EXAMPLE: How much heat is required to heat 10.0 g of ice at o C to steam at o C? q overall = q ice + q fusion + q water + q boil + q steam q= (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g) + (10.0g  4.18J/g o C  o C) + (10.0g  2260J/g) + (10.0g  2.03J/g o C  27.0 o C) q = ( × × × )J = 30.9 kJ

Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 isolated system  E=0 heat released by reaction raises the temperature of the solvent constant T, heat is released to the surroundings endothermic – reaction that absorbs energy q > 0

Expansion Type Work w = -P  V system does work P P V initial VV  V = V final - V initial q p = +2kJ

Do 250 J of work to compress a gas, 180 J of heat are released by the gas What is  E for the gas? J 2.70 J J J J

Enthalpy H  E = q + w at constant V, w expansion = 0  E = q v at constant P, w expansion = -P  V  E = q p - P  V Define  PV) =  P  V at constant P Hence  H = q p

Enthalpy Enthalpy  heat at constant pressure or the heat of reaction q p =  H = H products - H reactants Exothermic Reaction  H = (H products - H reactants ) < 0 2 H 2(g) + O 2(g)  2 H 2 O (l)  H < 0 Endothermic Reaction  H = (H products - H reactants ) > 0 2 H 2 O (l)  2 H 2(g) + O 2(g)  H > 0

State Functions  H and  E along with  P,  T,  V (or P, T, V) and many others are state functions. They are the same no matter what path we take for the change. q and w are not state functions, they depend on which path we take between two points. initial final EE q w q w  E=E final -E initial q and w can be anything

Path Independent Energy Changes

Which day would you like OWL quizzes due (4 AM) Monday 2.Tuesday 3.Wednesday 4.Thursday 5.Friday

Stepwise Energy Changes in Reactions

Laws of Thermochemistry 1. The magnitude of  is directly proportional to the amount of reaction.  H is for 1 mole of reaction as written 2 H 2(g) + O 2(g)  2 H 2 O (l)  H = kJ H 2(g) + ½ O 2(g)  H 2 O (l)  H = kJ Can have ½ mole O 2 just not ½ molecule

Laws of Thermochemistry 2.  H for a reaction is equal in magnitude but opposite in sign to  H for the reverse reaction. H 2(g) + ½ O 2(g)  H 2 O (l)  H = kJ H 2 O (l)  H 2(g) + ½ O 2(g)  H = kJ

Laws of Thermochemistry 3. The value of  H for the reaction is the same whether it occurs directly or in a series of steps.  H overall =  H 1 +  H 2 +  H 3 + · · · also called Hess’ Law

Enthalpy Diagram H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = kJ H 2 O(l)  H 2 O(g)  H = kJ H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = kJ

Given 3 CO + 3/2 O 2  3 CO 2  H = -849 kJ What is  H for CO 2  CO + ½ O 2 ? kJ kJ kJ kJ kJ

Energy and Stoichiometry Since  H is per mole of reaction we can relate heat to amount of reaction Given C 2 H 6 + 7/2 O 2  2 CO H 2 O  H= kJ If kJ are released to surroundings what mass of H 2 O is formed? kJ released means  H = kJ for this much H 2 O

Bomb Calorimeter measure q v q rxn + q cal = 0 q rxn = -q cal q rxn = - c cal  T  E rxn = q rxn /moles rxn  E rxn ≈  H rxn  H =  E +  (PV)  H =  E + RT  n RT = 2.5 kJ/mol

“Coffee Cup” Calorimeter q p Photo by George Lisensky

Measuring  H When 25.0 mL of 1.0 M H 2 SO 4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6  C the temperature rises to 33.9  C. What is  H for H 2 SO KOH  K 2 SO H 2 O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.  C) q soln = mc  T m = ( )mL×1.00g/mL = 75.0 g

Measuring  H cont. q=mc  T q soln = 75.0 g × 4.18 J/g.  C × ( )  C q soln = 2916 J q rxn + q soln = 0 q rxn = J  H rxn = q rxn /moles rxn

Measuring  H cont How many moles rxn? 1 mol rxn / 1 mol H 2 SO 4 1 mol rxn / 2 mol KOH Stoichiometric mixture so mol rxn

Measuring  H cont  H rxn = q rxn /moles rxn  H rxn = J / mol rxn  H rxn = J / mol rxn  H rxn = -117 kJ  H is per mole of reaction as written

If excess Al is added to 50 mL of M H 2 SO 4 how many moles of the following reaction occur? 2 Al + 3 H 2 SO 4  Al 2 (SO 4 ) H mol mol mol mol mol

Hess’s Law Can find  H for an unknown, or hard to measure, reaction by summing measured  H values of known reactions.

EXAMPLE  H for formation of CO cannot readily be measured since a mixture of CO and CO 2 is always formed. C (s) + ½ O 2 (g)  CO (g)  H = ? C (s) + O 2 (g)  CO 2 (g)  H 1 = kJ CO (g) + ½ O 2 (g)  CO 2 (g)  H 2 = kJ C (s) + ½ O 2 (g)  CO (g)  H =  H 1 -  H 2  H =  H 1 -  H 2 = – (-283.0) = kJ

Standard Enthalpy of Formation the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature For an element this is a null reaction O 2 (g)  O 2 (g)  H = 0  H  f = 0 for all elements in their standard states

For which one of these reactions is ΔH º rxn = ΔH º f ? 1.N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 2.C(graphite) + 2 H 2 (g)  CH 4 (g) 3.C(diamond) + O 2 (g)  CO 2 (g) 4.CO(g) + ½ O 2 (g)  CO 2 (g) 5.H 2 (g) + Cl 2 (g)  2 HCl(g)

Calculation of  H o  H o =  mols   H f o products –  mols   H f o reactants We can always convert products and reactants to the elements. Hess’s law says  H is the same whether we go directly from reactants to products or go via elements

Example What is the value of  H rxn for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = kJ/mol O 2(g)  H f o = 0 CO 2(g)  H f o = H 2 O (g)  H f o =  H rxn  mols   H f o  product –  mols  H f o  reactants

Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = kJ/mol; O 2(g)  H f o = 0 CO 2(g)  H f o = ; H 2 O (g)  H f o =  H rxn  mols  H f o  product -  mols  H f o  reactants  H rxn  ) + 6( )  product -  2( ) + 15(0)  reactants kJ/mol =  10 3 kJ

Fossil Fuels coal petroleum natural gas

Energy Resources in the U.S.

Caloric Value of Some Foods