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George Mason UniversityGeneral Chemistry 211

Chapter 6Thermochemistry: Energy Flow and Chemical

Change

Acknowledgements

Course Text: Chemistry: the Molecular Nature of Matter and Change, 7th edition, 2011, McGraw-

Hill Martin S. Silberberg & Patricia Amateis

The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

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Thermochemistry Whenever matter changes composition, such as in

a chemical reaction, the energy content of the matter changes also

In some reactions the energy that the reactants contain is greater than the energy contained by the products

This excess energy is released as heat

In other reactions it is necessary to add energy (heat) before the reaction can proceed

The energy contained by the products in these reactions is greater than the energy of the original reactants

Physical changes can also involve a change in energy such as when ice melts

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Thermochemistry Thermodynamics is the science of the

relationship between heat and other forms of energy

Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions

Energy is the potential or capacity to move matter (do work); energy is a property of matter

Energy can be in many forms:

Radiant Energy - Electromagnetic radiation

Thermal Energy - Associated with random motion of a molecule or atom

Chemical Energy - Energy stored within the structural limits of a molecule or atom

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Energy There are three broad concepts of energy:

Kinetic Energy (Ek) is the energy associated with an object by virtue of its motion,

Ek = ½mv2

Potential Energy (Ep) is the energy an object has by virtue of its position in a field of force,

Ep = mgh

Internal Energy (Ei or Ui) is the sum of the kinetic and potential energies of the particles making up a substance

E(i) = Ek + Ep

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Energy SI (metric) unit of energy is the Joule:

(J) = kgm2/s2

1 watt = 1 J/s

1 cal = amount of energy needed to raise 1 g of water 1 oC (common energy unit)

1 cal = 4.181 J

1 Btu = 1055 J (Btu - British Thermal Unit)

The Law of Conservation of Energy: Energy may be converted from one form to another, but the total quantities of energy remain constant

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Practice ProblemThermal decomposition of 5.0 metric tons of limestone (CaCO3) to Lime (CaO) and Carbon Dioxide (CO2) requires 9.0 x 106 kJ of heat (E)

Convert this energy to:

a. Joules b. calories c. British Thermal Units (Btu)

63 2(5 tons) CaCO (s) 9.0 x 10 kJ CaO(s) CO (g)

3

6 910 JΔE(J) 9.0 x 10 kJ) 9.0 x 10 J

1kJ

36 9 910 J 1 cal

ΔE(cal) 9.0 x 10 kJ 2.151105 x 10 2.2 x 10 cal1 kJ 4.181 J

3

6 6 610 J 1 BtuΔE(Btu) 9.0 x 10 kJ 8.5308 x 10 8.5x 10 Btu

1 kJ 1055 J

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Energy When a chemical system changes from reactants

to products and the products are allowed to return to the starting temperature, the Internal Energy (E) has changed (E)

The difference between the system internal energy after the change (Efinal) and before the change (Einitial) is:

E = Efinal - Einitial = Eproducts - Ereactants

If energy is lost to the surroundings, then

Efinal < Einitial E < 0

If energy is gained from the surroundings, then

Efinal > Einitial E > 0

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Heat of Reaction In chemical reactions, heat is often transferred

from the “system” to its “surroundings,” or vice versa

The substance or mixture of substances under study in which a change occurs is called the thermodynamic system (or simply system)

The surroundings are everything in the vicinity of the thermodynamic system

Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings

Heat flows from a region of higher temperature to one of lower temperature

Once the temperatures become equal, heat flow stops

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Heat of ReactionWhen heat (q) is released from the system (heat out) to the surroundings (q < 0 negative), the reaction is defined as an Exothermic reaction

When the surroundings deliver heat (heat in) to the system (q > 0 positive), the reaction is defined as an Endothermic reaction

Energy

System

SurroundingsSurroundingsEnergy

System

q<0

Exothermic Endothermic

q>0

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Heat of Reaction Heat of a system is denoted by the symbol “q”

