Chpt 6 - Thermochemistry Energy and framework (definitions) Internal energy, heat & work Enthalpy & Calorimetry Hess’s Law Standard Enthalpies Applications? HW: Chpt 6 - pg. 275-283, #s 14, 16, 24, 27, 28, 32, 34, 38, 41, 42, 44, 49, 54, 62, 72, 73, 80, 84, 112 Due Wed Oct. 7

Thermodynamic Framework Energy - capacity to do work or produce heat Law of conservation of energy PE (generally chemical potential energy) and KE Heat involves a transfer of energy - Very different from temperature (a measure of motions of particles) Heat is NOT a substance Work is force acting over a distance

State Functions Energy = work + heat Ball rolling down hill example - heat and work are different depending on pathway

State Functions - con’t State function or state property only depends on current state not pathway Energy is state function Heat and work are not state functions Another example, Elevation vs. Distance

Chemical Energy System and surroundings Exothermic and endothermic Chemical PE <--> thermal energy Recall energy reaction diagrams Only concerned with Energy of Reactants and Products (not pathway) Internal Energy E = q + w q=heat, w=work 1st Law of thermodynamics Energy of Universe is constant

Universe Energy is Constant Exothermic or Endothermic?

Chemical Energy - con’t Quantities have sign and magnitude System’s perspective for sign, thus endo is flow into system so q is positive (gaining heat) E <0 exothermic, E>0 endothermic Work done on system is positive, w>0 Work done by system is negative, w<0

PV Work Common types of work are expansion by a gas and compression on a gas PV work, w = - PV if volume is expanding w = PV if volume is compressing V = Vfinal - Vinitial,

PV Work - derivation Example that shows Pressure=force/area = F/A so F = P x A Work is force x dist = F x h so W = P x A x h volume of cylinder = A x h Thus W = PV The sign is (-) for expanding gas, since work is done by system on surroundings

Enthalpy, H H = E + PV, since E, P and V are state functions H is also a state function At constant pressure H = qp (qp is heat at const p) In general, for open laboratory chemical reactions pressure is constant, so the change in enthalpy is used interchangeably with the heat of a reaction. For a chemical reaction H = Hproducts -  Hreactants Exothermic means enthalpy, H < 0 Endothermic means enthalpy, H > 0

Calorimetry The science of measuring heat. Substances absorb heat differently; heat capacity, C, measures this C = heat absorbed / increase in temp Specific heat capacity is per gram substance A calorimeter measures heat change. q = C x m x T our text uses ‘s’ for C The AP test will use C, actually Cp

Calorimeter Simple styrofoam cup calorimeter is easily used for lab measurements and constant pressure measurements.

Bomb Calorimeter A “Bomb” calorimeter schematic - it only looks like a bomb. It is used when constant volume measurements are needed.

Hess’s Law Since enthalpy is a state function, heat of reactions can be calculated from a known set of simple chemical rxns combined together to get the final rxn. One step: N2(g) + 2O2(g) --> 2NO2(g) H = 68kJ Two distinct steps N2(g) + O2(g) --> 2NO(g) H = 180kJ 2NO(g) + O2(g) --> 2NO2(g) H = -112kJ Total these reactions

Hess’s Law schematic The overall reaction enthalpy is independent of pathway

Hess’s Law rules Characteristics of H for a rxn If a reaction is reversed, the sign of H is also reversed. The magnitude of H is directly proportional to the quantities of reactants and products in the rxn i.e. if the coefficients are multiplied by an integer, H is multiplied by same integer.

Example 6.8 pg. 254 We want to calculate the H for the synthesis of diborane, B2H6, from its elements. 2B(s) + 3H2(g) --> B2H6 (g) H = ? Use the following data: 2B(s) + 3/2O2(g) --> B2O3(s) H = -1273 kJ B2H6(g) + 3O2(g) --> B2O3(s) +3H2O(g) H = -2035 kJ H2(g) + 1/2O2(g) --> H2O(l) H = -286 kJ H2O(l) --> H2O(g) H = 44 kJ Hints - 1) work backward from the required/desired reaction, 2) reverse any reaction as needed to align reactants and products, 3)multiply any reaction as necessary to get correct coefficients. Recall H is a state function (independent of pathway)

Standard Enthalpies of Formation Hfo of a substance is the change in enthalpy that accompanies the formation of 1 mole of the substance from its elements in their standard states. Standard state: gas is 1 atm, liquid or solid is pure substance, solutions are 1M. Elements are state at 1atm and 25oC.

CalculatingH from Hfo Since enthalpies are state functions independent of pathway, in chemical reactions, the reactants can be taken apart into their elements and the products can be constructed from their elements. For a chemical reaction H = npHfo (products) -  nrHfo (reactants) Data found in Appendix 4 - pg A19 - A22

Example H from Hfo 4 NH3(g) + 7 02(g) --> 4 NO2(g) + 6 H2O(l) NH3(g) -46 kJ/mol NO2(g) 34kJ/mol H2O(l) -286kJ/mol O2(g) 0 kJ/mol H = Products - Reactants… 4x34kJ/mol + 6x(-286kJ/mol) - 0 - 4x(-46kJ/mol) H = -1396kJ/mol

Greenhouse Effect