Lecture 6. Thermochemistry (Part 1)

THERMOCHEMISTRY: Part 1. Energy Flow and Chemical Change

GENERAL CHEMISTRY

LECTURE 6

6.1 Define Thermochemistry

6.2 Define Energy, Heat and Work

6.3 Discuss the Principle of Heat Flow

6.4 Define Enthalpy

6.5 Calorimtery: heat capacity and specific heat

Scope

6.6 Stoichiometry of Thermal Equations

6.7 Hess’s Law

6.8 Standard Heats of Reaction (DHo rxn)

6.9 Bond Energies

Thermodynamics– the study of the energy transfers accompanying physical and chemical changes

Thermochemistry- part of thermodynamics which deals with heat involved in physical and chemical changes

What is Thermochemistry?

What is Energy?

• Energy is the capacity to do work or to produce heat.

• Law of conservation of energy

– Energy can be converted from one form to another but can be neither created nor destroyed

– The total energy content of the universe is constant.

• Potential energy – energy due to position or composition.

• Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.

Internal Energy

• Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

• The total amount of energy contained in a system

Components of Internal Energy

• Thermal Energy

– Kinetic energy associated with random molecular motion.

– In general proportional to temperature.

• Chemical Energy

– Potential energy associated with chemical bonds and intermolecular forces of attraction.

Internal Energy

• System – part of the universe on which we wish to focus attention.

• Surroundings – include everything else in the universe.

• Systems + Surroundings = Universe

Types of Systems

Energy Transfers

• Heat, q

- involves the transfer of energy between two objects due to a temperature difference.

• Work, w

– force acting over a distance.

• Change in Internal Energy:

• Energy is a state function; work and heat are not.

DE = q + w

– It only depends on the difference between the final and initial state.

– Independent of path

What is a State Function?

First Law of Thermodynamics

• Law of conservation of energy is often called the first law of thermodynamics.

• Energy gained by the surroundings must be equal to the energy lost by the system.

DEuniverse = DEsystem + DEsurrounding = 0

Energy Transfers

• Heat - involves the transfer of energy between two objects due to a temperature difference.

• Work – force acting over a distance.

DE = q + w

Heat

• Heat “flows” from hotter to colder.

• Sign reflects the system’s point of view.

• Endothermic Process:

– Heat flow is into a system.

– q is positive

• Exothermic Process: – Heat flows out of the system

– q is negative

Heat

heat heat

Exothermic, q < 0 Endothermic, q > 0

A system transferring energy as heat only

A. Heat flow out from a system B. Heat flow into a system

Concept Check

1. Is the freezing of water an endothermic or exothermic process? Explain.

Concept Check

2. Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.

– Your hand gets cold when you touch ice.

– The ice gets warmer when you touch it.

– Water boils in a kettle being heated on a stove.

– Water vapor condenses on a cold pipe.

– Ice cream melts.

Work

FIGURE 7-7 Illustrating work (expansion) during the chemical reaction

2 KClO3(s) 2 KCl(s) + 3 O2(g)

In addition to heat effects chemical reactions may also do work.

Gas formed pushes against the atmosphere.

The volume changes.

Pressure-volume work.

Work

• Work = P × A × Δh = PΔV

– P is pressure.

– A is area.

– Δh is the piston moving a distance.

– ΔV is the change in volume.

A system transferring energy as work only

1. Work done by system

2. Work done on a system

w < 0, DE < 0

w > 0, DE > 0

Exercise

Which of the following performs more work?

a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.

b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

c) They perform the same amount of work.

Summary

Exercise

Calculate the change in internal energy for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

ΔE = q + w

ΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ

Change in Enthalpy, ΔH

• Is the heat lost or gained during a chemical or physical change that occurs at constant pressure, qp.

• ΔH = q at constant pressure

• State function

• ΔHrxn = Hproducts – Hreactants

Heats of Reaction: DE and DH

Reactants → Products

Ei Ef

DE = Ef - Ei

DE = qrxn + w

In a system at constant volume:

DE = qrxn + 0 = qrxn = qv

But we live in a constant pressure world! How does qp relate to qv?

Two different paths leading to the same internal energy change in a system

Heats of Reaction

qV = qP + w

We know that w = - PDV and DU = qv, therefore:

DU = qP - PDV

qP = DU + PDV

These are all state functions, so define a new function.

Let enthalpy be H = U + PV

Then DH = Hf – Hi = DU + DPV

If we work at constant pressure and temperature:

DH = DU + PDV = qP

How do we measure the heat released (or absorbed)

by a physical or chemical change?

