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ThermochemistryThermochemistry-Energy of Chemical Reactions

-OR--The study of heat changes that occur during chemical reactions

and physical changes of state

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Heat Transfer and Heat Transfer and Changes of StateChanges of State

Heat Transfer and Heat Transfer and Changes of StateChanges of State

Changes of state involve energyChanges of state involve energy

Ice Water333 J/g333 J/g

(Heat of FusionHeat of Fusion)

Water Vapor 2260 J/g2260 J/g

(Heat of vaporization)(Heat of vaporization)

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Heating/Cooling Curve for WaterHeating/Cooling Curve for Water

11 22

33 44

Heat waterHeat water

Evaporate waterEvaporate water

Melt iceMelt ice

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Energy and ChemistryEnergy and Chemistry

ENERGYENERGY is the capacity to do is the capacity to do workwork or transfer or transfer heatheat..

can be: light, electrical, kinetic, potential, chemicalcan be: light, electrical, kinetic, potential, chemical

HEATHEAT (represented by q)(represented by q) is the form of energy that flows is the form of energy that flows between 2 samples because of a difference in temperature – between 2 samples because of a difference in temperature – will always go from a warmer object to a cooler one.will always go from a warmer object to a cooler one.

WORKWORK is the form of energy that results in a macroscopic is the form of energy that results in a macroscopic displacement of matter such as gas expansion or motion of displacement of matter such as gas expansion or motion of an object an object (force x distance)(force x distance)

LAW OF CONSERVATION OF ENERGYLAW OF CONSERVATION OF ENERGY energy is energy is neither created nor destroyedneither created nor destroyed

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Energy and ChemistryEnergy and Chemistry

CALORIECALORIE quantity of heat that raises the quantity of heat that raises the temperature of 1g of pure water 1temperature of 1g of pure water 1°C°C

1 Cal or 1 kcal = 1000 cal 1 Cal or 1 kcal = 1000 cal

JOULE JOULE SI unit of heat and energy SI unit of heat and energy

(1kcal = 4186J, 1 cal = 4.186 J and 1J = 0.239 cal)(1kcal = 4186J, 1 cal = 4.186 J and 1J = 0.239 cal)

HEAT CAPACITYHEAT CAPACITY the amount of heat it takes to the amount of heat it takes to change an object’s temperature by exactly 1°Cchange an object’s temperature by exactly 1°C

Depends on an object’s massEx. A cup of water has a greater heat capacity than a drop of

water.

CHEMICAL ENERGYCHEMICAL ENERGY

Chemical Potential Energy:Energy stored in chemicals because of their compositionsChemical bonds are a source of energy

• BOND BREAKING - requires energy• BOND MAKING - releases energy

In a chemical reaction:• if more energy is released in forming bonds than is used in breaking bonds then . . . reaction is EXOTHERMICEXOTHERMIC

• if more energy is used in breaking bonds than is released in

forming bonds then . . . reaction is ENDOTHERMIC

Energy is released as HEAT, LIGHT, SOUND, WORK

Energy can be provided by - LIGHT - photochemistry- WORK - electrochemistry- COOLING of surroundings

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Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat CapacityA.K.A. Specific Heat: (represented by C) the amount

of heat it takes to raise temperature of 1g of a substance 1°C (measured in J/(g x °C)1°C (measured in J/(g x °C)

A difference in temperature leads to energy transfer.

Specific heat capacity =

heat lost or gained by substance (J)

(mass, g) (T change, K)

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Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity

SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K)

HH22OO 4.1844.184

AlAl 0.9020.902

glassglass 0.840.84

AluminumAluminum

WaterWater

Specific Heat

• q = mCT

• q = heat (j)

• m = mass (g)

• C = specific heat (j/gC)T = change in temperature = Tf-Ti (C)

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Specific Heat Capacity - an exampleSpecific Heat Capacity - an exampleSpecific Heat Capacity - an exampleSpecific Heat Capacity - an example

If 25.0 g of Al cool from 310 K to 37 K, how many joules of heat energy are lost by the Al?

where T = Tfinal - Tinitial = 37 - 310 = -273 Kq = (0.902 J/g•K)(25.0 g)(-273 K)q = -6160 J

negative sign of q heat is “lost by” or transferred from Alnegative sign of q heat is “lost by” or transferred from Al

Specific heat capacity =

heat lost or gained by substance (J)

(mass, g)(T change, K)

0.902 J/g.K

=

heat gain/lost = q = (specific heat)(mass)(T)heat gain/lost = q = (specific heat)(mass)(T)

Specific Heat

• A 10.0 g sample of iron changes temperature from 25.0C to 50.4 C while releasing 114 joules of heat. Calculate the specific heat of iron.

