Chapter 5: Thermochemistry

Thermochemistry: – Energy Kinetic & Potential – First Law of Thermo internal energy, heat & work endothermic & exothermic processes state functions – Enthalpy – Enthalpies of Reaction

– Calorimetry heat capacity and specific heat constant-pressure calorimetry bomb calorimetry (constant-volume calorimetry) – Hess’s Law – Enthalpies of Formation for calculation of enthalpies of reaction – Foods and Fuels

Energy work is a form of energy w = F x d energy is the capacity to do work or transfer heat Kinetic Energy energy of motion E = ½ mv 2 potential energy energy of position applies to electrostatic energy applies to chemical energy (energy of bonds)

energy units one joule = energy of a 2 kg mass moving at 1 m/s E = ½ mv 2 (½)(2 kg) (m/s) 2 = kg m 2 /s 2 = 1 J 1 cal = J 1 kcal = 1 food calorie (Cal) Systems & Surroundings system -- chemicals in the reaction surroundings -- container & all outside environment closed system can exchange energy (but not matter) with its surroundings

2 H 2(g) + O 2(g)  2H 2 O (l) + energy (system) energy (as heat or work) no exchg of matter with surroundings Closed System

First Law of Thermo. Energy is always conserved any energy lost by system, must be gained by surroundings Internal Energy -- total energy of system combination of all potential and kinetic energy of system incl. motions & interactions of of all components we measure the changes in energy  E = E final - E initial

+  E = E final > E initial system has gained E from surroundings -  E = E final < E initial system has lost E to surroundings Relating  E to heat and work  E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings

system surroundings work heat +q +w system surroundings work heat - q - w

Endothermic system absorbs heat or heat flows into the system Exothermic system gives off heat or heat flows out of the system State Function a property of a system that is determined by specifying its condition or state (T, P, etc.) internal energy is a state function,   E depends only on E final & E initial

Enthalpy for most reactions, most of the energy exchanged is in the form of heat, that heat transfer is called enthalpy, H enthalpy is a state function like internal energy, we can only measure the change in enthalpy,  H  H = q p when the process occurs under constant pressure  H = H final - H initial = q p -  H  exothermic process +  H  endothermic process

system surroundings system surroundings  H > 0  H < 0

Enthalpies of Reaction  H rxn = H prod - H react enthalpy is an extensive property magnitude of  H depends directly on the amount of reactant C (s) + 2H 2(g)  CH 4(g)  H = kJ/mol 2C (s) + 4H 2(g)  2CH 4(g)  H = kJ/2mol

enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn CH 4(g)  C (s) + 2H 2(g)  H = kJ/mol C (s) + 2H 2(g)  CH 4(g)  H = kJ/mol enthalpy change for a reaction depends on the state of the reactants and products C (g) + 2H 2(g)  CH 4(g)  H = kJ/mol 2H 2(g) + O 2(g)  2H 2 O (g)  H = kJ/mol 2H 2(g) + O 2(g)  2H 2 O (l)  H = kJ/mol

H 2 O (g) H 2 O (l) Enthalpy 44 kJ kJ kJ  H = H final - H initial + -

Practice Ex. 5.2: Hydrogen peroxide can decompose to water and oxygen. Calculate the value of q when 5.00 g of H 2 O 2(l) decomposes at constant pressure. 2H 2 O 2(l)  2H 2 O (l) + O 2(g)  H = -196 kJ 5.00 g H 2 O 2(l) x 1 mol = mol H 2 O 2(l) 34.0 g H 2 O 2(l) mol H 2 O 2(l) x -196 kJ H 2 O 2(l) = kJ 2 mol

Calorimetry experimental determination of  H using heat flow heat capacity measures the energy absorbed using temperature change the amount of heat required to raise its temp. by 1 K molar heat capacity -- heat capacity of 1 mol of substance

specific heat heat energy required to raise some mass of a substance to some different temp. specific heat = quantity of heat trans. (g substance) (temp. change) = q. m  T S.H. = joule g K q = (S.H.) (g substance) (  T) remember: this is change in temp.

Practice Ex. 5.3: Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temp. increases by 12.0  C if the specific heat of the rocks is 0.82 J/gK. S.H. x g x  T = joules What unit should be in the solution? joules -- quantity of heat 0.82 J x 50.0 x 10 3 g x 12.0 K = 4.9 x 10 5 J g K

Constant-Pressure Calorimetry  H = q p at constant pressure as in coffee cup calorimeter heat gained by solution = q soln  q soln = (S.H. soln )(g soln )(  T) heat gained by solution must that which is given off by reaction  q rxn = - q soln = - (S.H. soln )(g soln )(  T) must be opposite in sign if  T is positive then q rxn is exothermic

Practice Ex. 5.4: When 50.0 mL of M AgNO 3 and 50.0 mL of M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from  C to  C. Calculate  H for this reaction, assuming that the combined solution has a mass of g and a S.H. = 4.18 J/g  C. AgNO 3(aq) + HCl (aq)  AgCl (s) + HNO 3(aq) q soln = 4.18 J x g soln x 0.81  C = 3.39 x 10 2 J g  C q rxn = - q soln = x 10 2 J = - 68,000 J or mol - 68 kJ/mol

rxn soln q insulating cup

Bomb Calorimetry (Constant-Volume) bomb calorimeter has a pre-determined heat capacity sample is combusted in the calorimeter and  T is used to determine the heat change of the reaction q rxn = - C calorimeter x  T because rxn is exothermic heat capacity of calorimeter

rxn water insulation thermometer

Practice Ex. 5.5: A g sample of lactic acid, HC 3 H 5 O 3, is burned in a calorimeter with C = kJ/  C. Temp. increases from  C to  C. Calculate heat of combustion per gram and per mole.  T =  C q rxn = - (4.812 kJ/  C) (1.85  C) = kJ per g lactic acid kJ = kJ/g g kJ x 90.1 g = kJ/mol 1 g 1 mol

Hess’s Law rxns in one step or multiple steps are additive because they are state functions eg. CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)  H = kJ CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)  H = kJ 2H 2 O (g)  2H 2 O (l)  H = - 88 kJ

Practice Ex. 5.6: Calculate  H for the conversion of graphite to diamond: C graphite  C diamond C graphite + O 2(g)  CO 2(g)  H = kJ C diamond + O 2(g)  CO 2(g)  H = kJ C graphite + O 2(g)  CO 2(g)  H = kJ CO 2(g)  C diamond + O 2(g)  H = kJ C graphite  C diamond  H = kJ

Enthalpies of Formation enthalpies are tabulated for many processes –vaporization, fusion, formation, etc. enthalpy of formation describes the change in heat when a compound is formed from its constituent elements,  H f standard enthalpy of formation,  H f o, are values for a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm

For elemental forms: eg. C (s) graphite, Ag (s), H 2(g), O 2(g), etc.  H f o, for any element is = 0 used for calculation of enthalpies of reaction,  H rxn  H rxn =  H f o prod -  H f o react

Practice Ex. 5.9: Given this standard enthalpy of reaction, use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO (s)( CuO (a) + H 2(g)  Cu (s) + H 2 O (l)  H o = kJ  H rxn =  H f o prod -  H f o react kJ = [(0) + (-285.8)] - [(CuO) + (0)]  f o CuO = kJ/mol