THERMOCHEMISTRYHEAT AND CHEMICAL CHANGE

THE FLOW OF ENERGY

ENERGY TRANSFORMATIONS

Thermochemistry is concerned with the heat changes that occurduring chemical reactions.

Energy is the capacity for doing work or supplying heat.

Energy stored within the structural units of chemical substancesis called chemical potential energy.

Heat, represented by q, is energy that transfers from one object to another because of a temperature difference between them. Heat always flows from the warmer object to a cooler object.

EXOTHERMIC AND ENDOTHERMIC PROCESSES

A system is the part of the universe on which you focus your attention.

The surroundings include everything else in the universe.

The system and the surroundings make up the universe.

A process which absorbs heat from the surroundings is called anendothermic reaction.

A process that releases energy to its surroundings is called an exothermicreaction.

The law of conservation of energy states that in any chemical orphysical process energy is neither created nor destroyed.

HEAT CAPACITY AND SPECIFIC HEAT

A calorie (c)is defined as the quantity of heat neededto raise the temperature of 1 gram of water 1oC.

A kilocalorie (C) is one thousand calories.

The SI unit for energy is the joule (J). One calorie is equal to4.184 J.

The amount of heat to increase the temperature of an object 1oC is the heat capacity of that object.

The specific heat of an object is the amount of energy necessary to raise one gram of the substance 1oC (J/goC).

At 4.184 J/goC, water has a very high specific heat. Most metals have arelatively low specific heat.

The vast amount of water that covers Earth (about 3/4ths it surface) results inslow heating of the Earth’s surface and,corresponding, slow cooling. Thismoderates the Earth’s temperature allowing life as we know it to exist.

CALCULATIONS:

The standard formula for the calculation for changes in energy is:

q = m (T)Cp

Where: q = heat of reaction (heat exchanged in joules)

m = mass of material

T = change in temperature (initial temperature (ti) - final temperature (tf).

Cp = specific heat of the material

Sample Problem 1.

A 37 mL (37 g) sample of water at an initial temperatureof 13oC is heated to 34o C. What is the amount of heatgained by the water?

q = m(T)Cp

q = 37 g (21oC)4.184 J/goC

q = 3251 JgoC/goC = 3251 J

m = 37 g ΔT = 34oC – 13oC = 21oC Cp water = 4.184 J/goC

Practice Problem 1:

What is the heat gained by 27 grams of a metal with aspecific heat of 0.177 J/goC, if its initial temperature is 44oC and its final temperature is 71oC?

Solving:

q = m(T)Cp

q = 27g(27oC)0.177J/goC

q = 129 J

Sample Problem 2:

What is the mass of zinc which, when heated goes from an initial temperature of 88OC to 131oC and gains 1678 J of heat? The Cp of zinc is 0.388 J/goC.

q = m(T)Cp

q = 1678 J m = g ? T = 131 – 88 = 43oC Cp = 0.388 J/goC

1678 J = m(43oC)0.388 J/goC

1678 J = m(16.68 oCJ/goC = 1678 J = m 16.68 J/g

1678 Jg16.68 J = m = 100.6 g

Practice Problem 2:

What is the mass of a metal with a specific heat of 1.44 J/goC which when heated begins at 25oC and has a final temperature of 32oC and gains 1145 J?

Solving:

q = m(T)Cp

1145 J = m(7oC)1.44 J/goC

1145 g = m10.08

m = 113.6 g

Sample Problem 3:

What is the specific heat of 50 g of a metal which, whenheated to 97oC and added to 30 g of water with an initialtemperature of 14oC, results in the final temperature of thewater being 27oC.

Since we are dealing with both water and a metal, we need an equation of eachone.

q = m(T)Cpwater q = m(T)Cpmetal

Entering what we know in each equation and solving were possible we have the following:The water equation: q = 30(13)4.184 q = 1631 J

Note: Because the hot metal will exchange heat with the cooler water until they both reach the same temperature (equilibrium), the final temperature ofthe metal and the water will be the same.

The metal equation: q = 50(70)CpThe equation appears to be unsolvable because we have two unknowns but remember the law of conservation of energy states that whatever energy wasgained by the water had to be lost by the metal, therefore the two q’s are equal.

