THERMOCHEMISTRY:

ENERGY FLOW AND CHEMICAL FLOW

In general…

All matter contains energy, so whenever matter undergoes a change, the quantity of energy that the matter contains also changeso When a burning candle melts a piece of ice

Higher energy reactants, wax and oxygen, form lower energy products, carbon dioxide and water

The difference is released as heat and light Some of the heat is absorbed when the lower

energy ice becomes higher energy watero In a thunderstorm

Lower energy N2 and O2 absorb energy from lightning and form higher energy NO, and the higher energy water vapor releases energy as it condenses to lower energy water that falls as rain

o Everyday examples Fuels such as oil and wood release energy to heat

our houses Fertilizers help convert solar energy into food Metal wires increase the flow of electrical energy Polymer fibers in winter clothing limit the flow

of thermal energy away from our bodies

Thermodynamics – the study of energy and its transformations (will be addressed in chapter 20 as well for students taking CHEM 102!)

Thermochemistry – the branch of thermodynamics that deals with heat in chemical and physical change

Forms of Energy and Their InterconversionsWhen energy is transferred from one object to another, it appears as work and/or heat

Defining the System and SurroundingsSystem – part of the universe we are focusing on

Surroundings – everything that is not the system

In the above picture, the contents of the flask is the system and everything thing else (even the flask) is the surroundings

Energy Transfer to and From a SystemInternal Energy, E, is the sum of the potential and kinetic energy found in each particle in a system

When the reactants in a chemical system change to products, the system’s internal energy changeso This is represented by ∆E

∆E = Efinal - Einitial = Eproducts - Ereactants

A change in the energy of the system must be accompanied by an equal and opposite change in the energy of the surroundings

A system can change its internal energy in one of two wayso By releasing some energy in a transfer to the

surroundings (diagram A below) Efinal < Einitial so that ∆E < 0

o By absorbing some energy in a transfer from the surroundings (diagram B below)

Efinal > Einitial so that ∆E > 0

∆E is a transfer of energy from system to surroundings, or vice versa…

Heat and Work: Two Forms of Energy TransferEnergy transferred from system to surroundings or vice versa appears in two forms…

Heat. Heat or thermal energy (symbolized as q) is the energy transferred as a result of a difference in temperature between the system and the surroundings. Movement from warmer system/surroundings to cooler system/surroundings.

Work. All other forms of energy transfer involve some type of work (w), the energy transferred when an object is moved by a force.

The total change in a system’s internal energy, E, is the sum of the energy transferred as heat and/or work.

∆E = q + wo The values of q and w can have either positive or

negative signs We define the sign (+ or -) of the energy change

from the system’s perspective Energy transferred into the system is

positive because the system ends up with more energy

Energy transferred out from the system is negative because the system ends up with less energy

Energy Transferred as Heat Only

Heat flowing out of a system… (example A)o q is negative, so ∆E is negative

Heat flowing into a system… (example B)

o q is positive, so ∆E is positive Thermodynamics in the kitchen…

o The air is your refrigerator (surroundings) has a lower temperature than a newly added piece of food (system)

Food releases energy to the surroundings, so q < 0 (system loses energy)

o The hot air in an oven (surroundings) has a higher temperature than a newly added piece of food (system)

Food is absorbing energy from the surroundings, so q > 0 (system gained energy)

o Just to clarify… your favorite beverage in a cooler full of ice makes the ice warmer (the ice does not make your beverage cooler)

The beverage is the system and the cooler is the surroundings. The beverage is releasing heat to the surroundings. So… q < 0 (system lost energy)

Energy Transferred as Work Only

Work done by a system… (in the above figure)o The piston is pushed out by the system, so w is

negative and ∆E is negative

Work done on a system… (no picture) so… imagine the piston being pushed in by the surroundingso w is positive and ∆E is positive

Law of Conservation of EnergyEnergy changes form (from chemical to kinetic to potential for example) but does not simply appear or disappear – energy cannot be created or destroyed.

