Chapter 6

Chemistry and Reactivity 4th Kotz

bulletThermodynamics is the study of energy changes that occur during physical and chemical processes
bulletThermochemistry is concerned with how we observe, measure, and predict energy changes for both physical changes and chemical reactions


bulletThe capacity to do work or transfer heat.
bulletPotential Energy + Kinetic Energy
bulletw = F x d

Energy and its Units

Energy: the capacity to do work

bulletKinetic Energy : the energy of motion
bulletEkinetic = 1/2mv2
bulletPotential Energy : energy by virtue of position or composition Ep = mgh

Other "Forms" of Energy

bulletRadiant Energy
bulletElectromagnetic Radiation
bulletThermal Energy
bulletAssociated with random motion of a molecule or atom
bulletChemical Energy
bulletEnergy stored within the structural limits of a molecule or atom

Units of Energy

bulletThe SI unit of energy is the joule(J) and is kg.m2/s2 .
bulletThe calorie (cal) is a non- SI unit and is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius.
bullet1 cal = 4.184 J (exact definition)


bulletJoule  E = 1/2 mv2 = kg m2/s2
bullet1 Joule = .239 calories Calorie = 1 g of water changing 1Co
bullet1 calorie = 3.965 x 10-3 btu
bulletHow many Joules = 1 btu?

Dimensional Analysis?

1 btu x 1 calorie / 3.965 x 103- btu x 1 Joule / .239 calories = 1060 Joules

Law of Conservation of Energy

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

Heat of Reaction

bulletThe energy that flows into or out of a system because of a difference between the thermodynamic system and its surroundings
bulletHeat is denoted by the symbol q

Heat of Reaction

bulletExothermic: overall process releases energy (q is negative)
bulletEndothermic: overall process absorbs energy (q is positive)

Heat of Reaction

bulletTherefore, the Heat of Reaction (at a given temperature) is the value of q required to return a system to the given temperature at the completion of the reaction.

First Law of Thermodynamics

bullet"The Energy in the Universe is Constant" DE = q + w
bulletq = heat ; w = work
bulletSystem : everything that is being included in the study
bulletSurroundings : everything in the environment around the system
bulletUniverse : the system plus its surroundings 


bullet"out of" a system
bulletDE < 0
bullet"into" a system
bulletDE > 0


bullet"out of" a system
bulletDH (-)
bullet"into" a system
bulletDH (+)

Enthalpy - H

bulletA chemist’s description for heat gained (+) or lost (-) by a closed system (constant pressure)      DH = H(final) - H(initial)
bulletA state function - doesn’t care where it has been or where it is going

Enthalpy is an Extensive Property

bulletIf one mole of isopropanol burns in air with DH = - 1987 kJ, what is the change in enthalpy for 50 g of isopropanol? (Hint MW = 60)
bullet2 C3H7OH (l) + 9 O2 (g) ==> 6 CO2 (g) + 8 H2O

DH = - 1987 kJ / 1 mol x 1mol / 60.0 g x 50.0 g = -1660 kJ

Heat Capacity

bulletThe heat absorbed by the system is a function of the mass and the ability to absorb heat.

Two Variations

bulletSpecific Heat Capacity
bulletJoules/ Co g
bulletMolar Heat Capacity
bulletJoules/ Ko mole

Calculating q

bulletWhen the temperature of a material is changed, the amount of heat involved in bringing about the change is represented by the equation
bulletq = s x m x Dt where
bullets = specific heat   m = mass  and    D t = tf - tI

Example 6.5 Ebbing pg 243

bulletCalculate the heat absorbed by 15.0g of water to raise its temperature from 20.0 oC to 50.0 oC (at constant pressure). The specific heat of water is 4.18 J/(g oC).
bulletq = s x m x D t
bulletq = 4.18J/goC x 15.0g x (50.0oC - 20.0oC)
bulletq = 1.88 x 103 J


bulletBy running a reaction in water, we can use the specific heat of water and the observed temperature change to measure DH for the system.
bulletThe device used to maintain an insulated closed system for measuring calories is called a calorimeter.


bulletApplying the Law of Conservation of Energy, it is noted that
bulletqwater = qrxn
bulletand that the heat needed to change the temperature of the calorimeter and water is usually the same for a given set up and is often given as the heat capacity (kJ/oC)

