Constant Pressure Calorimetry: Another Example

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© 2009 Brooks/Cole - Cengage

A piece of chromium metal weighing 24.26 g is A piece of chromium metal weighing 24.26 g is heated in boiling water to a temperature of 98.3heated in boiling water to a temperature of 98.3°C°C and then dropped into a coffee cup calorimeter and then dropped into a coffee cup calorimeter containing 82.3g of water at 23.3containing 82.3g of water at 23.3°C. When °C. When thermal equilibrium is reached, the final thermal equilibrium is reached, the final temperature is 25.6°C. Calculate the Ctemperature is 25.6°C. Calculate the Cpp of of

chromium.chromium.

Constant Pressure Calorimetry: Constant Pressure Calorimetry: Another ExampleAnother Example

Constant Pressure Calorimetry: Constant Pressure Calorimetry: Another ExampleAnother Example

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• Technique can be used to obtain the heat content Technique can be used to obtain the heat content of combustion of compoundsof combustion of compounds

• Used in the food, fuel and pharmaceutical Used in the food, fuel and pharmaceutical industries to know how much energy would be industries to know how much energy would be released by completely consuming the compoundreleased by completely consuming the compound

• Uses a Uses a BOMB CalorimeterBOMB Calorimeter

Constant Volume Calorimetry: It’s The Constant Volume Calorimetry: It’s The Bomb!Bomb!

Constant Volume Calorimetry: It’s The Constant Volume Calorimetry: It’s The Bomb!Bomb!

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Measuring Heats of Reaction

CALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

•Place sample of known mass inside the bomb

•Place oxygen in the sample chamber and immerse bomb into water

•Ignite the bomb and measure temperature of water

•Since the volume doesn’t change, no P-V work is done, so the qr is a measurement of the ΔU

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CalorimetCalorimetryry

Some heat from reaction warms waterqwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

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Calculate energy of combustion (∆U) of Calculate energy of combustion (∆U) of octane. octane. 2C2C88HH1818 + 25O + 25O22 -->--> 16CO16CO22 + 18H + 18H22OO

•• Burn 1.00 g of octaneBurn 1.00 g of octane

• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC

• Calorimeter contains 1200. g waterCalorimeter contains 1200. g water

• Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY


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Step 1Step 1 Calc. energy transferred from reaction to water. Calc. energy transferred from reaction to water.

q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J

Step 2Step 2 Calc. energy transferred from reaction to bomb. Calc. energy transferred from reaction to bomb.

q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T)

= (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J

Step 3Step 3 Total energy evolved Total energy evolved

41,200 J + 6860 J = 48,060 J41,200 J + 6860 J = 48,060 J

Energy of combustion (∆U) of 1.00 g of octane Energy of combustion (∆U) of 1.00 g of octane

= - 48.1 kJ= - 48.1 kJ



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The First Law of Thermodynamics

• Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat

• However, these changes rarely occur singly, so we can describe the change in internal energy as:

U = q + w (The 1st Law)

• The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

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The First Law: Put Another Way

11stst Law of Thermodynamics Law of Thermodynamics

A A systemsystem can store energy. A change in the can store energy. A change in the energy of a system means that there must be a energy of a system means that there must be a change in the heat or the work done change in the heat or the work done BYBY or or TOTO

the system.the system.

OROR

The Total Energy of the Universe is ConstantThe Total Energy of the Universe is Constant

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The First Law

• Internal energy is an example of a state function

• A state function is a property that only depends on the current state of the system and is independent of how that state was reached

• Pressure, Volume, Temperature and density are all examples of state functions

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State Functions and Things that Aren’t

• Work and heat ARE NOT state functions

• The amount of work done depends on how the change was brought about

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Change in Internal Energy (Implications of a State Function)

• It doesn’t matter what path we take to get to the final point, the change in internal energy is only dependent on where we started and where we finished

• Let’s think about this on a molecular level…– If we expand an ideal gas isothermally, the molecules will have the same

kinetic energy and will move at the same speed

– Despite the fact that the volume has increased, the potential energy of the system remains the same because there are no forces between molecules (KMT)

– Since neither the kinetic nor potential energy has changed, the change in internal energy is…

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Zero!