Sample Problem 3 continued:

1631 J = 50 g (70oC)Cp

1631 J = 3500 gOC Cp

1631 J/3500 goC = Cp

0.466 J/goC = Cpmetal

Practice Problem 3:

What is the specific heat of a metal if 42 g of the metalheated to 99oC is added to 25 g of water with an initial temperatureof 10.0oC? The water final temperature is 18.0oC.

q = m(ΔT)Cpwater q = m(ΔT)Cpmetal

q = 25(8)4.184

q = 836.8 J

836.8 J = 42(81)Cp

836.8 J = 3402 Cp

836.8 J = 3402 Cp3402 3402

Cp = 0.246 J/goC

Sample Problem 4:

Copper heated to 86oC is added to 78 g of water at an initial temperature of 14oC. The water rises to a final temperatureof 25oC. The specific heat of copper is 0.385 J/goC. What is the massof copper used in this experiment?

q = m(T)Cpwater q = m(T)Cpmetal

q = 78 g(11oC)4.184 J/goC

q = 3590 J

3590 J = m(61oC)0.385 J/goC

3590 g/23.5 = m

m = 152.8 g

Practice Problem 4:

Aluminum is heated to 100.0oC and then addedto 47 g of H2O with a Ti of 8.0oC. The water increasesto a Tf of 63.0oC. The Cp of aluminum is 0.9025 J/goC.What is the mass of aluminum used in this experiment?

q = m(ΔT)Cpwater q = m(ΔT)Cpaluminum

q = 47(55)4.184

q = 10 816 J

10 816 J = m(37)0.9025

10 816 J = 33.4 m

10 816 J = 33.4 m 33.4 33.4

323.8 g = m

Sample Problem 5:

What is the final temperature (Tf) of a systemin which 60 g of iron heated to 98oC is added to25 g of water with an initial temperature of 12oC?The Cp of iron is 0.449 J/goC.

Again we begin by writing a water and iron equation and inserting the know values.

q = m(T)Cpwater q = m(T)Cpiron

q = 25(Tf – 12)4.184 q = 60(98 –Tf)0.449

Each equation has two unknowns but we know that the q of each equation must be equal. We can therefore putthe two equations together as follows:

Sample Problem 5 continued:

25(Tf – 12)4.184 = 60(98 – Tf)0.449

Now simplify using distributive properties as follows:

(25Tf – 300)4.184 = (5880 – 60Tf)0.449

104.6Tf – 1255.2 = 2640 – 26.9Tf

Group like items on each side of the equation:

104.6Tf + 26.9Tf = 2640 + 1255.2

131.5Tf = 3895.2

131.5Tf

131.5=

3895.2131.5

Tf = 29.6oC

Practice Problem 5:

What is the Tf of a system in which 34 g ofnickel heated to 99oC is added to 40 g of water with a temperature of 10oC? The Cp of nickel is 0.444 J/goC.

q = m(T)Cpwater q = m(T)Cpnickel

q = 40(Tf – 10)4.184 q = 34(99 – Tf)0.444

40(Tf – 10)4.184 = 34(99 – Tf)0.444

(40Tf – 400)4.184 = (3366 – 34Tf)0.444

167.4Tf – 1673.6 = 1494.5 – 15.1Tf

167.4Tf + 15.1Tf = 1494.5 + 1673.6182.5Tf = 3168.1

182,5Tf

182,5= 3168,1

182.5Tf = 17.4oC

Sample Problems 6: J ↔ c and c ↔ C

a. 9.67 c = J

9.67 c 4.184 J 1 c

= 40.5 J

b. 3588 J = c

3588 J 1 c___ 4.184 J

= 857.6 c

c. 9.43 C = c

9.43 C 1000 c 1C

= 9430 c

Practice Problems 6:

a. 19.6 c = J 19.6 c 4.184 J 1 c

= 82.0 J

b. 3.44 J = c 3.44 J 1 c__ 4.184 J

= 0.82 c

c. 11 864 c = C 11 864 c 1 C__ 1 000 c

= 11.864 C

d. 24.1 C = c 24.1 C 1000 c 1 C

= 24 100 c

e. 378.2 J = C 378.2 J 1 c 1 C__ 4.184 J 1000 c

= 0.0904 C

CALORIMETRY

Calorimetry is the accurate and precisemeasurement of heat exchange for chemicaland physical processes.

