1

Thermochemistry

2

ThermochemistryIn every process, mass, charge,

and energy are conserved

2H2 + O2 2H2O + energy

3

Energy

Focus on two forms of energy

1. Thermal Energy: the energy of

a system associated with the

kinetic energy of the molecules.

Temperature in Kelvin measures the

average kinetic energy of a system.

4

Thermal Energy Heat

Heat: the transfer of thermal energy

between 2 objects at different temps.

Kinetic energy is transferred via

molecular collisions.

hot cold

Chemical Potential Energy

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2. Chemical Potential Energy: the

energy stored within a substance

due to the atoms, bonds, IM forces.

Which has greater potential energy,

one gram of solid water at 0oC or or

one gram of liquid water at 0oC? Why?

6

Exothermic Process

Gives off heat (transfer of

thermal energy to surroundings)

Where does the energy come from?

higher chem PE

lower chem PE

2H2 + O2 2H2O + energy

heat to

surroundings

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absorbs heat (transfer of

thermal energy from

surroundings to system)

energy + 2HgO 2Hg + O2

higher energylower energy

Endothermic Process

Where is the energy stored?

http://www.youtube.com/watch?v=_Y1alDuXm6A&feature=youtube_gdata_player

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Potential Energy DiagramA + B C + D

pote

ntial en

erg

y

pote

ntial en

erg

yreaction coordinate reaction coordinate

Ea

EaDH<0 DH>0

exothermic

reaction

endothermic

reaction

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1st Law of

ThermodynamicsTwo ways to change the

Energy (E) of a system.

1. System can exchange heat (q)

with surroundings:

+q if system absorbs heat

(endothermic)

-q if system emits heat

(exothermic)

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1st Law of

Thermodynamics

2. System can exchange work (w)

with surroundings

+w if work done on the system

-w if system does work

(Sign of both q and w is positive

if the system gains energy.)

PV Work

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We will only consider work

associated with a gas changing

volume at constant pressure.

w = -PDV

(note sign)

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1st Law of

Thermodynamics

DE = q + w(D is always final – initial)

DE = q –PDV

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DE = q - PDV

Why does the nozzle of a CO2fire extinguisher get cold?

DE = q – PDV = 0 – P(Vf – Vi)

Energy of system drops since the

CO2 gas is expanding against

atmospheric pressure.

Both E & T decrease.

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Chemical Reactions

and the 1st Law

DE = q + w

DE = DH - PDV

Enthalpy (H) is heat flow at constant

pressure (normal lab processes)

Chemical reactions that produce or

use up gases will do PDV work.

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DH and the 1st Law

DE = DH – PDV

= DH - RTDngas

Note: if Dngas = 0, no work &DH = DE

Since PV = nRT

PDV = RTDngas (R = 8.314 J/mol K)

where Dngas =

mol gasproduct – mole gasreactant

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DH and the 1st Law

DE = DH - PDV

What is the change in internal

energy for the following reaction

at 1 atm & 25oC?

2CO(g) + O2(g) 2CO2(g)

DH = -566 kJ

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2CO(g) + O2(g) 2CO2(g)

DH = -566 kJ

2-3 = -1molDE = DH – RTDn

= [-566 x 103J] –

[(8.314J/mol K)(298K)(–1)]

= -564 x 103J = -564 kJ

(per 2 moles of CO)

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Types of Heat Changes

1. Heating/cooling: exchange of

thermal energy

2. Phase change: exchange of

chem PE & thermal energy

3. Chem reactions: exchange of

chem PE & thermal energy

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1. Heating & Cooling

Specific heat: Amt. of heat to

raise 1 g of a substance by 1oC

s =q

m DT

(m = mass in g)

(DT = Tf – Ti) in oC or K

e.g. H2O s = 4.184 J/goC

Fe s = 0.444 J/goC

What are the units of s ?

or 1 K

why

different?

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Specific Heat (s)

s =q

m DT

rearranges to:

q = m s DT = C DT

heat capacity (C) = m s

(C has units of J/oC)

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Specific Heat (s)

466 g of water is heated from

8.50 to 74.60 oC.

Calculate the heat absorbed.

q = m s DT

q = 466g x 4.184J/goC x 66.10oC

= 1.29 x 105 J

Specific Heat: Try It

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How many grams of water

can be heated by 52.3oC

using 978 J of heat?

H2O: s = 4.184 J/goC

2. Phase Change (Change of State)

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For a substance at its melting or

boiling point, T does not change

during the phase change.

What does the added heat do

on the molecular level?

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Heat of Fusion

6.01 kJ + H2O(s) H2O(l)

Ice melts at 0OC

Chem. PE of system increases.

How much water does

6.01 kJ of heat refer to?

What is the “heat of solidification”

for water?