U = 0 for the isothermal expansion of an ideal gas

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Enthalpy

• In a constant volume system in which no work is done (neither expansion nor non-expansion), we can rearrange the first law to:

U = q + w (but w=0)

U = q• Most systems are constant pressure systems which can expand and

contract• When a chemical reaction takes place in such a system, if gas is

evolved, it has to push against the atmosphere in order to leave the liquid or solid phase

– Just because there’s no piston, it doesn’t mean that no work is done!

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Enthalpy

• Let’s look at an example:

• If we supply 100J of heat to a system at constant pressure and it does 20J of work during expansion, the U of the system is +80J (w=-20J)

– We can’t lose energy like this

• Enthalpy, H, is a state function that we use to track energy changes at constant pressure

H=U + PV

• The change in enthalpy of a system (H) is equal to the heat released or absorbed at constant pressure

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Enthalpy

• Another way to define enthalpy is at constant pressure:

H = q

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Enthalpy and Chemical Reactions

• Enthalpy is a tricky thing to grasp, but we can look at it this way:

Enthalpy is the macroscopic energy change (in the form of heat) that accompanies changes at the atomic level (bond formation or breaking)

• Enthalpy has the same sign convention as work, q and U– If energy is released as heat during a chemical reaction the enthalpy has a ‘-’

sign

– If energy is absorbed as heat from the surrounding during a reaction, the enthalpy has a ‘+’ sign

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Heat Transfers at Constant Pressure

H = U + PV

• The change in the enthalpy of a system is equal to the heat released or absorbed at constant pressure

WTF?

1. In a coffee cup calorimeter (constant pressure calorimeter), the heat, q, that is released or absorbed is equal to the change in enthalpy, H

2. When we add heat to a constant pressure system, the enthalpy increases by that amount

3. H<0 for exothermic reactions

H>0 for endothermic reactions

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Consider the formation of waterConsider the formation of water

HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) -->--> H H22O(g) + O(g) + 241.8 kJ241.8 kJ

USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY

Exothermic reaction — energy is a “product” andExothermic reaction — energy is a “product” and ∆H = – 241.8 kJ∆H = – 241.8 kJ

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Energy Transfer Energy Transfer withwith Change of State Change of State

Energy Transfer Energy Transfer withwith Change of State Change of State

Changes of state involve energy Changes of state involve energy (at constant T)(at constant T)Ice + 333 J/g (heat of fusion) Ice + 333 J/g (heat of fusion) ff Liquid water Liquid water

q = (heat of fusion)(mass)q = (heat of fusion)(mass)

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Energy Transfer and Energy Transfer and Changes of StateChanges of State

Energy Transfer and Energy Transfer and Changes of StateChanges of State

Requires energy (heat).Requires energy (heat).

This is the reasonThis is the reason

a)a)you cool down after you cool down after swimming swimming

b)b) you use water to put you use water to put out a fire.out a fire.

+ energy

Liquid Liquid -->--> Vapor Vapor

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Heating/Cooling Curve for Water

Heating/Cooling Curve for Water

Note that T is Note that T is constant as ice meltsconstant as ice melts

Note that T is Note that T is constant as ice meltsconstant as ice melts

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Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

What quantity of heat is required to melt What quantity of heat is required to melt 500. g of ice and heat the water to 500. g of ice and heat the water to steam at 100 steam at 100 ooC?C?

Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State

+333 J/g+333 J/g +2260 J/g+2260 J/g

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What quantity of energy as heat is required to melt 500. g What quantity of energy as heat is required to melt 500. g of ice and heat the water to steam at 100 of ice and heat the water to steam at 100 ooC?C?

1. 1. To melt iceTo melt ice

q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 1055 J J

2.2. To raise water from 0 To raise water from 0 ooC to 100 C to 100 ooCC

q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 10q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 1055 J J

3.3. To evaporate water at 100 To evaporate water at 100 ooCC

q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 J J

4. 4. Total energy = 1.51 x 10Total energy = 1.51 x 1066 J = 1510 kJ J = 1510 kJ

Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State

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Making Making liquidliquid H H22O from HO from H22 + +

OO22 involves involves twotwo exoexothermic thermic

steps. steps.

USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY

H2 + O2 gas

Liquid H2OH2O vapor

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Hess’s LawHess’s LawHess’s LawHess’s Law

The ∆The ∆rrHHoo for a reaction that is the sum of one or for a reaction that is the sum of one or

more reactions is the sum of the ∆more reactions is the sum of the ∆rrHHoo values for values for

all of the reactionsall of the reactions

Remember: Since ∆H is a state function, it Remember: Since ∆H is a state function, it doesn’t matter how we get to where we end up, doesn’t matter how we get to where we end up, all that matters is where we started and where all that matters is where we started and where we finished.we finished.

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Making HMaking H22O from HO from H22 involves two steps. involves two steps.

HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) ff H H22O(g) + 242 kJO(g) + 242 kJ

HH22O(g) O(g) ff H H22O(liq) + 44 kJ O(liq) + 44 kJ

-----------------------------------------------------------------------

HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) ff H H22O(liq) + 286 kJO(liq) + 286 kJ

Example of Example of HESS’S LAWHESS’S LAW——

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

USING ENTHALPYUSING ENTHALPY

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Hess’s Law Hess’s Law & Energy Level & Energy Level DiagramsDiagrams

Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

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Making CO from C (graphite) and OMaking CO from C (graphite) and O22 is a single step, but is a single step, but

the CO reacts with oxygen to form COthe CO reacts with oxygen to form CO22..

1) C(graphite) + ½O1) C(graphite) + ½O22(g) (g) -->--> CO (g) ? kJ/mole-rxn CO (g) ? kJ/mole-rxn

2) CO(g) + O2) CO(g) + O22(g) (g) -->--> CO CO22(g) + -283 kJ/mole-rxn (g) + -283 kJ/mole-rxn

-----------------------------------------------------------------------

3) C (graphite) + ½O3) C (graphite) + ½O22(g) (g) -->--> CO CO22(g) -393.5 kJ/mole-rxn(g) -393.5 kJ/mole-rxn

Example of Example of HESS’S LAWHESS’S LAW——If a rxn. is the sum of 2 or more others, the net If a rxn. is the sum of 2 or more others, the net

∆H is the sum of the ∆H’s of the other rxns.∆H is the sum of the ∆H’s of the other rxns.

∆∆HHoo33= = ∆H∆Hoo

11 + ∆H + ∆Hoo22

USING ENTHALPYUSING ENTHALPY

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Using EnthalpyUsing EnthalpyUsing EnthalpyUsing Enthalpy

∆ ∆HHoo33

= = ∆H∆Hoo11 + ∆H + ∆Hoo

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-393.5 kJ/mole-rxn = ∆H-393.5 kJ/mole-rxn = ∆Hoo1 1 + (-283.0 kJ/mole-rxn)+ (-283.0 kJ/mole-rxn)

∆∆HHoo11 = -110.5 kJ/mole-rxn = -110.5 kJ/mole-rxn

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Hess’s Law Hess’s Law & Energy Level & Energy Level DiagramsDiagrams

Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Active Figure 5.16

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Hess’s Law: Another Hess’s Law: Another ExampleExample


Use Hess’s Law to calculate the enthalpy change for the formation of Use Hess’s Law to calculate the enthalpy change for the formation of CSCS22(l) from C(s) and S(s).(l) from C(s) and S(s).

The overall equation is: C(s) + 2S(s) The overall equation is: C(s) + 2S(s) -->--> CS CS22(l)(l)

C(s) + OC(s) + O22(g) (g) -->--> CO CO22(g) (g) ∆∆HHoo = -393.5 kJ/mole-rxn = -393.5 kJ/mole-rxn

S(s) + OS(s) + O22(g) (g) -->--> SO SO22(g) (g) ∆∆HHoo = -295.8 kJ/mole-rxn = -295.8 kJ/mole-rxn

CSCS22(l) + 3O(l) + 3O22(g) (g) -->--> CO CO22(g) +2SO(g) +2SO22(g) (g) ∆∆HHoo = -1103.9 kJ/mole-rxn = -1103.9 kJ/mole-rxn