Law of conservation of energy (first law of thermodynamics) – the total energy of the universe is constanto ∆Euniverse = ∆Esystem + ∆Esurroundings

Units of Energy The SI unit of energy is the joule (J) The calorie is an older term originally defined as the

quantity of energy required to raise the temperature of 1 gram of water 1 oCo Now defined in terms of the joule…

1 calorie = 4.184 J 1 J = 0.2390 calories

1 kJ = 1000 J = 0.2390 kcal = 239.0 calories British thermal unit (Btu) is a unit used for energy output

of applianceso Quantity of energy required to raise the temperature of

1 lb. of water 1oF 1 Btu = 1055 J

Sample Problem 6.1 p. 234Determining the Change in Internal Energyq = -325 J (system releases 325 J to surroundings)

w = -451 J (system does 451 J on the pistons)

∆E = q + w

∆E = -325 J + (- 451 J)

∆E = - 776 J

- 776 J 1 kJ = - 0.776 kJ1000 J

- 0.776 kJ 1 kcal = - 0.185 kcal4.184 kJ

Follow-Up Problem 6.1 p. 23426.0 kcal 4.184 kJ = 108.784 kJ

1 kcal

15.0 Btu 1055 J 1 kJ = 15.825 kJ1 Btu 1000 J

q = -108.784 kJ (absorbed by surroundings)

w = 15.825 kJ (work done on the system)

∆E = q + w

∆E = -108.784 kJ + 15.825 kJ

∆E = - 92.959 kJ

∆E = - 93.0 kJ

State Functions & the Path Independence of the Energy ChangeThe internal energy (E) of a system is called a state function

A property dependent only on the current state of the system, not on the path the system takes to reach that state.

∆E is also a state function because it does not depend on how the change takes place, but only on the difference between the final and initial states.

q and w are not state functions because their values do depend on the path the system takes

In the above example, q and w for the two paths are different, ∆E is the same

Pressure (P), volume (V), and temperature are other state functions

Path independence means that changes in state functions - ∆E, ∆P, ∆V, and ∆T – depend only on the initial and final states.

Enthalpy: Chemical Change at Constant PressureTwo most important types of chemical work…

Electrical worko Work done by moving charged particles

Pressure-volume worko The mechanical work done when the volume of a

system changes in the presence of an external pressure

The quantity of PV work equals P times the change in V… w = - P∆V (it is negative because the gas is doing work on the surroundings

At constant pressure, a thermodynamic variable called enthalpy (H) eliminates the need to measure PV work. The

enthalpy of a system is defined as the internal energy plus the product of pressure and volume…

H = E + PV

The change in enthalpy (∆H) is the change in internal energy plus the product of the pressure, which is constant, and the change in volume

∆H = ∆E + P∆V

Most important point… the change in enthalpy (∆H) equals the heated absorbed or released at constant pressure.

∆H is more relevant than ∆E and easier to obtain

Comparing ∆E and ∆HKnowing the enthalpy change of a system tells us a lot about its energy change as well

Many reactions involve little (if any) PV work Most (or all) the energy change occurs as a transfer of heat For most reactions, ∆H equals or is very close to ∆E

Exothermic and Endothermic ProcessesBecause H is a combination of the three state functions E, P, and V, it is also a state function

∆H = Hfinal – Hinitial = Hproducts - Hreactants

Exothermic process

An exothermic process releases heat and results in a decrease in the enthalpy of the systemo Hproducts < Hreactants so… ∆H < Oo Heat is found on the product side of the equation

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat Diagram A below is showing an exothermic

reaction

Endothermic process

An endothermic process absorbs heat and results in an increase in the enthalpy of the systemo Hproducts > Hreactants so… ∆H > Oo Heat is found on the reactant side of the equationo Heat + H2O(s) H2O(l)

Sample Problem 6.2 p. 237Drawing Enthalpy Diagrams and Determining the Sign of ∆H