• Calorimetry

– Science of measuring heat

• Specific heat capacity

– The energy required to raise the temperature of one gram of a substance by one degree Celsius.

• Molar heat capacity

– The energy required to raise the temperature of one mole of substance by one degree Celsius.

q α ∆T or q = constant x ∆T or q = constant

∆T

Heat Capacity

The quantity of heat required by an objective to change its temperature by 1 ᵒC.

Heat capacity = q

∆T [in units of J/ᵒC]

Heat Capacity

q = C x DT

Specific Heat Capacity, c

The quantity of heat required by to change the temperature of 1 gram of a substance by 1 ᵒC

Specific heat capacity, c =

q

mass x ∆T [in units of J/g-ᵒC]

q = mass x c x DT

Molar Heat Capacity

The quantity of heat required by to change the temperature of 1 mole of a substance by 1 ᵒC

Molar heat capacity, C = q

mole x ∆T

[in units of J/mol-ᵒC]

Heat Capacity

Units of heat

• Calorie (cal)

– The quantity of heat required to change the temperature of one gram of water by one degree Celsius.

• Joule (J)

– SI unit for heat

1 cal = 4.184 J

A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25oC to 300.oC? The specific heat capacity (c) of Cu is 0.387 J/g•K.

SOLUTION: q = mass x c x DT

q = 125 g (300oC - 25oC) x x = 1.33 x 104 J 0.387 J/g•K

*DT in oC is the same as for K.

*q = mcDT

Finding the Quantity of Heat from Specific Heat

Capacity

Is used to measure the heat released (or absorbed) by a physical or chemical change.

Constant- pressure calorimeters

Constant- volume calorimeters

Assumptions: NO heat flows out of the calorimeter

(Adiabatic system)

Calorimeter

Calorimetry

• Energy gained/lost by the system (reaction) is equal to the energy lost/gain by the calorimeter.

• If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.

• An endothermic reaction cools the solution.

- qsys = qsurr

- qsolid = qcal + qH2O

qsolid = msolidcsolid DT qcal = Ccal DT qH2O = mH2OcH2O DT

- qsolid = qH2O

Coffee-cup Calorimeter

- qsys = qsurr

- qrxn = qcal + qH2O

qrxn = nLR (DHrxn) qcal = Ccal DT qH2O= mH2OcH2O DT

Measuring Heats of Chemical Reaction

You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00oC and carefully add 25.0 mL of 0.500 M HCl, also at 25.00oC. After stirring, the final temperature is 27.21oC. Calculate qsoln (in J) and DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specific heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g•K)

qsoln = masssoln x specific heatsoln x DT

DHrxn = qrxn

nLR

- qsol’n =

*[unit: J ]

*[unit: J /mol or kJ/mol]

qsoln = msoln x csoln x DT

nLR

Determining the Heats of a Reaction



Total volume after mixing = 50.0 mL + 25.0 mL = 75.0 mL

75.0 mL x 1.00 g/mL = 75.0 g of water

qsoln = mass x specific heat x DT

= 75.0 g x 4.184 J/g•oC x (27.21oC - 25.00oC)

qsoln = 693 J

Finding masssoln

qsoln = msoln x csoln x DT



- (693 J/0.0125 mol)(kJ/103 J) = -55.4 kJ/ mol

For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M x 0.0250 L = 0.0125 mol H+

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l)

HCl is the limiting reactant. 0.0125 mol of H2O will form during the reaction.

DHrxn = ? DHrxn = qrxn

nLR

- qsol’n nLR

=

DHrxn (kJ/mol)=

Bomb Calorimeter

- qsys = qsurr

- qsample = qcalorimeter

qsolid = msolidcsolid DT qcal = Ccal DT

Measuring Heats of Combustion

A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2 (the heat capacity of the calorimeter = 8.151 kJ/K). The temperature increases 4.937oC. Is the manufacturer’s claim correct?

- q sample = qcalorimeter

qcalorimeter = heat capacity x DT = 8.151 kJ/K x 4.937 K = 40.24 kJ

1 Calorie = 1 kcal = 4.184 kJ

The manufacturer’s claim is true.