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How do we start?How do we start?-Let’s set our overall reaction up:-Let’s set our overall reaction up:

C(s) + OC(s) + O22(g) (g) -->--> CO CO22(g) (g) ∆H ∆Hoo = -393.5 kJ/mole-rxn = -393.5 kJ/mole-rxnS(s) + OS(s) + O22(g) (g) -->--> SO SO22(g) ∆H(g) ∆Hoo = -295.8 kJ/mole-rxn = -295.8 kJ/mole-rxnCOCO22(g) +2SO(g) +2SO22(g) (g) -->--> CS CS22(l) + 3O(l) + 3O22(g) (g) ∆H∆Hoo = 1103.9 kJ/mole-rxn = 1103.9 kJ/mole-rxn----------------------------------------------------------------------------------------------------------------------------------------------------------------------C(s) + 2S(s) C(s) + 2S(s) -->--> CS CS22(l)(l) ∆H ∆Hoo = ? = ?

Note: We reversed reaction 3 and changed the signNote: We reversed reaction 3 and changed the signNow: We need to make sure that all of the molecules Now: We need to make sure that all of the molecules

balance and take care to multiply their ∆balance and take care to multiply their ∆rrHHoo values if values if necessarynecessary

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C(s) + OC(s) + O22(g) (g) -->--> CO CO22(g) (g) ∆H ∆Hoo = -393.5 kJ/mole-rxn = -393.5 kJ/mole-rxn

S(s) + OS(s) + O22(g) (g) -->--> SO SO22(g) (g) ∆H∆Hoo = -295.8 kJ/mole-rxn = -295.8 kJ/mole-rxn

COCO22(g) +2SO(g) +2SO22(g) (g) -->--> CS CS22(l) + 3O(l) + 3O22(g) ∆H(g) ∆Hoo = 1103.9 kJ/mole-rxn = 1103.9 kJ/mole-rxn

----------------------------------------------------------------------------------------------------------------------------------------------------------------------

C(s) + 2S(s) C(s) + 2S(s) -->--> CS CS22(l)(l) ∆H∆Hoo = ? = ?

We need 2 S(s) atoms, so let’s multiply reaction2 We need 2 S(s) atoms, so let’s multiply reaction2 by 2:by 2:

2S(s) + 2O2S(s) + 2O22(g) (g) -->--> 2SO 2SO22(g) ∆H(g) ∆Hoo = -591.6 kJ/mole-rxn = -591.6 kJ/mole-rxn

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Now let’s look at what we’ve got:Now let’s look at what we’ve got:

C(s) + OC(s) + O22(g) (g) -->--> CO CO22(g) (g) ∆H ∆Hoo11 = -393.5 kJ/mole-rxn = -393.5 kJ/mole-rxn

2S(s) + 2O2S(s) + 2O22(g) (g) -->--> 2SO 2SO22(g) ∆H(g) ∆Hoo22 = -591.6 kJ/mole-rxn = -591.6 kJ/mole-rxn

COCO22(g) +2SO(g) +2SO22(g) (g) -->--> CS CS22(l) + 3O(l) + 3O22(g) ∆H(g) ∆Hoo33 = 1103.9 kJ/mole-rxn = 1103.9 kJ/mole-rxn

----------------------------------------------------------------------------------------------------------------------------------------------------------------------

C(s) + 2S(s) C(s) + 2S(s) -->--> CS CS22(l)(l) ∆H∆Hoo = ∆H = ∆Hoo11 + ∆H + ∆Hoo

22 + ∆H + ∆Hoo33

Everything looks good, so let’s add it up:Everything looks good, so let’s add it up:∆∆HHoo

= (-393.5 kJ/mole-rxn)+(-591.6 kJ/mole-rxn)+(1103.95 kJ/mole-rxn)= (-393.5 kJ/mole-rxn)+(-591.6 kJ/mole-rxn)+(1103.95 kJ/mole-rxn)