Calorimetry: Example, Problem 6.49 page 266

bulletYou wish to make coffee. How much heat (in joules) must be used to raise the temperature of 0.180kg of tap water (enough for one cup of coffee) from 15oC to 96oC? Assume the specific heat is that of pure water.
bulletq = 4.18 J/goC x 180.g x (96oC - 15oC)
bulletq = 6.1 x 104 J

Calorimetry: Example, Problem 6.54 page 266

bulletWhen 23.6 g of calcium chloride, CaCl2, was dissolved in water in a calorimeter, the temperature rose from 25.0oC to 38.7oC. If the heat capacity of the solution and calorimeter is 1258 J/oC, what is the enthalpy change when 1 mol of calcium chloride dissolves in water?
bulletFirst, calculate the heat gained by the solution
bulletqsoln = (1258J/oC) (38.7oC-25.0oC) =   1.723 x 104 J [this is equal to the heat lost by the dissolving of 23.6 g of CaCl2] . This ratio can then be used as a conversion factor.
bullet1.723 x 104 J this is equal to the heat lost by the dissolving of 23.6 g of CaCl2 . This ratio can then be used as a conversion factor.
bulletSecond, calculate the number of grams in one mole :
bullet1.000 mol CaCl2 x (110.98g CaCl2 / 1 mol CaCl2)   = 110.98g
bulletThen use the conversion factor from the first step to convert from grams to Joules
bullet110.98g x (1.723 x 104 J / 23.6 g CaCl2) =  8.102 x 104 J written as -8.102 x 104 J                                                                                                                   (heat lost)

Bond Energies/Strengths

bulletThe Bond Dissociation Energy is the DH for the bond breaking reaction in the gas phase.
bulletCH4 (g) ==> C (g) + 4 H (g) DH = 1660 kJ

                        Bond Energy             kJ/mole

                        C-H                              413

                        C-C                              348

                        C=C                              614

                        C-Br                             276

Hess’s Law

bullet"The change in enthalpy is the same regardless of the steps taken to get there."

Enthalpy Diagrams

(not shown)

Oxidation of Nitrogen

bulletWhat is the enthalpy change for the reaction of oxygen and nitrogen?
bulletN2 (g) + 2 O2 (g) => 2 NO2 (g)
bulletN2 (g) + O2 (g) => 2 NO (g)DH = 180 kJ
bullet2 NO (g) + O2 (g) => 2 NO2 (g) DH = -112 kJ
bulletN2 (g) + 2 O2 (g) => 2 NO2 (g) DH = 68 kJ

Path Independent

(drawing not shown)

Hess’s Law

bulletHydrogen peroxide, H2O2, is a colorless liquid. Solutions of H2O2 are used as bleach and antiseptics. It can be prepared in a process whose overall change is
bulletHydrogen peroxide, H2O2, is a colorless liquid. Solutions of H2O2 are used as bleach and antiseptics. It can be prepared in a process whose overall change is
bulletH2(g) + O2(g) H2O2(l)    target equation
bulletCalculate the enthalpy change using the following data:
bulletH2O2(l) H2O(l) + 1/2O2(g) ; D H = - 98.0 kJ
bullet2H2(g) + O2(g) 2H2O(l) ; D H = -571.6 kJ

Change equations so that the reactants of the target equation are on the reactant side and the products of the target equation are on the product side. Note that reversing the direction of the reaction requires changing the sign on the D H value. When it is necessary to divide the coefficients in order to get them the same as the target equation, the D H value must be divided by the same number.

        H2O(l) + 1/2O2(g) + H2O2(l) ;                         D H =     (- 98.0 kJ x (-1) )

bulletH2(g) + 1/2O2(g) H2O(l) ;                          D H    =     (-571.6 kJ) / 2

Finally, add up the equations canceling all species that are the same on both sides and add up the heat values.

H2(g) + O2(g) H2O2(l) ;                                        D H = -187.8 kJ

Calculating Enthalpy; Problem 6.60 page 267

bulletGiven target equation:
bulletCH4(g) + NH3 --> HCN(g) + 3H2(g)
bulletUse the given equations and heat information to calculate D H for the target equation. Remember: when the given equations are combined, like values on opposite sides of the equation should cancel until only the target equation remains.