The sign of “q” is positive (q > o) if heat is absorbed by the system, i.e., system temperature increases

The sign of “q” is negative (q < o) if heat is evolved by the system, i.e., system temperature decreases

Heat of Reaction (at a given temperature) is the value of “q” required to return a system to the given temperature when the reaction stops at the completion of the reaction

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Heat of Reaction Consider a chemical reaction that begins with the

system and surroundings temperature at 25oC

If the temperature of a system decreases during the reaction, heat flows from the surroundings into the system

When the reaction stops, heat continues to flow until the system temperature returns to the temperature of the surroundings at 25oC

Heat has been absorbed by or added to the system from the surroundings

The value of “q” is positive, that is: q > 0

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Heat of Reaction If the temperature of the system rises, heat flows

from the system to the surroundings

When the reaction stops, heat continues to flow to the surroundings until the system returns to the temperature of its surroundings (at 25oC)

Heat has flowed out of the system; it has evolved (lost) heat; thus, “’q” is negative, that is q < 0

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Heat Flow and Phase ChangesPredict the sign of q for each of the processes below

1. H2O(g) H2O(l)

2. CO2(s) CO2(g)

3. CH4(g) + O2(g) CO2(g) + H2O(g)

Condensation - Energy (heat) is lost by water vapor

Exothermic reaction - q is negative

Evaporation – Energy (heat) is absorbed (added) by the system from its surroundings Endothermic reaction - (q is positive)Combustion – Burning (oxidation) of Methane releases (evolves) heat to surroundings Exothermic reaction - (q is negative)

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Work & Internal Energy Internal Energy

The Internal Energy of a system, E, is precisely defined as the heat at constant pressure (qp) plus any work (w) done by the system

Work is the energy transferred when an object is moved by a force

pΔE = q + w

pΔE = q + (-P ΔV)

pq = ΔE + P ΔV

final initialw = - P Δ V = - P (V - V ) Internal Energy used to expand volume by increasing pressure is lost to the surroundings, thus the negative sign

Adiabatic Process – Thermodynamic Process Without the Gain or Loss of Heat (∆q = 0)

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Practice ProblemA system delivers 225 J of heat to the surroundings while delivering 645 J of work.Calculate the change in the internal energy, ∆E, of the system

pΔE = q + w

pq = heat delivered to surroundings from system = - 225 J

w = work delivered to surroundings from system = - 645 J

ΔE = change in internal energy

ΔE = - 225 J + (-645 J) = - 870 JHeat is lost

to the surroundings

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Pressure-Volume Work

Sign conventions for q, w, and ∆E:

q + w (-P∆V) = ∆E + + + + — Depends on sizes of q and w — + Depends on sizes of q and w — — —

q: + system gains heat q: — system loses heatw: + work done on systemw: — work done by system

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Practice Problem A system expands in volume from 2.00 L to

24.5 L at constant temperature.

Calculate the work (w), in Joules (J), if the expansion occurs against a constant pressure of 5.00 atm

4w = -1.14×10 J

w = - pΔV

5 -3 32

kg1

1.01325×10 Pa 10 mm •sw = - 5.00 atm × × × (24.5 L - 2.0 L) atm Pa L

2 24

2 2 2

2

kg •m kg •m 1Jw = -1,1399.1 = -1.14×10 ×

s s kg •m

s

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Practice ProblemA system that does no work but which transfers heat to the surrounding has:

a. q < 0, ∆E > 0 b. q < 0, ∆E < 0 c. q > 0, ∆E > 0 d. q > 0, ∆E < 0 e. q < 0, ∆E = 0

A system that does no work but receives heat from the surroundings has:

a. q < 0, ∆E > 0 b. q > 0, ∆E < 0 c. q = ∆E d. q = - ∆E e. w = ∆E

A system which undergoes an adiabatic change (i.e., ∆q = 0) and does work on the surroundings has:

a. w < 0, ∆E = 0, b. w > 0, ∆E > 0 c. w > 0, ∆E < 0 d. w < 0, ∆E > 0 e. w < 0, ∆E < 0