Example

• q = mcT

• 114 = (10.0)c(50.4-25)

• c = 0.449 j/g C

Another example

• If the temperature of 34.4 g of ethanol increases from 25.0 C to 78.8 C how much heat will be absorbed if the specific heat of the ethanol is 2.44 j/g C

Another example

• q = mcT

• q = (34.4)(2.44)(78.8-25)

• q = 4.53 x 10 -3 j

Yet another example

• 4.50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25.0 C & the specific heat of the gold is 0.129j/g C

Yet another example

• q = mcT

• 276 = (4.50)(0.129)TT = 475C

T = Tf-Ti

• 475 = Tf-25

• Tf = 500 C

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H = HH = Hfinalfinal - H - HinitialinitialH = HH = Hfinalfinal - H - Hinitialinitial

If Hfinal > Hinitial then H is positive

Process is ENDOTHERMIC

If Hfinal > Hinitial then H is positive

Process is ENDOTHERMIC

If Hfinal < Hinitial then H is negative

Process is EXOTHERMIC

If Hfinal < Hinitial then H is negative

Process is EXOTHERMIC

ENTHALPYENTHALPYENTHALPYENTHALPYFor systems at constant pressure, the heat content is the

same as the Enthalpy, or H, of the system.

Heat changes for reactions carried out at constant pressure are the same as changes in enthalpy (∆H)

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Standard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesMost H values are labeled Ho

• P = 1 atmosphere ( = 760 torr = 101.3 kPa)

• Concentration = 1 mol/L

• T = usually 25 oC

• with all species in standard states

e.g., C = graphite and O2 = gas

o means measured under standard conditions

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Endo- and ExothermicEndo- and ExothermicEndo- and ExothermicEndo- and Exothermic

SurroundingsSurroundings

HeatSystem

ENDOTHERMICENDOTHERMICENDOTHERMICENDOTHERMIC

HeatHeat

SurroundingsSurroundings

System

EXOTHERMICEXOTHERMIC

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But the reverse reaction, the decomposition of water :

H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)Endothermic reaction — heat is a “reactant”, H = +242 kJ. This does not occur spontaneously.

Consider the combustion of H2 to form water . .

H2(g) + 1/2 O2(g) ---> H2O(g) 242 kJExothermic reaction — heat is a “product”. H = -242 kJ. This is spontaneous and proceeds readily once initiated.

USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY

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Making H2 from liquidliquid H2O involves two steps.

H2O(liq) + 44 kJ H2O(g)

H2O(g) + 242 kJ H2(g) + 1/2 O2(g)---------------------------------------------------H2O(liq) + 286 kJ H2(g) + 1/2 O2(g)

This is an example of HESS’S LAW —

If a reaction is the sum of 2 or more others, the net H is the sum of the H’s of the other rxns.

If a reaction is the sum of 2 or more others, the net H is the sum of the H’s of the other rxns.

Heat and Changes of State• Heat of combustion (∆H)= the heat of reaction for the complete

burning of one mole of a substance

• Molar heat of fusion (∆Hfus)= the heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature

• Molar heat of solidification (∆Hsolid)= heat lost when one mole of a liquid freezes to a solid at a constant temperature (equal to the negative heat of fusion)

• Molar heat of vaporization (∆Hvap)= the heat absorbed by one mole of a substance in vaporizing from liquid to a gas

• Molar heat of condensation (∆Hcond)= heat released by one mole of a vapor as it condenses

Example(Heat of combustion)

• The standard heat of combustion (∆H°rxn) for glucose (C6H12O6) is -2808 kJ/mol. If you eat and burn 70.g of glucose in one day, how much energy are you getting from the glucose?– Step one: convert g of glucose to moles

• 70. g glucose 1 mol = 0.28 mol glucose

246 g

– Step two: Use (∆H°rxn) to find amount of kJ gained• 0.28 mol glucose x 2808 kJ = 790 kJ gained (+ b/c gained not lost)

1 1 mol

Example(Other ∆H’s)• You have a sample of H2O with a mass 23.0 grams at a

temperature of -46.0 °C. ΔHfus= 6.01 kJ/mol ΔHvap= 40.7 kJ/mol• How many kilojoules of heat energy are necessary to carry

out each step? • Heat the ice to 0 °C?

– Which equation do you need?• q = mCΔT• q = (23g)(4.184 J/g °C)(0 °C – (-46) °C)• q = 4427 J = 4.43 kJ

• Boil the water?– Which equation do you need?

• ΔH = mol x ΔHvap• ΔH = 23 g x 1 mol x 40.7kJ

18.02g 1 mol• ΔH = 52 kJ