The insulated device used to measure the absorption or release of heat in a chemical or physical processesis called a calorimeter.

Calorimeters may be as simple as a styrofoam cup andlid or a complex device such as a bomb calorimeter.

Calorimetry continued:

Enthalpy (H) is the heat content ofa system at constant pressure.

Because most chemical reactions and physical changescarried out in the laboratory are open to the atmosphere, these changes occur at constant pressure.

Therefore:

q = H = m(T)Cp

THERMOCHEMICAL EQUATIONS

If you mix calcium oxide with water, an exothermic reaction takes place and the water becomes warm.

When 1 mol of calcium oxide reacts with 1 mol of water, 1 mol of calcium hydroxide forms and 65.2 kJ of heat isreleased.You can show this in the chemical equation by including heat change as a product of the reaction.

CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

An equation that includes the heat change is called a thermochemical equation.

Thermochemical Equations continued:

A heat of reaction is the heat change for the equation exactly as it is written.

It is usually reported as H, the heat change at constant pressure (the change in enthalpy).

In a thermochemical equation, the physical state ofthe reactants and products must also be given.

Thermochemical Reactions continued:

In an exothermic reaction, heat is given off.This means, according to the law of conservation of energy, that the chemical potential energy of the resulting product(s) isless than the energy of the reactant(s). Because of thisloss of energy, the heat of reaction (H) is negative.

CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

H = -65.2 kJ

Thermochemical Equations continued:

The decomposition of baking soda, sodiumhydrogen carbonate, by heat is an endothermic reaction.The products have more chemical potential energy than the reactants therefore the H is positive.

2NaHCO3(s) + 129 kJ → Na2CO3(s) + H2O(g) + CO2(g)

H = 129 kJ

Mass – Energy Stoichiometry

The mass – energy relationships within achemical equation can be determined in a mannersimilar to mass – mass stoichiometry.

The steps to mass – energy stoichiometry are:

1. Balance the equation

2. Convert grams of given substance to moles.

3. Give the ratio of energy to moles of given substance.

4. Solve

Mass – Energy Stoichiometry continued:

Using the reaction below, what amount of energywill be produced by the combination of 155 g of carbon monoxide (CO) with a excess of iron (III) oxide?

Fe2O3(s) + CO(g) → Fe(s) + CO2(g) + 26.3 kJ

Step 1. Balance the equation

3 2 3

Step 2. Convert grams of given to mols

155 g CO 1 mol CO

28.0 g CO

Step 3. Give ratio of mols given substance to energy

26.3 kJ 3 mol CO

Step 4. Solve

= 48.5 kJ

Mass – Energy Stoichiometry continued:

Practice Problem:

Gasohol contains ethanol (C2H5OH)(l), which when burnedreacts with oxygen to produce CO2(g) and H2O(g). How much heat is released when 12.5 g of ethanol burns?

C2H5OH(l) + O2(g) → CO2(g) + H2O(g) + 1235 kJ3 2 3

12.5 g C2H5OH 1 mol C2H5OH 46.1 g C2H5OH

1235 kJ 1 mol C2H5OH

= 334.9 kJ

HEATS OF FUSION AND SOLIDIFICATION

All solids absorb heat as they melt to becomeliquids.

The heat absorbed by one mole of a substance inmelting from a solid to a liquid at constant temperatureis the molar heat of fusion (Hfus).

The heat lost when one mole of liquid solidifies at a constant temperature is the molar heat of solidification (Hsolid).

Hfus = - Hsolid

Heats of Fusion and Solidification of Water

The melting of one mole of ice to one mole of water at 0oC requires the absorption of 6.01 kJ of heat.

Hfus = 6.01 kJ

The solidification of one mole of water at 0oC to one mole ofice at 0oC releases 6.01 kJ.

Hsolid = - 6.01 kJ

Heat of Fusion – Sample Problem:

How many grams of ice at 0oC and 101.3 kPacould be melted by the addition of 2.25 kJ of heat?