Follow-Up Problem 6.2 p. 237C3H5(NO3)3( l ) (reactant)

∆H = - 5.72x103 kJ Exothermic

3CO2( g ) + 5/2 H 2O( g ) + ¼ O 2( g ) + 3/2 N 2( g) (products)

Calorimetry: Measuring the Heat of a Chemical or Physical ChangeTemperature change depends on the object…

Every object has its own heat capacity, the quantity of heat required to change its temperature 1 Ko Specific heat capacity (c), the quantity of heat

required to change the temperature of 1 gram of an object by 1 K

To calculate the heat absorbed or released…

q = c x mass x ∆To q = heat absorbed or releasedo c = specific heat capacity (J/g K)o mass in gramso ∆T in K

Molar heat capacity – the quantity of heat required to change the temperature of 1 mole of a substance 1 K

Enthalpy

The combination of high specific heat of water and the amount of water that covers the Earth play an important role maintaining an environment that is favorable to life as we know it…

If the Earth had no oceans, it would take one-sixth as much energy for the Sun to heat the rocky surface, and the surface would five off its heat six times as much heat after sundown. Water has a high specific heat which means it takes a lot of energy input to change the temperature and it takes a

considerable loss for the temperature to go down. People that live in Chicago experience this with Lake Michigan as well. (The lake stays cold until well into the summer and stays warm well into the fall and early winter)

Sample Problem 6.3 p. 239Finding the Quantity of Heat from a Temperature Change∆T = Tfinal - Tinitial ∆T = 300. oC – 25oC = 275 oC = 275 K

q = c x mass x ∆T

q = 0.387 J/g K x 125 g x 275 K

q = 1.33x104 J

Follow-Up Problem 6.3 p. 239∆T = Tfinal - Tinitial ∆T = 25.0 oC – 37.0oC = 12.0 oC = -12.0 K

5.50 L 1000 mL 1.11 g = 6050 g1 L 1 mL

q = c x mass x ∆T

q = 2.42 J/g K x 6050 g x -12.0 K

q = - 175,692 J = -176,000 J

-176,000 J 1 kJ = -176 kJ1000 J

The Two Major Types of CalorimetryCalorimeter – a device used to measure the heat released or absorbed by a physical or chemical process

Two types…o One designed to measure the heat at constant pressureo One designed to measure the heat at constant volume

Constant Pressure Calorimetry

Often measured in a coffee-cup calorimeter

Can be used to determine the specific heat capacity of a solid (as long as it does not react with or dissolve in the water)

The solid (system) is weighed, heat

to some known temperature, and added to a known mass and temperature of water (surroundings) in the calorimeter.

After stirring, the final water temperature is measured, which is also the temperature of the solid. Assuming no heat escapes the calorimeter, the heat released by the system (-qsystem or –qsolid) is equal in magnitude but opposite in sign to the heat absorbed by the surroundings (+qsurroundings or +qH2O)

So… -qsolid = qH2O

Substituting for q on both sides of the equation…

-(csolid x masssolid x ∆Tsolid) = cH2O x massH2O x ∆TH2O

Solving for csolid

csolid = - (cH2O x massH2O x ∆TH2O / masssolid x ∆Tsolid)

Sample Problem 6.4 p. 240Determining the Specific Heat Capacity of a SolidFinding ∆Tsolid and ∆Twater

∆Twater = Tfinal - Tinitial

∆Twater = 28.49oC – 25.10oC = 3.39oC = 3.39 K

∆Tsolid = Tfinal - Tinitial

∆Tsolid = 28.49oC – 100.00oC = - 71.51oC = - 71.51K

Solving for csolid

-(csolid x masssolid x ∆Tsolid) = cH2O x massH2O x ∆TH2O

csolid = - (cH2O x massH2O x ∆TH2O / masssolid x ∆Tsolid)

csolid = - (4.184 J/g K x 50.00 g x 3.39 K / 22.05 g x – 71.51K)