10 Calorie = 41.84 kJ

Calculating the Heat of a Combustion

Reaction

A 30.5 g sample of an alloy at 93.0 ᵒC is placed into a 50.0 g of water at 22.0 ᵒC in an insulated coffee cup with a heat capacity of 9.2 J/K. If the final temperature of the system is 31.1 ᵒC, what is the specific heat capacity of the alloy? (cH2O = 4.184 J/g•K)

- qalloy = qcal + qH2O

- (malloy x calloy x DT) = (Ccal x DT) + (mH2O x cH2O x DT )

- [30.5 g x calloy x (31.1 ᵒC - 93.0 ᵒC ) = [( 9.3 J/K) x (31.1 ᵒC - 22.0 ᵒC )] +

[(50.0 g) x (4.184 J/g-K) x (31.1 ᵒC - 22.0 ᵒC )]

calloy = (84.63 J ) + (1903.72 J)

1887.95 g-ᵒC

calloy = 1.053 J/g-ᵒC

Determining the Specific Heat Capacity

High purity benzoic acid (C6H5COOH; ∆Hrxn for combustion = 3227 kJ/mol) is used as a standard for calibrating bomb calorimeter. A 1.221-g sample burns in a calorimeter (heat capacity = 1365 J/C) that contains exactly 1.200 kg of water. What temperature change is observed?

- qbenzoic acid = qcal + qH2O

- (DHrxn x nLR) = (Ccal x DT) + (mH2O x cH2O x DT )

[Ccal + (mH2O x cH2O)]x DT

- (3227 kJ/mol x ) = 1.221 g 122.1 g/mol

[1.365 kJ/C + (1200 g x 4.184 x10-3kJ/g-ᵒC)]x DT

- (DHrxn x ) = g

MW

DT = [1.365 kJ/ᵒC + (5.0208 kJ/ᵒC)]

- 32.27kJ

DT = - 5.05 ᵒC

Determining Changes in Temperature

Thermochemical Equations

• balanced chemical equation that includes the heat of reaction (DHrxn)

• Examples:

A + B C + D DHrxn= 100 kJ/mol

X + Y XY DHrxn= -120 kJ/mol

A + B + 100 kJ C + D

X + Y XY + 120 kJ

1. Sign – will depend on whether the reaction is exothermic (-) or endothermic (+). The forward reaction has the opposite sign of the reverse reaction.

2H2O(l) 2H2(g) + O2(g) DHrxn = 572 kJ

2H2(g) + O2(g) 2H2O(l) DHrxn = -572 kJ

2. Magnitude – DH is proportional to the amount of substance reacting.

286 kJ is thermochemical equivalent to 1 mol of H2(g)

286 kJ is thermochemical equivalent to 1/2 mol of O2(g)

286 kJ is thermochemical equivalent to 1mol of H2O (l)

H2(g) + O2(g) H2O(l) DHrxn = -286 kJ 1

2

Properties of DHrxn

In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic.

(a) H2(g) + O2(g) H2O(l) + 285.8 kJ

(b) 40.7 kJ + H2O(l) H2O(g)

1

2

Solution:

(a) Heat is in the product side , heat is produce in the reaction.

DHrxn < 0 Exothermic reaction

(b) Heat is in the reactant side , heat is gained in the reaction.

DHrxn > 0 Endothermic reaction

Determining the Sign of DHrxn

SOLUTION:

Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production.

S8(s) + 8H2(g) 8H2S (g) DHrxn = -20.2 kJ

a. Is this an exothermic or endothermic reaction?

b. What is the DHrxn for the reverse reaction?

c. What is the DH when 2.6 mol of S8 reacts?

d. What is the DH when 25.0 g of H2 reacts?

a. Exothermic reaction

b. DHrxn = 20.2 kJ

c. 2.6 mol S8

- 20.2 kJ

1 mol S8

= - 52.52 kJ x

d. 25.0 g H2

- 20.2 kJ

8 mol H2

= - 505 kJ x

Stoichiometry Involving Heats of Reactions

The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is transferred?

Al2O3(s) 2Al(s) + O2(g) DHrxn = 1676 kJ

1.000 x 103 kJ x 2 mol Al

1676 kJ

26.98 g Al

1 mol Al = 32.20 g Al x

3

2

Solution:

Stoichiometry Involving Heats of Reactions

* Indirect Determination of DH:

Hess’s Law

• DH is an extensive property.

–Enthalpy change is directly proportional to the amount of substance in a system.

N2(g) + O2(g) → 2 NO(g) DH° = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) DH° = +90.25 kJ

• DH changes sign when a process is reversed

NO(g) → ½N2(g) + ½O2(g) DH° = -90.25 kJ

Hess’s Law

• the enthalpy change, DH of an overall process is the sum of the enthalpy changes of its individual steps

• In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

DH total= DHrxn1 + DHrxn2 + DHrxn3 + …

Hess’s Law Schematically

Hess’s law of constant heat summation

If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

½N2(g) + O2(g) → NO2(g) DH° = +33.18 kJ

½N2(g) + O2(g) → NO(g) + ½ O2 (g) DH° = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) DH° = -57.07 kJ

Problem-Solving Strategy

• Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.

