolid to liquid to gas Lesser # particles to greater #particlesLesser # particles to greater #particles Solid dissolving Solid dissolving

Possible reactionsPossible reactions: : RESULTS:RESULTS:

1. Exothermic and Entropy increase GO2. Endothermic and Entropy decrease NO GO 3. Exothermic and Entropy decrease ?4. Endothermic and Entropy increase ?

#3 & #4 may or may not “Go” depending on which of the two driving forces is dominant .

#3 example: water freezes. In this case, enthalpy(∆H) is the dominant factor.

#4 example : Ice melts. Therefore, the entropy (∆S) is the dominant factor.

Gibb’s Free Energy (G)Gibb’s Free Energy (G)Predicts whether ∆H or ∆S will dominate in a Predicts whether ∆H or ∆S will dominate in a

reactionreaction Natural processes proceed in the Natural processes proceed in the

direction that lowers the free energy (G) direction that lowers the free energy (G) of a system.of a system.

Only Only ΔΔG can be measured: G can be measured: ∆G=∆H-∆G=∆H-(T∆S)(T∆S)… … Predicts whether a reaction will be Predicts whether a reaction will be spontaneous (at constant T & P).spontaneous (at constant T & P).

Negative ΔG = spontaneous reaction Negative ΔG = spontaneous reaction Positive ΔG = nonspontaneous reaction Positive ΔG = nonspontaneous reaction

Sample problem:Sample problem:Calculate the free energy change for the following reaction at 250 C:

Ca(s) + 2H2O(l) Ca(OH)2 + H2

Data: ∆H = -411.6 kJ ∆S = 31.8 J/mol.K

∆G = ∆H – (T∆S) -411.6 - (298 x .0318 kJ/mol.K)

ANSWER: ∆G = -421.1 kJ

Summary Summary Favorable conditions for a spontaneous reaction:

Free Energy (ΔG) = Negative

Enthalpy (ΔH) = Negative

Entropy (ΔS) = Positive

When enthalpy and entropy are at odds, temperature determines whether the reaction will proceed: ΔΔG = G = ΔΔH – TH – TΔΔSS

17.3 The Reaction Process17.3 The Reaction ProcessCollision TheoryCollision Theory: Explanation of : Explanation of

reactions as a result of collisions.reactions as a result of collisions.– Particles must collideParticles must collide– Particles must collide in the Particles must collide in the

correct spatial orientation correct spatial orientation – Collision must be energetic Collision must be energetic

enough to disrupt bonds enough to disrupt bonds

Reaction PathwaysReaction Pathways Activation EnergyActivation Energy: minimum energy : minimum energy

required to transform reactants required to transform reactants activated complex.activated complex.

Activated ComplexActivated Complex: transitional : transitional structure formed from an effective structure formed from an effective collision.collision. A= Energy of Reactants

B= Activation Energy

C= Energy of Activated Complex

D= Energy of Products

Is this an exothermic or endothermic reaction?

17.4 Reaction Rates and 17.4 Reaction Rates and KineticsKinetics

Rate-Influencing FactorsRate-Influencing Factors: Anything that : Anything that will influence collision frequency and/or will influence collision frequency and/or efficiency.efficiency.– Nature of ReactantsNature of Reactants– Surface Area (increased surface area)Surface Area (increased surface area)– Temperature (increased temperature)Temperature (increased temperature)– Concentration (increased conc. of reactants)Concentration (increased conc. of reactants)– Catalyst (addition of)Catalyst (addition of)

Rate LawsRate Laws…Relate reaction rate to the [concentration] of …Relate reaction rate to the [concentration] of reactants, at a given temperature.reactants, at a given temperature.

As concentration increases, rate increases.As concentration increases, rate increases.

1)1)Elementary Reactions Elementary Reactions (single step)(single step)

2A + 3B A2A + 3B A22BB33

Rate Law: R = k[A]Rate Law: R = k[A]22 [B] [B]3 3

3D E + F3D E + F

Rate Law: R = k [D]Rate Law: R = k [D]33

2) 2) Multi-step ReactionsMulti-step Reactions

Rate is determined from the slowest step.example: Net equation: NO2 + CO NO + CO2

step 1: 2NO2 NO3 + NO (slow) step 2: NO3 + CO NO2 + CO2 (fast)

Rate Law?

ANSWER: Rate Law = k[NO2]2

Rate law determination Rate law determination by experimental by experimental

datadata::

In a one step reaction: •Doubling [A]: the rate increases by 2X•Doubling [B]: the rate increases by 4X•Reduce [B] to 1/3: the rate decreases to 1/9

What is the rate law?

ANSWER: Rate= k [A] [B]2