A system which undergoes an adiabatic change (i.e., ∆q = 0) and has work done on it by the surroundings has:

a. w = ∆E b. w = -∆E c. w > 0, ∆E < 0 d. w < 0, ∆E > 0 e. w > ∆E

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Enthalpy and Enthalpy Change Enthalpy, denoted H, is an extensive property of a

substance that can be used to obtain the heat absorbed or evolved in a chemical reaction

An extensive property is one that depends on the quantity of substance

Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system

Enthalpy represents the heat energy tied up in chemical bonds

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Enthalpy and Enthalpy Change The change in Enthalpy for a reaction at a given

temperature and pressure, called the Enthalpy of Reaction, is obtained by subtracting the Enthalpy of the reactants from the Enthalpy of the products.

rxn (products) (reactants)ΔH = H - H

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Enthalpy and Enthalpy Change Enthalpy is defined as the internal energy plus the

product of the pressure and volume (work)

The change in Enthalpy is the change in internal energy plus the product of constant pressure and the change in Volume

H E PV

ΔH = ΔE + P ΔV

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Enthalpy and Enthalpy Change Recall:

The change in Enthalpy equals the heat gained or lost (heat of reaction, Hrxn) at constant

pressure

This represents the entire change in internal energy (DE) minus any expansion “work” done by the system (PV would have negative sign)

pq = ΔE + P ΔV

ΔH = ΔE + P ΔV

pΔH = q (At Constant Pressure)Thus :

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Practice ProblemAn ideal gas (the system) is contained in a flexible balloon at a pressure of 1 atm and is initially at a temperature of 20.0oC.

The surrounding air is at the same pressure, but its temperature is 25oC. When the system is equilibrated with its surroundings, both systems and surroundings are at 25oC and 1 atm.

In changing from the initial to the final state, which of the following relationships regarding the system is correct?

a. ∆E = 0

b. ∆E < 0

c. ∆H = 0

d. w > 0

e. q > 0

Heat is added, internal energy increases

Heat is added, internal energy increases

∆E increases and P∆V work is done by system

P∆V work is done by system (volume increase)

Temperature (heat) in system increases

∆E > 0

∆E > 0

∆H > 0

W < 0

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Practice ProblemIn which of the following processes is ∆H = ∆E, i.e. P∆V = 0?

a. 2HI(g) H2(g) + I2(g) at atmospheric pressure

(P∆V = 0 no change in moles, volume)

b. Two moles of Ammonia gas are cooled from 325oC to 300oC at 1.2 atm

(P∆V ≠ 0 Vol decreases)

c. H2O(l) H2O(g) at 100oC at atmospheric pressure

(P∆V ≠ 0 Vol increases)

d. CaCO3(s) CaO(s) + CO2 (g) at 800oC at atmospheric

pressure

(P∆V ≠ 0 Vol increases)

e. CO2(s) CO2(g) at atmospheric pressure

(P∆V ≠ 0 Vol increases)

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Comparing E & H Reactions that do not involve gases

Reactions such as precipitation, acid-base, many redox, etc., do not produce gases

Since the change in volumes of liquids and solids are quite small:

V 0 P V 0 H E Reactions in which the amount (mol) of gas does

not change(Vol of Gaseous Reactants = Vol Gaseous Products V = 0 P V = 0 H = E

Reactions in which the amount (mol) of gas does change PV 0 However, qp is usually much greater than PV

Therefore: H E

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Comparing E & H Example:

2H2(g) + O2(g) 2H2O(g)

Change in moles: 3 mol 2 mol PV 0

H = -483.6 kJ and PV = -2.5kJ

E = H - PV = -483.6 kJ - (-2.5 kJ) = -481.1 kJ

Most of E occurs as Heat (H = qp)

\ H E

For many reactions, even when PV 0, H is close to E

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Comparing E & HFor which one of the following reactions will ∆H be approximately (or exactly) equal to ∆E?