• Reverse any reactions as needed to give the required reactants and products.

• Multiply reactions to give the correct numbers of reactants and products.

Example: Hess’s law of constant heat

summation

• Consider the following data:

• Calculate ΔH for the reaction

3 2 2

2 2 2

1 3

2 2NH ( ) N ( ) H ( ) H = 46 kJ

2 H ( ) O ( ) 2 H O( ) H = 484 kJ

D

D

g g g

g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

1 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

D

D

g g g

g g g

3 2 2

2 2 2

1 3

2 2NH ( ) N ( ) H ( ) H = 46 kJ

2 H ( ) O ( ) 2 H O( ) H = 484 kJ

D

D

g g g

g g g


4 ( ) 4( )

3 ( ) 3( )

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

1 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

D

D

g g g

g g g

2 2 3

2 2 2

2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

D

D

g g g

g g g


• Final reactions:

ΔH = +1268 kJ

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

D

D

g g g

g g g

Oxidation of sulfur to sulfur trioxide

S(s) + O2(g) SO2(g) DH1 = -296.8 kJ Equation 1:

2SO2(g) + O2(g) 2SO3(g) DH2 = -198.4 kJ Equation 2:

S(s) + O2(g) SO3(g) DH3 = ? Equation 3: 3

2

S(s) + O2(g) SO2(g) DH1 = -296.8 kJ Equation 1:

SO2(g) + O2(g) SO3(g) ½(DH2) = -99.2 1/2(Equation 2):

S(s) + O2(g) + SO2(g) + O2(g) SO2(g) + SO3(g)

Equation 3:

1

2

1

2

S(s) + O2(g) SO3(g) DH3 = ? 3

2

DH 3= DH1 + ½(DH2) = -296.8 kJ + (-99.2 kJ) = -396.0 kJ

Using Hess’s Law to Calculate an Unknown ∆H

SOLUTION:

PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + N2(g) DH = ?

Given the following information, calculate the unknown DH:

Equation A: CO(g) + O2(g) CO2(g) DHA = -283.0 kJ

Equation B: N2(g) + O2(g) 2NO(g) DHB = 180.6 kJ

DHB = - 90.3 kJ

CO(g) + O2(g) CO2(g) DHA = -283.0 kJ

NO(g) N2(g) + O2(g)

DHrxn = -373.3 kJ CO(g) + NO(g) CO2(g) + N2(g)

1

2

1

2

1

2

1

2

1

2

1

2


PROBLEM: Calculate the DH rxn for:

2NOCl(g) N2(g) + O2(g) + Cl2(g) DHrxn = ?

Given the following set of reactions:

N2(g) + O2(g) NO(g) DH1 = 90.3 kJ 1

2

1

2

NO(g) + Cl2(g) NOCl(g) DH2 = 38.6 kJ 1

2

2NOCl(g) 2NO(g) + Cl2(g)

2NO(g) N2(g) + O2(g)

2NOCl(g) N2(g) + O2(g) + Cl2(g)

-2 (DH2) = - 77.2 kJ

-2 (DH1 )= - 180.6 kJ

DHrxn = - 257.8 kJ


2. Diamond and graphite are two crystalline forms of carbon. At 1 atm and 250C, diamond changes to graphite so slowly that the ethalpy change of the process must be obtained indirectly. Determine ∆Hrxn for C(diamond) C(graphite)

With equations from the following list:

C(diamond) + O2(g) CO2(g) ∆H= -395.4kJ 2CO2(g) 2CO(g) + O2(g) ∆H= 566.0 kJ C(graphite) + O2(g) CO2(g) ∆H= -393.5kJ 2CO(g) C(graphite) + CO2(g) ∆H= -172.5kJ

1. A coffee-cup calorimeter is used to determine the specific heat of a metallic sample. The calorimeter is filled with 50.0mL of water at 25.0C (density = 0.997 g/mL). A 36.5-gram sample of metallic material is taken from water boiling at 100.0C and placed in the calorimeter. The equilibrium temperature of the water and sample is 32.5C. The calorimeter constant is known to be 1.87 J/C. Calculate the specific heat of the metallic material. The specific heat of water is 4.184 J/g-C.

Homework

Standard Enthalpies of Formation, DHf°

•The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.

The standard enthalpy of formation of a pure element in its reference state is 0.

Conventional Definitions of Standard States

• For a Compound

– For a gas, pressure is exactly 1 atm.

– For a solution, concentration is exactly 1 M.

– Pure substance (liquid or solid)

• For an Element

– The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.

– Heat of formation is zero.

