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Chap 6: Thermochemistry Bushra Javed

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Contents1. Introduction to Thermochemistry2. Energy and it’s units3. Understanding Heats of Reaction2. Enthalpy and Enthalpy Changes3. Thermochemical Equations4. Applying Stoichiometry to Heats of Reaction5. Measuring Heats of Reaction6. Using Heats of Reaction; Hess’s Law7. Standard Enthalpies of Formation

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ThermodynamicsThe science of the relationship between heat and other forms of energy.

ThermochemistryAn area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction.

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Energy

EnergyThe potential or capacity to move matter.One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical.

Next, we’ll study kinetic energy, potential energy, and internal energy

Kinetic Energy, EK

The energy associated with an object by virtue of its motion.

m = mass (kg)v = velocity (m/s)

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2mvE21

K

Energy & it’s units

The SI unit of energy is the joule, J, pronounced “jewel.”

The calorie is a non-SI unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. The exact definition is given by the equation:

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2

2

smkg

J

(exact)J4.184cal1

Kinetic Energy, EK

Example 1A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy?

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2

K sm

1.78kg)(75.021

E

m = 75.0 kgv = 1.78 m/s

EK = ½ mv2 = 119kg.m2 /s2

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Kinetic Energy KEExample 2What is the kinetic energy of a 2100-lb car traveling at 48 miles per hour? (1 lb = 0.4536 kg, 1 mi = 1.609 km)

a) 3.3 × 10–8 Jb)2.2 × 105 Jc) 3.7 × 1019 Jd)1.1 × 106 J

Potential Energy, EP

The energy an object has by virtue of its position in a field of force, such as gravitational, electric or magnetic field.

Gravitational potential energy is given by the equation

m = mass (kg)

g = gravitational constant (9.80 m/s2)h = height (m)

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mghE P

Internal Energy

Internal Energy, UThe sum of the kinetic and potential energies of the particles making up a substance.

Total EnergyEtot = EK + EP + U

In the laboratory, assuming flasks and test tubes are at rest, both Ek and Ep are equal to zero.  Etotal = U(internal energy due to molecular motion) 

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Law of Conservation of Energy

Energy may be converted from one form to another, but the total quantity of energy remains constant.

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Thermodynamic System and Surroundings

Thermodynamic system: the substance or mixture of substances under study in which a change occurs surroundings. Thermodynamic surroundings: Everything outside the system is surroundings

A system is separated from its surroundings by a boundary across which matter and/or energy is transferred.

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Heat, q

The energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings.

Heat flows spontaneously from a region of higher temperature to a region of lower temperature.

• q is defined as positive if heat is absorbed by the system (heat is added to the system)• q is defined as negative if heat is evolved by a

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Heat of ReactionThe value of q required to return a system to the given temperature at the completion of the reaction (at a given temperature).

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Heat of Reaction

Endothermic ProcessA chemical reaction or process in which heat is absorbed by the system (q is positive). The reaction vessel will feel cool.

Exothermic ProcessA chemical reaction or process in which heat is evolved by the system (q is negative). The reaction vessel will feel warm.

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Heat of ReactionIn an endothermic reaction:

The reaction vessel cools.Heat is absorbed.Energy is added to the system.q is positive.

In an exothermic reaction:The reaction vessel warms.Heat is evolved.Energy is subtracted from the system.q is negative.

Enthalpy of Reaction

The change in enthalpy for a reaction at a given temperature and pressure:

DH = H(products) – H(reactants)

Note: D means “change in.”

Enthalpy change is equal to the heat of reaction at constant pressure:

DH = qP

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Enthalpy of Reaction

Enthalpy, HAn extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.

Extensive PropertyA property that depends on the amount of substance. Mass and volume are extensive properties.

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Enthalpy change ∆HExample 3The phrase “the heat absorbed or released by a system undergoing a physical or chemical change at constant pressure” is

a) the definition of a state function.b) the change in enthalpy of the system.c) a statement of Hess’s law.d) the change in internal energy of the system.

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Enthalpy change ∆H

Example 4If ∆ H = –31 kJ for a certain process, that process

a) occurs rapidly.b) is exothermic.c) is endothermic.d) cannot occur.