a. H2(g) + Br2(g) 2HBr(g)

(No change in volume; no change in work, PV = 0)

b. H2 O(l) H2O(g)

(Change in volume; change in work due to gas expansion, PV 0)

c. CaCO3(s) CaO(s) + CO2(g)

(Change in volume; change in work due to gas expansion, PV 0

d. 2H(g) + O(g) H2O(l)

(Change in volume; condensation, heat (q) released, PV 0)

e. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

(Change in volume; condensation, heat (q) released, PV 0)

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Exothermic and Endothermic Processes

Energy (E), Pressure (P), and Volume (V) are “state” functions

Enthalpy (H) is also a state function, which means that H depends only on the difference between Hfinal & Hinitial

The Enthalpy change of a reaction, also called the Heat of Reaction (Hrxn), always refers to

Hrxn = Hfinal - Hinitial = Hproducts - Hreactants

Hproducts can be either more or less than Hreactants

The resulting sign of H indicates whether heat is absorbed from the surroundings (heat in) or released to the surroundings (heat out) in the process

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Exothermic and Endothermic Processes

An Exothermic reaction releases heat (heat out) to surroundings with a decrease in system Enthalpy

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

Exothermic: Hfinal < Hinitial H < 0 (negative)

An Endothermic reaction absorbs heat (heat in) from the surroundings resulting in an increase in system Enthalpy

Heat + H2O(s) H2O(l)

Endothermic Hfinal > Hinitial H > 0 (positive)

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Types of Enthalpy Changes When a compound is produced from its elements,

the Enthalpy change (Heat of Reaction) is called:

Heat of Formation (∆Hf)

K(s) + ½Br2()l) KBr(s) ∆H = ∆Hf

When a substance melts, the Enthalpy change is called:

Heat of Fusion (∆Hfus)

NaCl(s) NaCl(l) ∆H = ∆H(fus)

When a substance vaporizes, the Enthalpy change is called:

Heat of Vaporization

C6H6(l) C6H6(g) ∆H = ∆H(vap)

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Thermochemical Equations A Thermochemical Equation is the chemical

equation for a reaction (including phase labels) in

which the equation is given a molar interpretation, and the Enthalpy of Reaction (∆Hrxn) for these molar

amounts is written directly after the equation.

2 2 3N (g) + 3 H (g) 2 NH (g)

H is negative; heat is lost to surroundings

1 mol N2 + 3 mol H2 yields 91.8 kJ of heat

rxnΔH = - 91.8 kJ

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Practice ProblemSulfur, S8, burns in air to produce Sulfur Dioxide. The

reaction evolves (releases) 9.31 kJ of heat per gram of Sulfur

at constant pressure. Write the thermochemical equation for

this reaction.

8 2 2S O SO + Heat

8 2 2S 8 O 8 SO ΔH - 9.31 kJ

Exothermic Reaction

Balance the Reaction

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Practice ProblemIn a phase change of water between the liquid and the gas phases, 770.1 kJ of energy was released by the system. What was the product, and how much of it was formed in the phase change.

(Data: H2O(l) H2O(g) ∆H = 44.01 kJ/mol)

a. 315 g of water vapor was produced

b. 17.5 g of water vapor was produced

c. 17.5 mol of water vapor was produced

d. 17.5 mol of liquid water was produced

∆H is positive (endothermic reaction)Since energy was released, the gas condensed to

liquid

e. 17.5 g of liquid water was produced770.1 kJ / 44.01 kJ / mol = 17.5 mols

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Thermochemical Equations The following are two important rules for

manipulating Thermochemical equations:

When a thermochemical equation is multiplied

by any factor, the value of H for the new

equation is obtained by multiplying the H in the

original equation by that same factor

When a chemical equation is reversed, the

value of H is reversed in sign

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Practice ProblemWhen White Phosphorus burns in air, it producesPhosphorus (V) Oxide (Change in Oxidation state)

P4(s) + 5O2(g) P4O10(s) H = -3010 kJ

What is H for the following equation?