Standard Enthalpy of Formation

Some standard enthalpies of formation at 298.15 K

number of carbons

Determining the ∆H ᵒrxn from ∆H ᵒf of reactants

and products

• The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

DH ᵒrxn = Σ m DHᵒf (products) - Σ n DHᵒf (reactants)

Step 1: Each reactants decomposes to its element, the standard enthalpy change is - DHᵒf

Step 2: Each product forms from its elements, the standard enthalpy change is DHᵒf

TiCl4(l) + 2H2O(g) TiO2(s) + 4HCl (g) DHᵒrxn = ?

TiCl4(l) Ti(s) + 2Cl2 (g) -DHᵒf [TiCl4(l)]

2H2O(g) 2H2(g) + O2(g) -2DHᵒf [2H2O(g)]

Ti(s) + O2(g) TiO2(s) DHᵒf [TiO2(s)]

2H2(g) + 2Cl2(g) 4HCl(g) 4 DHᵒf [HCl(g]

TiCl4(l) + 2H2O(g) TiO2(s) + 4HCl (g) DHᵒrxn = ?

DHᵒrxn = {DHᵒf [TiO2(s)] + 4 DHᵒf [HCl(g]} + {-DHᵒf [TiCl4(l)]} + {-2DHᵒf [2H2O(g)]}

DH ᵒrxn = Σ m DHᵒf (products) - Σ n DHᵒf (reactants)

Determining the ∆H ᵒrxn from ∆H ᵒf of reactants and products.

Exercise

Calculate ∆H°rxn for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Given the following information: ∆Hf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 ∆H°rxn = –368 kJ/mol

Calculate the DH orxn for the combustion of C3H8 using the following values for DH of.

C3H8 (g) + 5O2(g) 3CO2(g) + 4H2O(g)

C3H8 (g): DH of = -103.8 kJ CO2(g): DH of = -393.5 kJ H2O (l): DH of = -285.8 kJ

DHrxn = 3 DHᵒf [CO2(g)] + 4 DHᵒf [H2O(g] + -DHᵒf [C3H8(g)] + -5DHᵒf [O2(g)]

DHrxn = 3 (-393.5 kJ) + 4 (-285.8 kJ) + [- (-103.8 kJ) -5(0 kJ)]

DHrxn = - 2219.9 kJ

Calculating the Heat of Reaction from Heats of

Formation

SOLUTION:

PROBLEM: Nitric acid, whose worldwide annual production is about 10 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate DHorxn from DHo

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

DHrxn = S mDHof (products) - S nDHo

f (reactants)

DHorxn = {4[DHo

f NO(g)] + 6[DHof H2O(g)]} - {4[DHo

f NH3(g)] + 5[DHof O2(g)]}

= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] DHo

rxn = -906 kJ/mol

Calculating the Heat of Reaction from Heats of

Formation

NH3 (g): DH of = -45.9 kJ/mol NO(g): DH of = 90.3 kJ/mol H2O (l): DH of = -241.8 kJ/mol

DHorxn = DHo

reactant bonds broken + DHoproduct bonds formed

Using Bond Energies to Calculate ∆Horxn

DHorxn = Σ m BEreactants - Σ n BE products

Using Bond Energies to Calculate ∆Horxn

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

Exercises Using Bond Energies to Calculate DHorxn

SOLUTION:

PROBLEM: Calculate DHorxn for the chlorination of methane to form chloroform:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

H

C H

H

H + Cl Cl3

Cl

C Cl

Cl

H + H Cl3

bonds broken bonds formed

Bond Energy, kJ/mol C-H : 413 kJ/mol Cl-Cl: 243 kJ/mol C-Cl: 339 kJ/mol H-Cl: 427 kJ/mol

Answer: DHorxn = - 330 kJ/mol

Exercises Using Bond Energies to Calculate DHorxn

PROBLEM: Calculate DHorxn for the combustion of ethanol.

CH3CH2OH(g) + 3O2(g) 2CO2(g) + 3H2O(g)

Bond Energy, kJ/mol C-H : 413 kJ/mol C-O: 358 kJ/mol C-C: 347 kJ/mol O-H: 467 kJ/mol C=O: 799 kJ/mol O=O: 498 kJ/mol

2. Solving enthalpy of reaction (heat of reaction), DHrxn

a. Calorimetry

Skills that you should have learned

b. Hess’s Law

c. Standard Enthalpy of formation, DH of.

1. Solving heat generated (or absorbed) by a chemical reaction, q.

d. Bond energy

Additional Exercises

Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l); ΔH = –2221 kJ

Assume that all of the heat comes from the combustion

of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

Documents

Lecture 6. Thermochemistry (Part 1)