csolid = 0.450 J/g K

Follow-Up Problem 6.4 p. 240Finding ∆Tsolid and ∆Twater

∆Twater = Tfinal - Tinitial

∆Twater = 27.25oC – 25.55oC = 1.70oC = 1.70 K

∆Tsolid = Tfinal - Tinitial

∆Tsolid = 27.25oC – 65.00oC = - 37.75oC = - 37.75K

Solving for csolid

-(csolid x masssolid x ∆Tsolid) = cH2O x massH2O x ∆TH2O

csolid = - (cH2O x massH2O x ∆TH2O / masssolid x ∆Tsolid)

csolid = - (4.184 J/g K x 25.00 g x 1.70 K / 12.18 g x – 37.75K)

csolid = 0.387 J/g K [unknown is Copper, Cu]

Sample Problem 6.5 p. 240Determining the Enthalpy(a)

Finding masssolution and ∆Tsolution

masssolution = (25.0 mL + 75.0 mL) x 1.00 g/mL = 75.0 g

∆Tsolution = 27.21oC – 25.00oC = 2.21oC = 2.21K

qsolution = csolution x masssolution x ∆Tsolution

qsolution = 4.184 J/g K x 75.0 g x 2.21K

qsolution = 693 J

(b)

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

25.0 mL HCl 1 L 0.500 mol HCl = 0.0125 mol HCl1000 mL 1L

50.0 ml NaOH 1L 0.500 mol NaOH

= 0.0250 mol NaOH

1000 mL

1L

**HCl is the limiting reactant, so 0.0125 mol HCl is produced

Heat absorbed by the solution was released by the reaction

qsoln = -qrxn = 693 J so qrxn = -693 J

-693 J 1 kJ = -55.4 kJ / mol H2O0.0125 mol H2O 1000 J

Follow-Up Problem 6.5 p. 241Step 1

Calculate moles of reactants

50.0 mL 1L 0.500 mol Ba(OH)2 = 0.025 mol Ba(OH)2

1000 mL 1L

50.0 mL 1L 0.500 mol HCl = 0.025 mol HCl1000 mL 1L

Step 2

Determine the limiting reactant…

0.025 mol Ba(OH)2 2 mol H2O = 0.500 mol H2O1 mol Ba(OH)2

0.025 mol HCl 2 mol H2O = 0.025 mol H2O2 mol HCl

**HCl is the limiting reactant (produces 0.025 mol H2O)

- 1.386 kJ / 0.0250 mol H2O = - 55.4 kJ/mol

Constant-Volume Calorimetry

Constant-volume Calorimetry is often carried out in a bomb calorimeter (a device commonly used to measure the heat of combustion reactions, such as for fuels and foods)

With the much more precise bomb calorimeter, the heat capacity of the entire

calorimeter is known (or can be determined)

Sample Problem 6.6 p. 242Calculating the Heat of a Combustion ReactionStep 1

Calculate ∆T

∆T = ∆Tfinal - ∆Tinitial = 26.799oC – 21.862oC = 4.937oC = 4.937K

Step 2

Calculate the heat absorbed by the calorimeter

When the dessert (system) burns, the heat released is absorbed by the calorimeter

-qsystem = qcalorimeter

qcalorimeter = heat capacity x ∆T

qcalorimeter = 8.151 kJ/K x 4.937 K

qcalorimeter = 40.24 kJ

Step 3

Compare answer to 10 Calories

10 Calories 1kcal 4.184 kJ =41.84 kJ1 Calorie 1 kcal

40.24 kJ < 41.84 kJ The claim is correct…

Follow-Up Problem 6.6 p. 242

Step1

Calculate the amount of heat released (qsample)

0.8650 g C 1 mol C 393.5 kJ = 28.34 kJ12.01 g C 1 mol C

Step 2

Calculate heat capacity of bomb calorimeter

-qsample = qcalorimeter

qcalorimeter = heat capacity x ∆T

28.34 kJ = heat capacity x 2.613 K

Heat capacity = 10.85 kJ/K

Stoichiometry of Thermochemical EquationsThermochemical equation – a balanced equation that includes the enthalpy change of the reaction (∆H)