Calculating Enthalpy; Problem 6.60 page 267

bulletGiven target equation:
bulletCH4(g) + NH3 --> HCN(g) + 3H2(g)
bulletUsing the following:
bulletN2(g) + 3H2(g) ---> 2NH3(g); D H = -91.8kJ
bulletC(graphite)+ 2H2(g) ---> CH4(g); D H = -74.9kJ
bulletH2(g) + 2C(graphite) +N2(g) ---> 2HCN(g) ; D H = +270.3kJ
bulletNote which molecules are reactants and those that are products in the target equation.

Calculating Enthalpy; Problem 6.60 page 267

bulletGiven target equation:
bulletCH4(g) + NH3 --> HCN(g) + 3H2(g)
bulletRearrange the equations so that the target reactants are on the reactant side and the target products are on the product side.
bullet2NH3(g) ---> N2(g) + 3H2(g) ;                                         D H = +91.8kJ
bulletCH4(g) ---> C(graphite)+ 2H2(g);                                  D H = +74.9kJ
bulletH2(g) + 2C(graphite) +N2(g) ---> 2HCN9g) ;               D H = +270.3kJ
bulletWhen R&P are reversed the D H changes too!!

Calculating Enthalpy; Problem 6.60 page 267

bulletNext divide or multiply as needed to have the same number of moles as the target equation.
bullet1CH4(g) + 1NH3 --> HCN(g) + 3H2(g)
bullet1NH3(g) ---> 1/2N2(g) + 3/2H2(g) ;                                      D H = +45.9kJ
bullet1CH4(g) --->C(graphite)+2H2(g);                                          D H = +74.9kJ
bullet1/2H2(g) + 1C(graphite) + 1/2N2(g) ---> 1HCN9g) ;        D H = +135.2kJ

Calculating Enthalpy; Problem 6.60 page 267

bulletCombine equations after canceling like terms and add up the D H values.
bullet1NH3(g) ---> 1/2N2(g) + 3/2H2(g) ;                              D H = +45.9kJ

1CH4(g) -->1C(graphite)+2H2(g);              D H = +74.9kJ

bullet        leaves 1 H2
bullet1/2H2(g) + 1 C(graphite) + 1/2N2(g) ---> 1HCN9g) ;              D H = +135.2kJ

Calculating Enthalpy; Problem 6.60 page 267

bulletAfter canceling like terms, the molecules left should match the target equation.
bullet1NH3(g) ---> 1/2N2(g) + 1 H2(g) ; D H = +45.9kJ
bullet1CH4(g) -->1C(graphite)+2H2(g); D H = +74.9kJ
bullet1/2H2(g) + 1 C(graphite) + 1/2N2(g) ---> 1HCN9g) ; D H = +135.2kJ
bulletPutting the parts together and adding up D H
bulletCH4(g) + NH3(g) --> HCN(g) + 3H2(g);                  D H = 256.0kJ

Enthalpies of Formation
-Heat of Formation

bulletIn order to standardize a set of data, we have established a set of conditions under which DHfo are reported.
bulletStandard State - most stable form of an element at STP
bulletSTP - standard temperature and pressure
bulletP = 105 Pascals = 1 atm
bulletT = 273 K (0 Co)

Elemental Heats of Formation

bulletFor an element, the heat of formation is defined as 0.0 under STP conditions.

To Calculate DHo Reaction

bulletDHorxn = S nDHfo(prod) - S mDHfo(react)
bulletRemember to change the sign of DHfo if the reaction is reversed.
bulletIf the formation reaction must be multiplied by an integer, DHfo must be as well.
bulletElements in their standard states don’t contribute to DHo.

An example -
Alternative Automobile Fuels

bulletMethanol is sometimes used as a fuel alternative to octane (gasoline) in high performance race cars. Determine the enthalpy of combustion per gram of each.
bulletDHfo CH3OH (l) = -239 kJ/mole
bulletDHfo C8H18 (l) = -269 kJ/mole
bulletDHfo CO2 (g) = -394 kJ/mole
bulletDHfo H2O (l) = -286 kJ/mole

Methanol Combustion

bulletCH3OH (l) + O2 (g) => CO2 (g) + H2O (l)
bulletBalance the equation:

Methanol Combustion

bullet2CH3OH (l) + 3O2 (g) => 2CO2 (g) +4H2O(l)


bulletDHoreaction = S nDHfo(prod) - S mDHfo(react)
bullet=      2 x DHfo (CO2) + 4 x DHfo (H2O)
bullet     - 2 x DHfo(CH3OH) - 3 x DHfo (O2)