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The diagram illustrates the enthalpy change for the reaction

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

The reactants are at the top. The products are at the bottom. The products have less enthalpy than the reactants, so enthalpy is evolved as heat. The signs of both q and DH are negative.

Thermochemical Equations

The 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 for these molar amounts is written directly after the equation.

For the reaction of sodium metal with water, the thermochemical equation is:

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g); DH = –368.6 kJ

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Thermochemical Equations 

A thermochemical equation expresses the enthalpy of the reaction next to a balanced chemical equation.  1. N2(g) + 3H2(g) → 2NH3(g); ΔH = –91.8kJ

 2. C6H12O6(s) + 6O2(g) → CO2(g) + H2O(l); ΔH = –2803kJ 

3. H2(g) + O2(g) → H2O (g); ΔH = –242kJ 

Thermochemical Equation

Manipulating a Thermochemical Equation• When the equation is multiplied by a factor,

the value of DH must be multiplied by the same factor.

• When a chemical equation is reversed, the sign of DH is reversed.

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Manipulating Thermochemical Equations

Example 5 When sulfur burns in air, the following reaction occurs:S8(s) + 8O2(g) 8SO2(g);

DH = – 2.39 x 103 kJ

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Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements.

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S8(s) + 8O2(g) 8SO2(g); DH = –2.39 × 103 kJ

We want SO2 as a reactant, so we reverse the given reaction, changing the sign of DH:

8SO2(g) S8(g) + 8O2(g) ; DH = +2.39 × 103 kJ

We want only one mole SO2, so now we divide every coefficient and DH by 8:

SO2(g) 1/8S8(g) + O2(g) ; DH = +299 kJ

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Manipulating Thermochemical Equations

Example 6You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is S8(s) + 8O2(g) 8SO2(g); DH = -2.39 × 103 kJ

Manipulating Thermochemical Equations

S8(s) + 8O2(g) 8SO2(g); DH = -2.39 103 kJ

Molar mass of S8 = 256.52 g

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88 8

1mol S 2.39 10 kJ15.0 g S256.5 g S 1mol S

q

q = –1.40 102 kJ

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Manipulating Thermochemical Equations

Example 7Given:4AlCl3(s) + 3O2(g) → 2Al2O3(s) + 6Cl2(g); ∆H = –529.0 kJdetermine ∆H for the following thermochemical equation.Cl2(g) + ⅓Al2O3(s) → ⅔AlCl3(s) + ½O2(g)

a) +529.0 kJb) +88.2 kJc) +176.3 kJd) +264.5 kJ

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Applying Stoichiometry to Heats of Reaction

Example 8How much heat is evolved upon the complete oxidation of 9.41 g of aluminum at 25°C and 1 atm pressure? (∆H for Al2O3 is –1676 kJ/mol.)4Al(s) + 3O2(g) → 2Al2O3(s)

a) 146 kJb) 1169 kJc) 292 kJd) 585 kJ

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Measuring Heats of ReactionWe will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction.

Measuring Heats of Reaction

Heat Capacity, C, of a Sample of Substance The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one Kelvin).

Molar Heat Capacity The heat capacity for one mole of substance.

Specific Heat Capacity, s (or specific heat)The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure.

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Measuring Heats of ReactionThe heat required can be found by using the following equations.Using specific heat capacity:

q = s x m x Dt

 s = specific heat of the substancem = mass in grams of substance∆ t = tfinal – tinitial (change in temperature)

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Specific HeatExample 9The units for specific heat are

a)J/(g · °C).b)(J · °C).c) (J · g).d)J/°C.

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Example10A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C).

m = 35.8 gs = 0.388 J/(g°C)Dt = 28.00°

q = m s Dt

C8.00Cg

J0.388g35.8

q

q = 111 J

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Specific Heat

Example 11 How much heat is gained by nickel when 29.2 g of nickel is warmed from 18.3°C to 69.6°C? The specific heat of nickel is 0.443 J/(g · °C).

a) 2.37 × 102 Jb)9.00 × 102 Jc) 22.7 Jd)6.64 × 102 J

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Hess’s Law of Heat Summation

• For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps

• ∆Hreaction = ∆H1 + ∆H2 + ∆H3

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Hess’s Law of Heat SummationSuppose we want DH for the reaction

2C(graphite) + O2(g) 2CO(g)It is difficult to measure directly. However, two other reactions are known:

C(graphite) + O2(g) CO2(g); DH = -393.5 kJ 2CO2(g) 2CO(g) + O2(g); DH = – 566.0 kJ• In order for these to add to give the reaction we

want, we must multiply the first reaction by 2. • Note that we also multiply DH by 2.