18.0 g ice 1 mol ice

2.25 kJ 1 mol ice 6.01 kJ

= 6.74 g ice

Heat of Fusion – Practice Problem:

How many grams of ice at 0oC and 101.3 kPacould be melted by the addition of 0.400 kJ of heat?

18.0 g ice 1 mol ice

0.400 kJ 1 mol ice 6.01 kJ

= 1.2 g ice

HEAT OF VAPORIZATION AND CONDENSATION

The amount of heat necessary to vaporize one mole of a given liquid is called its molar heat of vaporization (Hvap).

The molar heat of vaporization of water is 40.7 kJ/mol.

This means, to vaporize one mole of water, 40.7 kJ of energymust be provided.

The amount of heat released when one mole of vaporcondenses is called the molar heat of condensation,Hcond.

The molar heat of condensation for water vapor is – 40.7 kJ.

Heat of Vaporization Practice Problem:

How much heat (in kJ) is absorbed when 24.8 g H2O(l)at 100oC is converted to steam at 100oC?

24.8 g H2O 1 mol H2O 18.0 g H2O

40.7 kJ 1 mol H2O

= 56.1 kJ

Heat of Vaporization Practice Problem:

How much heat (in kJ) is absorbed when 63.7 g H2O(l)at 100oC is converted to steam at 100oC?

63.7 g H2O 1 mol H2O 18.0 g H2O

40.7 kJ 1 mol H2O

= 144.0 kJ

HEAT OF SOLUTION

The heat change caused by dissolution of one mole of substance in the molar heat of solution (Hsoln).

Sodium hydroxide is a good example of an exothermicmolar heat of solution. When 1 mol of sodium hydroxide(NaOH)(s) is dissolved in water, the solution can become so hot that it steams. The heat is released as the sodium ions and the hydroxide ions separate and interact with the water.

The temperature of the solution increases, releasing 445.1 kJ of heat as the molar heat of solution (Hsoln = - 445.1 kJ/mol)

A practical application of exothermic heat of solution is thepopular hot packs to relieve pain.

Heat of Solution continued:

The dissolution of ammonium nitrate (NH4NO3)(s) is an example of an endothermic process. When ammonium nitrate dissolves in water, the solutionbecomes very cool.

The heat absorbed as the ammonium and nitrate ionsof one mole of ammonium nitrate separate is the molarheat of solution (Hsoln = 25.7 kJ/mol)

Heat of Solution Sample Problem:

How much heat (in kJ) is released when 100.0 g ofNaOH(s) is dissolved in water?

100.0 g NaOH 1 mol NaOH 40.0 g NaOH

-445.1 kJ 1 mol NaOH

= - 1112.8 kJ

Heat of Solution Practice Problem:

How many grams of NH4NO3(s) must be dissolved in water so that 88.0 kJ of heat is released from the water? (The Hsoln of NH4NO3 = 25.7 kJ/mol)

88.0 kJ 1 mol NH4NO3

25.7 kJ 80.0 g NH4NO3

1 mol NH4NO3

= 273.9 g NH4NO3

CALCULATING HEAT CHANGES

HESS’S LAW

Hess’s Law makes it possible to measure a heat of reactionindirectly.

Hess’s law of heat of summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reactions togive the final heat of reaction.

Hess’s Law continued:

Carbon in the diamond form is less stable thanthe graphite form therefore you would expect the followingreaction to take place:

C(diamond) → C(graphite)

This enthalpy change is too slow to be measured directly.

You can use Hess’s law to find the enthalpy changes for the conversion of diamond to graphite by using thefollowing combustion reactions:

Hess’s Law continued:

a. C(s, graphite) + O2(g) → CO2(g) H = -393.5 kJb. C(s, diamond) + O2(g) → CO2(g) H = -395.4 kJ

Write equation a in reverse to give:

c. CO2(g) → C(s, graphite) + O2(g) H = 393.5 kJ

Note: When you reverse the reactions the sign for H changes.

Adding the equations b and c you get the equation for theconversion of diamond to graphite.