P4O10(s) P4(s) + 5O2(g) H = ?

Ans:

The original reaction is reversed

Change the Sign !!

H = + 3010 kJ

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Practice ProblemCarbon Disulfide (CS2(l)) burns in air, producing Carbon Dioxide and Sulfur Dioxide

CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g) H = -1077 kJ

What is H for the following equation?

1/2 CS2(l) + 3/2 O2(g) 1/2 CO2(g) + SO2(g)

Ans: The new reaction uses ½ the original amounts

Divide H by 2

H = (-1077 / 2) = - 538.5 kJ

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Applying Stoichiometry andHeats of Reactions

Consider the reaction of Methane, CH4, burning in the presence of Oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of10.0 grams CH4?

4

4 4

1 mol CH-890.3 kJ

4 16.0 g CH 1 mol CH10.0 g CH × × = - 556 kJ

o4 2 2 2CH (g) + 2 O (g) CO (g) + 2 H O(l) ΔH = - 890.3 kJ

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Measuring Heats of Reaction To see how Heats of Reactions (Enthalpy change of

reaction, Hrxn) are measured, look at the heat

required to raise the temperature of a substance

A thermochemical measurement is based on the

relationship between heat and temperature change

The heat required to raise the temperature of a substance

is its:

Heat Capacity

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Measuring Heats of Reaction Heat Capacity and Specific Heat

The molar heat capacity, C, of a sample of substance is the quantity of heat required to raise the temperature of one mole of substance one degree Celsius

C is in units of J/mol oC, n = moles of substance

The specific heat capacity, S, (or “specific heat”) is the heat required to raise the temperature ofone gram of a substance by one degree Celsius

S is in units of J/g oC m = grams of sample

q = n C ΔT

q = m S ΔT

T = Tfinal - Tinitial

T = Tfinal - Tinitial

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Measuring Heats of Reaction

Specific Heats and Molar Heat Capacities of some substances

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Practice ProblemSuppose you mix 20.5 g of water at 66.2 oC with 45.4 g of water at 35.7 oC in an insulated cup. What is the maximum temperature of the solution after mixing?

Ans: The heat lost by the water at 66.2 oC is balanced by the heat gained by the water at 35.7 oC

ofT = 45.2 C

q = m S ΔT o o1 f 2 fΔT = T - 66.2 C ΔT = T - 35.7 C

1 2m = 20.5 g m = 45.4 g

lost gained 1 1 2 2-q = q = - m SΔT = m SΔT

o o1 f 2 f-m * T - 66.2 C = m * T - 35.7 C

2f f

1

mT - 66.2 = T - 35.7

- m

f f

45.4T - 66.2 = T - 35.7

-20.5

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Measuring Heats of Reaction Bomb Calorimeter used to measure heats of

combustion

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Practice ProblemHow much heat is gained by Nickel when 500 g of Nickel is warmed from 22.4oC to 58.4°C?

[The specific heat of Nickel is 0.444 J/(g • °C)]

a. 2000 J b. 4000 J c. 6000 J

d. 8000 J e. 10000 J

Ans: d

q = 7992 = 8000 J

q = m s ΔTo o o

f iΔT = T (58.4 C) - T (22.4 C) = 36.0 Cos = Specific Heat Nickel = 0.444 J / g • C

m = 500 go oq = 500 g * 0.444 J / g • C * 36.0 C

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Practice ProblemWhen 25.0 mL of 0.5 M H2SO4 is added to 25.0 mL of 1.00 M