∆H refers only to the amounts (mol) of substances and their states of matter in that equation

The sign of ∆H depends on whether the reaction is exothermic (-) or endothermic (+)o A forward reaction has the opposite sign of the reverse

reaction 2H2O(l) 2H2(g) + O2(g) ∆H = 572 kJ

Because ∆H is positive, this reaction is endothermic

2H2(g) + O2(g) 2H2O(l) ∆H = -572 kJ

Because ∆H is negative, this reaction is exothermic

The magnitude of ∆H is proportional to the amount of substanceo H2(g) + ½ O2(g) H2O(l) ∆H = -286 kJ

Notice that this is ½ the above reaction

Two key points to understand about thermochemical equations…

When necessary, we use fractional coefficients to balance the equation, because we are specifying the magnitude of ∆H for a particular amount, often 1 mol of substanceo ⅛ S8(s) + O2(g) SO2(g) ∆H = -296.8 kJ

For a particular reaction, a certain amount of substance is thermochemically equivalent to a certain quantity of energy. For example…o 296.8 kJ is thermochemically equivalent to ⅛ mol of

S8

o 286 kJ is thermochemically equivalent to ½ mol of O2

o 286 kJ is thermochemically equivalent to 1 mol H2O

Sample Problem 6.7 p. 243Using the Enthalpy Change of a Reaction to Find Amounts of Substance1.000x103 kJ 2 mol Al 26.98 Al = 32.20 g Al

1676 kJ 1 mol Al

Follow-Up Problem 6.7 p. 243C2H4(g) + H2(g) C2H6(g) ∆H = -137 kJ

15 kg C2H6 1000 g 1 mol C2H6 137 kJ = 6.83x104 kJ1 kg 30.07 g C2H6 1 mol C2H6

Hess’s Law: Finding ∆H of Any ReactionIn some cases, a reaction is difficult, even impossible, to carry out…

May be part of a complex biochemical process May take place under extreme conditions May need a change of conditions to take place

Even if we can’t run a reaction in the lab, we can still find its enthalpy change…

We can find the ∆H for any reaction for which we can write an equation

Hess’s law – the enthalpy change of an overall process is the sum of the enthalpy changes of the individual steps

∆Hoverall = ∆H1 + ∆H2 + …. + ∆Hn

We apply Hess’s Law by…

Imagining that an overall reaction occurs through a series of individual reaction steps, whether or not it actually does. Adding the steps must give the overall reaction.

Choosing individual reaction steps that each have a known ∆H

Adding the known ∆H values for the steps to get the unknown ∆H of the overall reaction. We can also find the unknown ∆H of any step by subtraction (if we know the ∆H values for the overall reaction and all the other steps.

Example of the use of Hess’s law – the oxidation of sulfur to sulfur trioxide

Equation 1: S(s) + O2(g) SO2(g)∆H1 = -296.8 kJ

Equation 2: 2SO2(g) + O2(g) 2SO3(g) ∆H2 = -198.4 kJ

Equation 3: S(s) + 3/2 O2(g) SO3(g) ∆H3 = ????

Need to manipulate equations 1 and/or 2 so that they add up to equation 3

We need to have the same coefficients for SO2 so that they will cancel (this will mean that we have to divide equation 2 by 2)

Equation 2: SO2(g) + ½ O2(g) SO3(g) ∆H2 = -99.2 kJ

Add equations 1 and 2 together…

Equation 1: S(s) + O2(g) SO2(g)∆H1 = -296.8 kJ

Equation 2: SO2( g ) + ½ O 2( g ) SO 3( g ) ∆ H 2 = -99.2 kJ

Equation 3: S(s) + O2(g) + SO2(g) + ½ O2(g) SO2(g) + SO3(g)