Methanol Combustion

bullet2CH3OH (l) + 3O2 (g) => 2CO2 (g) +4H2O (l)
bulletDHoreaction = S nDHfo(prod) - S mDHfo(react)
bullet=    2 x DHfo (CO2) + 4 x DHfo (H2O)
bullet     - 2 x DHfo (CH3OH)    note that DHfo (O2) = 0 by definition and is not shown here
bullet=    2 x (-394 kJ)  +   4 x (-286 kJ)
bullet    - 2 x (-239 kJ)
bullet=  -1454 kJ/ 2 moles CH3OH

Conversion to DHfo/g

bulletDHoreaction/g = -1454 kJ/ 2 moles CH3OH
bullet= -1454 kJ/[2 moles x (32.0 g/mol)]
bullet= -22.7 kJ/g

Standard Enthalpies of Formation

bulletHydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. The gas burns in oxygen as follows:2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g)
bulletCalculate the standard enthalpy change for this reaction using standard enthalpies of formation.

Standard Enthalpies of Formation

2H2S(g)   +    3O2(g)     2H2O(l)   +      2SO2(g)

2(-20)          3(0)           2(285.8)       2(-296.8) kJ

Standard Enthalpies of Formation

DHo = Sn DHo (products) - Sm DHo (reactants)

= [2(-285.8) + 2(-296.8)]-[2(-20) + 3(0)] kJ

=-1125.2 = -1125 kJ

The Benefits of Thermochemistry

bulletA means of quantifying chemical energy
bulletThe first law of thermodynamics
bulletAll energy must be accounted for
bulletHess’s Law
bulletKnown reactions can be used to determine energetics of composite, unknown reactions

Heat of Phase Transition

Heat of Fusion

"The heat necessary to melt one mole of a solid."

DHfus = 6.01 kJ/mol for ice

Heat of vaporization

bulletThe heat needed for the vaporization of a liquid. Also called the enthalpy of vaporization.
bulletFor water at 100oC the heat of vaporization is 40.7 kJ/mol

Heat of a Phase Change


A Question of Energy

bulletWhat parameters do you need to calculate the energy necessary to raise the temperature of water from -20 oC to 120 oC?
bulletg H2O, Cwater, Cice, Csteam, b.p. of water, m.p. of water, DHfus, DHvap


A particular refrigerator cools by evaporating liquified dichlorodifluoromethane, CCl2F2 . How many kilograms of this liquid must be evaporated to freeze a tray of water at OoC to ice at OoC ? The mass of the water is 525 g, the heat of fusion of ice = 6.01kJ/mol and the heat of vap. of CCl2F2 = 17.4 kJ/mol.

Example Solution (a)

bulletThe heat that must be removed to freeze 525 g of water at OoC is
bullet525g H2O      x     1 mol H2O   x      -6.01kJ    
bullet                              18.0 g H2O          1 mol H2O
bullet= -175kJ
bulletNote: the negative sign is used because the water is losing heat.


Solution (b)

Since the water must lose - 175 kJ , the CCl2F2 must gain +175 kJ. The mass of CCl2F2 that must be vaporized to absorb this quantity of heat is

bullet175 kJ    x      1 mol CCl2F2   x      121 g CCl2F2
bullet                           17.4 kJ                  1 mol CCl2F2
bullet= 1.22 x 103 g CCl2F2     =    1.22 kg CCl2F2


The heat of vaporization of ammonia is 23 .4 kJ/mol. How much heat is require to vaporize 1.00 kg of ammonia ? How many grams of water at OoC could be frozen to ice at OoC by the evaporation of this amount of ammonia ?

Exercise Solution (a)

bullet1.00 kg NH3       x     1000 g    x    1 mol NH3
bullet                                   1 kg              17.04 gNH3
bullet= 58.68 mol NH3
bullet58.68 mol NH3       x      23.4 kJ
bullet                                    1 mol NH3
bullet= 1373.11 kJ needed for the vaporization

Exercise  Solution (b)

bulletThe mass of water that could be frozen by 1373.11 kJ of heat removed by the vaporization of 1.00 kg NH3 is
bullet1373.11 kJ      x      1 mol H2O    x    18.02 g H2O
bullet                                  6.01 kJ               1 mol H2O
bullet= 4120 g H2O could be frozen to ice