THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6

Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. We have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery

Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of their difference in temperature. Other forms of energy — light electrical nuclear kinetic potential

Law of Conservation of Energy Energy can be converted from one form to another but can neither be created nor destroyed. (Euniverse is constant)

(m = mass, g = acceleration of gravity, and h = height) The Two Types of Energy Potential: due to position or composition - can be converted to work PE = mgh (m = mass, g = acceleration of gravity, and h = height) Kinetic: due to motion of the object KE = 1/2 mv2 (m = mass, v = velocity)

Kinetic and Potential Energy Potential energy — energy a motionless body has by virtue of its position.

Kinetic and Potential Energy Kinetic energy — energy of motion. • Translation • Rotation • Vibration

Units of Energy 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = 4.184 joules James Joule 1818-1889

Temperature v. Heat Temperature reflects random motions of particles, therefore related to kinetic energy of the system. Heat involves a transfer of energy between 2 objects due to a temperature difference

Extensive & Intensive Properties Extensive properties depends directly on the amount of substance present. mass volume heat heat capacity (C) Intensive properties is not related to the amount of substance. temperature concentration pressure specific heat (s)

State Function Depends only on the present state of the system - not how it arrived there. It is independent of pathway. Energy change is independent of the pathway (and, therefore, a state function), while work and heat are dependent on the pathway.

System and Surroundings System: That on which we focus attention Surroundings: Everything else in the universe Universe = System + Surroundings

Exo and Endothermic Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). Endothermic: Heat flows into the system (from the surroundings).

ENERGY DIAGRAMS Exothermic Endothermic

Endo- and Exothermic qsystem > 0 w > 0 ENDOTHERMIC heat Surroundings System heat qsystem > 0 w > 0 E goes up ENDOTHERMIC

Endo- and Exothermic qsystem > 0 w > 0 qsystem < 0 w < 0 Surroundings Surroundings System System heat heat qsystem > 0 w > 0 qsystem < 0 w < 0 E(system) goes down E(system) goes up ENDOTHERMIC EXOTHERMIC

First Law First Law of Thermodynamics: The energy of the universe is constant.

Enthalpy DH = Hfinal - Hinitial If Hfinal > Hinitial then DH is positive Process is ENDOTHERMIC If Hfinal < Hinitial then DH is negative Process is EXOTHERMIC

First Law E = q + w E = change in system’s internal energy q = heat w = work

Piston moving a distance against a pressure does work.

Work P = F/A F = PA w = F h w = PA h w = P V w & V must have opposite signs, since work is being done by the system to expand the gas. wsystem = -P V 1 Latm = 101.3 J 1 J = kgm2/s2

Enthalpy Enthalpy = H = E + PV E = H  PV H = E + PV At constant pressure, qP = E + PV, where qP = H at constant pressure H = energy flow as heat (at constant pressure)

Some Heat Exchange Terms specific heat capacity (s) heat capacity per gram = J/°C g or J/K g molar heat capacity (s) heat capacity per mole = J/°C mol or J/K mol

Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Al 0.902 glass 0.84 Aluminum

Ho = - qp qp = mst q = heat (J) m = mass (g) Simple Calorimeter s = specific heat (j/gCo) t = “change” in temperature (Co) Ho = “change in” enthalpy (kJ) Simple Calorimeter

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? where DT = Tfinal - Tinitial heat gain/lost = q = m s DT

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? where DT = Tfinal - Tinitial q = (0.902 J/g•K)(25.0 g)(37 - 310)K q = - 6160 J heat gain/lost = q = m s DT

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? q = - 6160 J Notice that the negative sign on q signals heat “lost by” or transferred out of Al.

Bomb Calorimeter

Heat Capacity E = qv & qv = -(C t + ms t) E = “change in” internal energy (J) qv = heat at constant volume (J) C = heat capacity (J/Co) t = “change”in temperature (Co)

In bomb calorimetry, volume is constant so: REMEMBER!!! In regular calorimetry pressure is constant, but the volume will change so: qp = -H qp = E + p  V In bomb calorimetry, volume is constant so: qv = E since p  V = zero.

Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane Temp rises from 25.00 to 33.20 oC Calorimeter contains 1200 g water Heat capacity of bomb = 837 J/K Hcm = -(Ct + mst) where Hc is heat of combustion.

Measuring Heats of Reaction CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = C t = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ

Hess’s Law Reactants  Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

Standard States Compound For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.

Calculations via Hess’s Law 1. If a reaction is reversed, H is also reversed. N2(g) + O2(g)  2NO(g) H = 180 kJ 2NO(g)  N2(g) + O2(g) H = 180 kJ 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g)  3N2(g) + 3O2(g) H = 540 kJ

Using Enthalpy Consider the decomposition of water H2O(g) + 243 kJ ---> H2(g) + 1/2 O2(g) Endothermic reaction — heat is a “reactant” DH = + 243 kJ

Using Enthalpy Making H2 from H2O involves two steps. H2O(l) + 44 kJ ---> H2O(g) H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) ----------------------------------------------------------------- H2O(l) + 286 kJ --> H2(g) + 1/2 O2(g) Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the other rxns.

Using Enthalpy Calc. DH for S(s) + 3/2 O2(g) --> SO3(g) S(s) + O2(g) --> SO2(g) -320.5 kJ SO2(g) + 1/2 O2(g) --> SO3(g) -75.2 kJ _______________________________________ S(s) + 3/2 O2(g) --> SO3(g) -395.7 kJ

 DH along one path =  DH along another path energy S solid direct path DH = 1 +O 2 -320.5 kJ + 3/2 O 2 DH = SO gas 2 -395.7 kJ + 1/2 O 2 SO gas 3 DH = -75.2 kJ 2  DH along one path =  DH along another path

This equation is valid because DH is a STATE FUNCTION DH along one path = DH along another path This equation is valid because DH is a STATE FUNCTION These depend only on the state of the system and not how it got there. Unlike V, T, and P, one cannot measure absolute H. Can only measure DH.

Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of DHof = standard molar enthalpy of formation This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. DHof is always stated in terms of moles of product formed. See Appendix A21-A24.

DHof, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) DHof = -241.8 kJ/mol By definition, DHof = 0 for elements in their standard states.

Using Standard Enthalpy Values Use DH°’s to calculate enthalpy change for H2O(g) + C(graphite) --> H2(g) + CO(g) (product is called “water gas”)

Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find H2(g) + 1/2 O2(g) --> H2O(g) DH°f of H2O vapor = - 242 kJ/mol C(s) + 1/2 O2(g) --> CO(g) DH° f of CO = - 111 kJ/mol

Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) DHo = +242 kJ C(s) + 1/2 O2(g) --> CO(g) DHo = -111 kJ ----------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) DHonet = +131 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic.

Change in Enthalpy Can be calculated from enthalpies of formation of reactants and products. Hrxn° = npHf(products)  nrHf(reactants) H is an extensive property--kJ/mol For the reaction: 2H2 (g) + O2 (g) ---> 2H2O(g) Enthalpy would be twice the H value for the combustion of hydrogen.

Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., DHorxn for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn =  DHof (prod) -  DHof (react)

Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn =  DHof (prod) -  DHof (react) DHorxn = DHof (CO2) + 2 DHof (H2O) - {3/2 DHof (O2) + DHof (CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} DHorxn = -675.6 kJ per mol of methanol DHorxn is always in terms of moles of reactant.

Pathway for the Combustion of Methane

Schematic diagram of the energy changes for the combustion of methane.

Greenhouse Effect -- a warming effect exerted by the earth’s atmosphere due to thermal energy retained by absorption of infrared radiation. Greenhouse Gases: CO2 H2O CH4 N2O