∆∆HHoo = 118.9 kJ/mole-rxn= 118.9 kJ/mole-rxn

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Standard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesMost ∆H values are labeled Most ∆H values are labeled ∆H∆Hoo

Measured under Measured under standard conditionsstandard conditionsP = 1 bar P = 1 bar Concentration = 1 mol/LConcentration = 1 mol/LT = usually 25 T = usually 25 ooCC

with all species in standard stateswith all species in standard states

e.g., C = graphite and Oe.g., C = graphite and O22 = gas = gas

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Standard Enthalpy ValuesStandard Enthalpy Values

NIST (Nat’l Institute for Standards and NIST (Nat’l Institute for Standards and Technology) gives values ofTechnology) gives values of

∆∆HHffoo = standard molar enthalpy of = standard molar enthalpy of

formationformation

— — the enthalpy change when 1 mol of the enthalpy change when 1 mol of compound is formed from elements under compound is formed from elements under standard conditions.standard conditions.

See Appendix See Appendix

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∆∆HHffoo, standard molar , standard molar

enthalpy of enthalpy of formationformation

Equations for Standard Molar Enthalpy of Formation:Equations for Standard Molar Enthalpy of Formation:

KNOKNO33

K(s) + ½ NK(s) + ½ N22(g) + 3/2O(g) + 3/2O22(g) (g) -->--> KNO KNO33

FeClFeCl33

Fe(s) + 3/2 ClFe(s) + 3/2 Cl22(g) (g) -->--> FeCl FeCl33(s)(s)

CC1212HH2222OO1111

12C(s) + 11H12C(s) + 11H22(g) + 11/2 O(g) + 11/2 O22(g) (g) -->--> C C1212HH2222OO1111(s)(s)

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Using Standard Enthalpy Using Standard Enthalpy ValuesValues


In general, when In general, when ALLALL enthalpies enthalpies

of formation are known, of formation are known,

∆∆HHoo = = ∆ ∆ffHHoo (products) - (products) - ∆ ∆ffHHoo

(reactants)(reactants)∆∆HHoo = = ∆ ∆ffHHoo (products) - (products) - ∆ ∆ffHHoo

(reactants)(reactants)

Remember that ∆ always = final – initial

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Calculate the heat of combustion of Calculate the heat of combustion of

methanol, i.e., ∆Hmethanol, i.e., ∆Hoo for for

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) (g) -->--> COCO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoo = = ∆H ∆Hffoo

(prod) - (prod) - ∆H ∆Hffoo

(react)(react)

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∆∆HHrroo = ∆H = ∆Hff

oo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O) O)

- {3/2 ∆H- {3/2 ∆Hffoo

(O(O22) + ∆H) + ∆Hffoo

(CH(CH33OH)} OH)}

= (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ)

- {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}

∆∆HHrroo = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) (g) -->--> COCO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆rrHHoo = = ∆H ∆Hffoo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

41




Calculate the heat of reaction of nitric oxide, i.e., ∆HCalculate the heat of reaction of nitric oxide, i.e., ∆Hrroo for for

4NH4NH33 (g) + 5 O (g) + 5 O22(g) (g) -->--> 4NO(g) + 6 H4NO(g) + 6 H22O(g)O(g)

∆∆HHoo = = ∆H ∆Hffoo

(prod) - (prod) - ∆H ∆Hffoo

(react)(react)

∆∆HHoo (prod) = 4(∆H(prod) = 4(∆Hff

o o NO) + 6(∆HNO) + 6(∆Hffo o HH22O)O)

= 4(90.29) + 6(-241.830) = 4(90.29) + 6(-241.830)

= -1089 kJ/mole-rxn= -1089 kJ/mole-rxn

∆∆HHffoo

(react) = 4(∆H(react) = 4(∆Hffo o NHNH33) + 5(∆H) + 5(∆Hff

o o OO22))

= 4(-45.90) + 5(0)= 4(-45.90) + 5(0)

= -183.6 kJ/mole-rxn= -183.6 kJ/mole-rxn

∆∆HHrroo = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

= -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = = -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = -906.2 kJ/mole-rxn-906.2 kJ/mole-rxn

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Constant Pressure Calorimetry: Another Example

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