Heat of Vaporization

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DHvap(H2O) = 40.7 kJ/mol

What is the heat change when

218 g water condenses?

Does the chem. PE increase or

decrease?

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3. Chemical Reactions

Methane burns

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l) + 890.4 kJ

DH = -890.4 kJ/molrxn

(exothermic)

Chem PE decreases -

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DH: Rules

Coefficients are always moles.

If reaction reversed, DH sign

changes.

If coefficients in balanced

equation x n, DH x n.

Equation must specify states.

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Give it a Shot !!

Calculate heat evolved when

266 g of white phosphorus burns.

P4(s) + 5O2(g) P4O10(s)

DH = -3013 kJ/molrxn

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Constant Pressure

Calorimetry

Calorimetry: the measurement

of heat changes.

qP = DH (constant pressure,

lab conditions)

In an isolated system, the sum of

the heat changes equals zero.

Sq = 0

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Constant P Calorimetry

Coffee-cup

calorimeter

For a process or reaction in solution,

heat given off by the reaction warms

the water and the cup.

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qsoln is the heat gained by the

water/solution, assumed to have

thermal properties like water.

qcup is the heat gained by

calorimeter cup. If it is a perfect

insulator, qcup = 0

qrxn is the heat given off by the

process/reaction

qsoln + qcup + qrxn = 0

Calorimetry: Try It

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85 g of iron is heated to 143oC and

put in a calorimeter (Ccup = 2.1J/oC)

containing 120 g water at 24.2oC.

The iron + water in the cup reaches

a maximum T= 32.5oC.

What is specific heat of iron?

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Heat of Neutralization

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Try It !!!

100. mL of 0.500 M HCl is mixed

with 100. mL 0.500 M NaOH, both

at 22.50oC, in a calorimeter with

Ccup = 335 J/oC. Final T = 24.90 oC.

Calculate the reaction heat and the

molar heat of neutralization.

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Different “Heats” (T.6.2)

Heat of Neutralization

Heat of Solution

Heat of Fusion

Heat of Vaporization

Heat of Reaction

Heat of Formation

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Standard

Enthalpies

DH = Hprod - Hreact

Since absolute value of H can’t

be measured (only changes),

need arbitrary reference point.

(sea level analogy)

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Std. H of Formation

DHfo 1 mol of cmpd formed

from its elements

1 atm, 25oC

DHfo of most stable form of an

element is zero.

DHfo are listed in App. 3 in text.

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Std. H of Reaction

DHrxno = [c DHf

o(C) + d DHfo(D)]

– [a DHfo(A) + b DHf

o(B)]

Reaction: aA + bB cC + dD

Thus need to know DHfo

DHrxn = Hprod - Hreact

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Determining DHfo

1. Direct Method: synthesize a

compound from its elements.

Cgraphite + O2(g) CO2(g)

DHrxno = -393.5 kJ

Since DHfo(C) and DHf

o(O2) = 0,

DHfo(CO2) = -393.5kJ

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Determining DHfo

2. Indirect Method:

(Hess’s Law)

DH is same whether

reaction takes place

in one step or many.

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Hess’s Law

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Hess’s Law

(for simplicity, I’ll ignore states !!)

C + 2H2 CH4 DHfo(CH4)= ?

C + O2 CO2 -393.5

2H2 + O2 2H2O -571.6

CH4 + 2O2 CO2 + 2H2O -890.4

DHrxn (kJ)

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Hess’s Law cont.

C + O2 CO2 -393.5

2H2 + O2 2H2O -571.6

CO2 + 2H2O CH4 + 2O2 +890.4

DHrxn (kJ)

C + 2H2 CH4 -74.7

reversed reaction

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Hess’s Law: Try It!

Calculate DHfo of acetylene:

2C(graph) + H2(g) C2H2(g)

Given: DHrxno kJ

C(graph) + O2(g) CO2(g) -393.5

H2(g) + O2(g) H2O(l) -285.8

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

-2598.8

1

2

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Heat of Solution: DHsoln

heat change upon dissolution

e.g. NaCl(s) Na+(aq) + Cl-(aq)H2O

DHsoln + endothermic

- exothermic

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Solution: 2 Step Process

Solid

crystal

Ions in

gas state

SolutionHeat of

Solution = Hsoln

U + Hhydr = Hsoln

+ -

Heat of Solution: DHsoln

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Na+Cl- solid

lattice

energy (+)

hydration(-)

heat of

solution

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Solution: 2 Step Process

NaCl(s) Na+(g) + Cl-(g) 788

Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) -784

NaCl(s) Na+(aq) + Cl- (aq) +4

kJ

Thus DHsoln = 4 kJ

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Heat of Solution: DHsoln

How about for DHsoln

ammonium chloride?

Compare relative

values of U and Hhydr