Chapter 5:Chapter 5:

ThermochemistryThermochemistry

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 mv2 • 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 mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J•1 cal = 4.184 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

2H2(g) + O2(g)

2H2O(l)+ energy

(system)

energy(as heat or work)

no exchg of matterwith 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 = Efinal - Einitial

• + E = Efinal > Einitial system has gained E from surroundings

• - E = Efinal < Einitial 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

workhe

at

+q +w

system

surroundingsw

orkhea

t

- q - w

Endothermic•system absorbs heat or heat flows into the

systemExothermic

•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 Efinal & Einitial

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 = qp when the process occurs under constant pressureH = Hfinal - Hinitial = qp

• - H exothermic process • +H endothermic process

system

surroundings

system

surroundings

H > 0

H < 0

Enthalpies of Reaction Hrxn = Hprod - Hreact

• enthalpy is an extensive property

•magnitude of H depends directly on the amount of reactant

C(s) + 2H2(g) CH4(g) H = -74.8

kJ/mol

2C(s) + 4H2(g) 2CH4(g) H = -149.6

kJ/2mol

• enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn

CH4(g) C(s) + 2H2(g) H = +74.8 kJ/mol

C(s) + 2H2(g) CH4(g) H = - 74.8 kJ/mol

• enthalpy change for a reaction depends on the state of the reactants and products

C(g) + 2H2(g) CH4(g) H = -793.2 kJ/mol

2H2(g) + O2(g) 2H2O(g) H = -486.6 kJ/mol

2H2(g)+ O2(g) 2H2O(l) H = -571.7 kJ/mol

H2O(g)

H2O(l) Ent

halp

y 44 kJ

-285.8 kJ

-241.8 kJ

H = Hfinal - Hinitial

+

-

Practice Ex. 5.2:• Hydrogen peroxide can decompose to water and oxygen .

Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.

2H2O2(l) 2H2O(l) + O2(g) H = -196 kJ

5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l)

34.0 g H2O2(l)

0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ

2 mol

Calorimetry•experimental determination of H using

heat flowheat 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 103 g x 12.0 K = 4.9 x 105 Jg K

Constant-Pressure Calorimetry• H = qp at constant pressure as in coffee cup

calorimeter

•heat gained by solution = qsoln

qsoln = (S.H.soln)(gsoln)(T)•heat gained by solution must that which is

given off by reaction

qrxn = - qsoln = - (S.H.soln)(gsoln)(T)

must be opposite in signmust be opposite in sign

if T is positive then qrxn is exothermic

Practice Ex. 5.4:• When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M

HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30C to 23.11C. Calculate H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g C.

AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)

qsoln = 4.18 J x 100.0 g soln x 0.81C = 3.39 x 102 J g C

qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 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

• qrxn = - Ccalorimeter x T

because rxn is exothermic

heat capacity of calorimeter

rxn

water

insulation

thermometer

Practice Ex. 5.5:• A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a

calorimeter with C = 4.812 kJ/C. Temp. increases from 23.10C to 24.95C. Calculate heat of combustion per gram and per mole. T = +1.85C

qrxn = - (4.812 kJ/C) (1.85C) = - 8.90 kJ per 0.5865 g lactic acid -8.90 kJ = - 15.2 kJ/g 0.5865 g

- 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol

Hess’s Law• rxns in one step or multiple steps are additive

because they are state functions

eg.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = - 802 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = - 890 kJ

2H2O(g) 2H2O(l) H = - 88 kJ

Practice Ex. 5.6:• Calculate H for the conversion of graphite to

diamond:

Cgraphite Cdiamond

Cgraphite + O2(g) CO2(g) H = -393.5 kJ

Cdiamond + O2(g) CO2(g) H = -395.4 kJ

Cgraphite + O2(g) CO2(g) H = -393.5 kJ

CO2(g) Cdiamond + O2(g) H = 395.4 kJ

Cgraphite Cdiamond H = + 1.9 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, Hf

• standard enthalpy of formation, Hfo, 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) , H2(g) , O2(g) , etc.

Hfo, for any element is = 0

• used for calculation of enthalpies of reaction, Hrxn

• Hrxn = Hfo prod - Hf

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) + H2(g) Cu(s) + H2O(l) Ho = -130.6 kJ

Hrxn = H f o

prod - H f o

react

-130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]

fo CuO = -155.2 kJ/mol