∆H1 = -396.0 kJ

Delete SO2 because it is common to both sides Add O2 together on reaction side

Final answer…

S(s) + 3/2 O2(g) SO3(g) ∆H1 = -396.0 kJ

Summarize calculating an unknown ∆ H

1. Identify the target equation, the step whose ∆H is unknown, and note the amount (mol) of each reactant and product

2. Manipulate each equation with known ∆H values so that the target amount (mol) of each substance is on the correct side of the equation. Remember to:

a. Change the sign of ∆H when you reverse an equationb. Multiply amount (mol) and ∆H by the same factor

Sample Problem 6.8Using Hess’s Law to Calculate an Unknown ∆HEquation A: CO(g) + ½ O(g) CO2(g) ∆H = -283.0 kJ

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

Equation C: CO(g) + NO(g) CO2(g) + ½ N2(g) ∆H = ?

Reverse equation B so that N2 is on the correct side of equation and divide by 2 so that you have the correct number of mol of N2 , O2 and NO

Equation B: NO(g) 1/2 N2(g) + ½ O2(g) ∆H = -90.3 kJ

Add equations A & B together…

Equation A: CO(g) + ½ O(g) CO2(g) ∆H = -283.0 kJ

Equation B: NO( g ) 1 / 2 N2( g ) + ½ O 2( g) ∆ H = -90.3 kJ

Equation C: CO(g) + ½ O(g) + NO(g)

CO2(g) + 1/2 N2(g) + ½ O2(g) ∆H = -373.3 kJ

Equation C:

CO(g) + NO(g) CO2(g) + 1/2 N2(g) ∆H = -373.3 kJ

Follow-Up Problem 6.8 p. 245Equation 1: N2O5(s) 2NO(g) + 3/2 O2(g) ∆H = 223.7 kJ

Equation 2: NO(g) + ½ O2(g) NO2 (g) ∆H = -57.1 kJ

Equation 3: 2NO2(g) + ½ O2(g) N2O5(s) ∆H = ???

Reverse equation 1…

Equation 1: 2NO(g) + 3/2 O2(g) N2O5(s) ∆H = -223.7 kJ

Multiply equation 2 x 2 and reverse…

Equation 2: 2NO2( g ) 2NO ( g ) + O 2( g ) ∆ H = 114.2 kJ

Equation 3: 2NO(g) + 3/2 O2(g) + 2NO2(g)

2NO (g) + O2(g) + N2O5(s) ∆H = -109.5 kJ

Equation 3: 2NO2(g) + ½ O2(g) N2O5(s) ∆H = -109.5 kJ

Standard Enthalpies of Reaction - ∆H orxn

We can use Hess’s Law to determine the ∆H values of an enormous number of reactions…

∆H varies somewhat with conditions… so chemist have established a set of specific conditions called standard stateso For a gas, it is 1 atm and ideal behavior

o For a substance that aqueous solution, it is 1 M concentration

o For a pure substance (element or compound), it is usually the most stable form of the substance at 1 atm and the temperature of interest (usually 25oC or 298K)

The standard state symbol (shown as a degree sign) indicates that the variable has been measured with all the substances in their standard stateso Standard enthalpy of formation, ∆H o

f is the enthalpy change if measured at the standard state

Also called the standard heat of formationo C (graphite) + 2H2(g) CH4(g) ∆H o

f = -74.9 kJo Fractional coefficients are often used with reactants to

obtain 1 mol of product Na(s) + ½ Cl2(g) NaCl(s) ∆H o

f = -411.1 kJo Table 6.3 p. 246 shows selected standard enthalpy of

formations at 298K For an element in its standard state, ∆H o

f = 0 The standard state for molecular elements, such

as the halogens, is the molecular form, not separate atoms. An example would be… Cl2 = ∆H o

f = 0 Some elements exist in different forms, but only

one is the standard state… Carbon ∆H o

f = 0, not diamond ∆H of = 1.9

kj/mol Oxygen ∆H o

f = 0, not ozone ∆H of = 143

kJ/mol

Sulfur (S8) ∆H of = 0, not S ∆H o

f = 0.3 kJ/mol

Most compounds have a negative ∆H of

Most compounds have exothermic formation reactions (under standard conditions, heat is released when most compounds form from their elements)