KOH in a calorimeter at 23.5 oC, the temperature rises to 30.17oC

Calculate Hrxn for each reactant. Assume density (d) and

specific heatof the solution (s) are the same as water

2 4 2 4 22 KOH(aq) + H SO (aq) K SO (aq) + 2 H O(l)o

soln f iq = m s ΔT (Δ T = T - T ) d = 1 g / mL s = 4.184 J / g • C

osoln o

1.00 g 4.184 J 1 kJq = 25.0 + 25.0 mL 30.17 - 23.5 C = 1.395354 kJ

mL 1000 Jg • C

1.00 mol KOH 1 L25.0 mL = 0.0250 mol KOH

L 1000 mL

2 4 2 4Both KOH or H SO are limiting (2 moles KOH / 1 mol H SO ) Con’t

Calculate moles

2 42 4

0.500 mol H SO 1 L25.0 mL = 0.0125 mol H SO

L 1000 mL

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Practice Problem (Con’t)When 25.0 mL of 0.5 M H2SO4 is added to 25.0 mL of 1.00 M

KOH in a calorimeter at 23.5 oC, the temperature rises to 30.17oC.

Calculate Hrxn for each reactant. Assume density (d) and

specific heat of the solution are the same as water.

rxn-1.395364 kJ

H (KOH) - 55.81456 - 55.8 kJ / mol KOH0.0250 mol KOH

Temperature of water increased (23.5oC 30.17oC)

The Reaction is Exothermic (heat released to surroundings (water))

Thus, qrxn is negative

soln rxnq = q = -1.395634 kJ

rxn 2 4 2 42 4

-1.395364 kJH (H SO ) -111.62912 -112 kJ / mol H SO

0.0125 mol H SO

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Hess’s Law Hess’s law of Heat Summation

For a chemical equation that can be writtenas the sum of two or more steps, theEnthalpy change for the overall equationis the sum of the Enthalpy changes for the individual steps

In coupled reactions, the Enthalpy change for the overall reaction is the sum of the Enthalpy changes for the coupled reactions

Note: It is often necessary to reverse chemical equations to couple them so chemical species are on the correct side of yield sign, or multiply through by a coefficient to cancel common chemical species

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Hess’s LawFor example, suppose you are given the following data:

o2 2S(s) + O (g) SO (g) ΔH = -297 kJ

o3 2 22 SO (g) 2 SO (g) + O (g) ΔH = 198 kJ

o2 32 S(s) + 3 O (g) 2 SO (g) ΔH = ?

Could you use these data to obtain the Enthalpy change for the following reaction?

Con’t

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Hess’s LawIf we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third

o2 22 S(s) + 2 O (g) 2 SO (g) ΔH = (-297 kJ)×(2)

o2 2 32 SO (g) + O (g) 2 SO (g) ΔH = (198 kJ)×(-1)

o2 32 S(s) + 3 O (g) 2 SO (g) ΔH = (-792 kJ)

Note the change in H values with the changes in the molar coefficients to balance equation 1 and the reversal of equation 2

Note : ( - 297 × 2) + (198 × - 1) = - 594 kJ - 198 kJ = - 792 kJ

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Practice ProblemGiven the following data,

A(s) + O2(g) AO2(g) H° = – 105 kJ/mol

A(g) + O2(g) AO2(g) H° = – 1200 kJ/mol

Find the heat required for the reaction converting:

A(s) to A(g) at 298 K and 1 atm pressure.

A(s) A(g) + 1095 kJ

2 2A(s) + O (g) AO (g) - 105 kJ

2 2AO (g) A(g) + O (g) + 1200 kJ

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Standard Enthalpies of Formation The term standard state refers to the standard

thermodynamic conditions chosen for substances when listing or comparing thermodynamic data:

Pressure - 1 atmosphere (760 mm Hg)

Temperature - (usually 25oC).