C(s,diamond) + O2(g) → CO2(g) H = - 395.4 kJ

CO2(g) → C(s, graphite) + O2(g) H = 393.5 kJ

C(s, diamond) → C(s, graphite) H = - 1.9 kJ

Hess’s Law continued:

If you wished to determine the enthalpy changefor the formation of carbon monoxide from its elements,the reaction would be:

C(s, graphite) + ½ O2(g) → CO(g) H = ?

During this reaction carbon dioxide (CO2) is also produced sothat the measured heat of reaction relates to both.Using Hess’s law and two reactions which can be carried outin the laboratory we have:a. C(s, graphite) + O2(g) → CO2(g) H = - 393.5 kJ

b. CO(g) + ½O2(g) → CO2(g) H = - 283.0 kJ Writing the reverse reaction of equation b and changing the sign of H gives:

c. CO2(g) → CO(g) + ½O2(g) H = 283.0 kJ

Hess’s Law continued:

Adding equations a and c yields:

C(s, graphite) + O2(g) → CO2(g) H = - 393.5 kJ

CO2(g) → CO(g) + ½O2 H = 283.0 kJ

C(s, graphite) + ½O2 → CO(g) H = - 110 .5 kJ

The formation of CO(g) is exothermic; 110.5 kJ of heatis given off when 1 mol of CO(g) is formed from its elements.

STANDARD HEATS OF FORMATION

Sometimes it is hard to measure the heatchange for a reaction. In such cases you can calculate the heat of reaction from standard heats of formation.

The standard heat of formation (Hf0) of a compound

is the change in enthalpy that accompanies the formationof one mole of a compound from its elements with allsubstances in their standard state at 25oC.

Standard Heats of Formation continued:

The Hf0

of a free element in its standard state is arbitrarily set at zero. For example, the Hf

0 = 0 for the

diatomic molecules H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and I2(s).Many values of Hf

0 have been measured. The following table lists Hf

0 for some common substances. Standard heatsof formation of compounds are handy for calculating heats of reaction at standard conditions.

The standard heat of reaction (H0) is the difference betweenthe standard heats of formation of all reactants and products. This relationship can be expressed by the following equation.

H0 = Hf0(products) - Hf

0(reactants)

Standard Heats of Formation (Hf

0) at 25oC and 101.3 kPa

Hf0 Hf

0 Hf0

Substance (kJ/mol) Substance (kJ/mol) Substance (kJ/mol)

Al2O3(s) - 1676.0 Fe(s) 0.0 NO(g) 90.37

Br2(g) 30.91 Fe 2O3(s) - 822.1 NO2(g) 33.85

Br2(l) 0.0 H2(g) 0.0 Na2CO3(s) - 1131.1

C(s, diamond) 1.9 H2O(g) - 241.8 NaCl(s) - 411.2

C(s, graphite) 0.0 H2O(l) - 285.8 O2(g) 0.0

CH4(g) - 74.86 H2O2(l) - 187.8 O3(g) 142.0

CO(g) - 110 .5 HCl(g) - 92.31 P(s, white) 0.0CO2(g) - 393.5 H2S(g) - 20.1 P)s, red) - 18.4

CaCO3(s) -1207.0 I2(g) 62.4 S(s, rhombic) 0.0

CaO(s) - 635.1 I2(s) 0.0 S(s, monoclinic) 0.30

Cl2(g) 0.0 N2(g) 0.0 SO2(g) - 296.8

F2(g) 0.0 NH3(g) - 46.19 SO3(g) - 395.7

   

Standard Heats of Reaction continued:

What is the standard heat of reaction (ΔH0)for the reaction of gaseous carbon monoxide withoxygen to form gaseous carbon dioxide?

1. Write and balance the equation.

CO + O2 → CO2 2 2

2. Find and sum the ΔHf0 of all reactants, taking into

account the number of moles of each.

2 mol CO(g) - 110.5 kJ 1 mol CO(g)

= - 221.0 kJ

1 mol O2(g) 0 kJ 1 mol O2(g)

= 0 kJ

- 221.0 kJ

3. Find the ΔHf0 of the product

2 mol CO2 - 393.5 kJ 1 mol CO2

= - 787.0 kJ

4. Find the difference between ΔHf0(products)

and ΔHf0(reactants).

ΔH0 = (- 787.0 kJ) – ( - 221.0 kJ)

ΔH0 = - 566.0 kJ