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Hess’s Law of Heat Summationwe can add the reactions and the DH values.

Cancel the species that appear on both sides.2C(graphite) + 2O2(g) 2CO2(g); DH = -787.0 kJ

2CO2(g) 2CO(g) + O2(g); DH = – 566.0 kJ

2 C(graphite) + O2(g) 2 CO(g); DH = –1353.0 kJ

2C(graphite) + O2(g) 2CO(g)

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Hess’s Law of Heat Summation

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Hess’s lawExample 12Consider the following changes:H2O(s) → H2O(l); ∆H1 H2O(l) → H2O(g); ∆H2 H2O(g) → H2O(s); ∆H3

Using Hess’s law, the sum ∆H1 + ∆H2 + ∆H3 isa) greater than zero.b) equal to zero.c) less than zero.d) sometimes greater than zero and sometimes less

than zero.

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Hess’s lawExample 13 Given:Pb(s) + PbO2(s) + 2H2SO4(l) → 2PbSO4(s) + 2H2O(l); ∆H° = –509.2 kJ

SO3(g) + H2O(l) → H2SO4(l); ∆H° = –130. kJ

determine ∆H° for the following equation.Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s)

a) –3.77 × 103 kJb) 3.77 × 103 kJc) –639 kJd) –521 kJe) –769 kJ

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Hess’s lawExample 14 Given:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ∆H = –26.8 kJFeO(s) + CO(g) → Fe(s) + CO2(g); ∆H = –16.5 kJdetermine ∆H for the following thermochemical equation.Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g)

a) –43.3 kJb) –10.3 kJc) 6.2 kJd) 10.3 kJe) 22.7 kJ

Standard Enthalpies of Formation

The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C). These standard conditions are indicated with a degree sign (°).

When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction, DH°. (DH° is read “delta H zero.”)

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Allotropes

Elements can exist in more than one physical state

Some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O2 or as O3 (ozone).

These different forms of an element in the same physical state are called allotropes.

The reference form is the most stable form of the element (both physical state and allotrope).

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Standard enthalpy of formation, DHf°

is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. DHf° for an element in its reference and standard state is zero.

For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction

H2(g) + 1/2O2(g) H2O(l)DHf° = –285.8 kJ

Other DHf° values are given in Table 6.2 and Appendix C.

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Finding heat of reaction from Standard enthalpy of formation, DHf°

Example 15What is the heat of vaporization of methanol, CH3OH, at 25°C and 1 atm?Use standard enthalpies of formation (Appendix C).

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We want DH° for the reaction:CH3OH(l) CH3OH(g)

molkJ

200.7Δ:methanolgaseousFor

molkJ

238.7Δ:methanolliquidFor

f

f

H

H

reactants

fproducts

freaction ΔΔΔ HnHnH

molkJ

238.7 mol 1molkJ

200.7 mol 1Δ vapH

DHvap= +38.0 kJ

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Finding ΔH from Standard Heats of Formation

Example 16All of the following have a standard enthalpy of formation value of zero at 25°C except

a) CO(g).b) Fe(s).c) C(s).d) F2(g).e) Ne(g).

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Finding ΔH from Standard Heats of Formation

Example 17What is ∆H° for the following phase change?NaI(s) → NaI(l)

∆H°f (kJ/mol)NaI(s) –287.86 NaI(l) –266.51

a) 554.37 kJb) –554.37 kJc) 0 kJd) –21.35 kJe) 21.35 kJ

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Finding ΔH from Standard Heats of Formation

Example 18What is ∆H° for the following reaction?2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)

∆H°f (kJ/mol)C2H2(g) +226.7 CO2(g) –393 H2O(l) –285.8

a) +1692.2 kJb) –2599.0 kJc) +2599.0 kJd) –1692.2 kJe) –452.6 kJ