The Enthalpy change for a reaction in which reactants are in their standard states is denoted as the

“Standard Heat of Reaction” orxnΔH

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Standard Enthalpies of Formation Standard Enthalpy of Formation of Substance

The Enthalpy change for the formation of

one mole of a substance in its standard state from its

component elements in their standard states

Note: The standard Enthalpy of Formation for a

“Pure Element” (C, Fe, Au, N, etc.)

in its standard state is zero

ofΔH

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Standard Enthalpies of FormationLaw of Summation of Heats of Formation

The Enthalpy of a reaction i.e., the “Standard Heat of Reaction:

(∆Horxn)”

is equal to the total formation energy of the products minus that of the reactants

Where is the mathematical symbol meaning

“the sum of”

and m and n are the coefficients of the substances in the chemical equation, i.e., the relative number of moles of each substance

o o orxn f fΔH = nΔH (products) - mΔH (reactants)

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Standard Enthalpies of Formation

Formula Hof

(kJ/mol)Formula Ho

f(kJ/mol)

Formula Hof

(kJ/mol)

CalciumCa(s)CaO(s)CaCO3(s)

0-635.1

-1206.9

HydrogenH2(g)H(g)

0218.0

SilverAg(s)AgCl(s)

0-127.0

CarbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CS2

01.9

-110.5-393.5

-74.9-238.6

13587.9

OxygenO2(g)O3(g)H2O(g)H2O(l)

0143

-241.8-285.8

SodiumNa(s)Na(g)NaCL(s)

0107.8

-411.1

ChlorineCl2(g)Cl(g)Cl-(aq)Cl-(g)HCl(g)

0121.0167.2

-234.0-92.31

NitrogenN2(g)NH3(g)NO(g)

0-45.990.3

BromineBr2(l)Br(g)Br2(g)Br-(ag)Br-(g)HBr(g)

0111.930.91

-121.5-219.0-36.44

SulfurS8(rhombic)S8(monoclinic)SO2(g)SO3(g)

00.3

-296.8-396.0

Selected Standard

Heats of Formation

(Enthalpies)

At 25oC (298oK)

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Practice ProblemCalculate the Heat of Reaction, H°rxn, for the combustion of C3H6(g):

C3H6(g) + 9/2 O2(g) 3 CO2(g) + 3 H2O(l)

Hof values in kilojoules per mole are as follows:

C3H6(g) = 21 CO2(g) = –394 H2O(l) = –286

a. –2061 kJ b. –2019 kJ c. –701 kJ

d. 2019 kJ e. 2061 kJ

Ans: a

orxnΔH = - 2061 kJ

o o orxn f fΔH = n ΔH (products) - m ΔH (reactants)

orxnΔH = 3× -394 + 3× -286 - 21 + 0

orxnΔH = -1182 - 858 - 21

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Practice ProblemAcetylene burns in air according to the equation below.

Given: Hof CO2(g) = -395.5 kJ/mol

Hof H2O(g) = -241.8 kJ/mol

o2 2 2 2 2 rxnC H (g) + 5 / 2 O (g) 2 CO (g) + H O(g) ΔH = -1255.8 kJ

Calculate Hof of C2H2(g)

o o orxn f 2 f 2

o of 2 2 f 2

ΔH = 2 mol (Δ H , CO (g)) + 1 mol (Δ H , H O(g)) -

1 mol (Δ H , C H (g)) + 5 / 2 mol (Δ H , O (g))

of 2 2

-1255.8 kJ = 2 mol (-393.5 kJ / mol) + 1 mol (-241.826 kJ / mol) -

1 mol ( (Δ H , C H (g) + 5 / 2 mol (0.0) of 2 2- 1255.8kJ = - 787.0kJ - 241.8kJ - 1 mol (Δ H , C H (g))

of 2 2

-227.0 kJΔH , C H (g) = = 227.0 kJ / mol

-mol

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Summary – Equations & Relationships2

kE = 1 / 2 m v

final initial products reactantsΔE = E - E = E - E

pΔE = q + w pΔE = q + (-P ΔV)

pq = ΔE + P ΔV

ΔH = ΔE + P ΔV pΔH = q (at Constant P)

q = sm ΔT

o o orxn f fΔH = nΔ H (products) - mΔ H (reactants)

lost gained-q (exothermic) = q (endothermic)

final initialw = - P Δ V = - P (V - V )