Sample Problem 6.9 p. 246Writing Formation Equations(a)

Ag(s) + ½ Cl2(g) AgCl(s)

∆H of = -127.0 kJ

(b)

Ca(s) + C(graphite) + 3/2 O2 CaCO3(s)

∆H of = -1206.9 kJ

(c)

½ H2(g) + C(graphite) + ½ O2(g) HCN(g) ∆H o

f = 135.0 kJ

Follow-Up Problem 6.9 p. 247

(a)

C(graphite) + H2(g) + ½ O2(g) CH3OH(l) ∆H of = -238.6 kJ

(b)

Ca(s) + ½ O2(g) CaO(s) ∆H of = -635.1 kJ

(c)

C(graphite) + 1/4S8(rhombic) CS2 ∆H of = 87.9 kJ

Determining ∆H orxn from ∆H of Values for Reactants & ProductsWe can use ∆H o

f values to determine ∆H orxn for any reaction

Step 1

Each reactant decomposes to its elements. This is the reverse of the formation reaction for the reactant, so the standard enthalpy change is -∆H o

f

Step 2

Each product forms from its elements. This step is the formation reaction for the product, so the standard enthalpy change is ∆H o

f

Text example… p. 247

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

TiCl4(l) Ti(s) + 2Cl2(g) -∆H of [TiCl4(l)]

2H2O(g) 2H2(g) + O2(g) -2∆H of [H2O(g)]

Ti(s) + O2(g) TiO2(s) ∆H of [TiO2(s)]

2H2( g ) + 2Cl 2( g ) 4HCl( g ) 4∆ H o f [HCl( g )]

TiCl4(l) + 2H2O(g) + Ti(s) + O2(g) + 2H2(g) + 2Cl2(g)

Ti(s) + 2Cl2(g) + 2H2(g) + O2(g) + TiO2(s) + 4HCl(g)

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

It is important to realize that when titanium (IV) chloride and water react, the reactions don’t actually decompose to their elements, which then recombine to form the products. ∆H o

rxn is the difference between two state functions, H o

products minus H o

reactants, so it doesn’t matter how the reaction actually occurs. We simply need to add the individual enthalpy changes to find ∆H o

rxn

∆H orxn = ∆H o

f [TiO2(s)] + 4∆H of [HCl(g)] + {-∆H o

f [TiCl4(l)] + {-2∆H o

f [H2O(g)]}

OR

∆H orxn = {∆H o

f [TiO2(s)] + 4∆H of [HCl(g)]} -

{∆H of [TiCl4(l)] + 2∆H o

f [H2O(g)]}

The standard enthalpy of reaction (∆H orxn) is the sum of the

standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants

∆H orxn = ∑m∆H o

f(products) - ∑n∆H of(reactants)

Sample Problem 6.10 p. 248Calculating ∆H orxn from ∆H of Values4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

∆H orxn = ∑m∆H o

f(products) - ∑n∆H of(reactants)

∆H orxn = {4∆H o

f [NO(g)] + 6∆H of [H2O(g)]} –

{4∆H of [NH3(g)] + 5∆H o

f [O2(g)]}

∆H orxn = {(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)} –

{(4 mol)(-45.9 kJ/mol) + (6mol)(0 kJ/mol)

∆H orxn = [361 kJ + (-1451 kJ)] – [(-184 kJ) + 0 kJ]

∆H orxn = - 906 kJ

Follow-Up Problem 6.10 p. 248

CH3OH(l) + 3/2O2(g) CO2(g) + 2H2O(g) ∆H orxn = -683.5 kJ

∆H orxn = ∑m∆H o

f(products) - ∑n∆H of(reactants)

-683.5 kJ = {(1 mol)(-393.5 kJ) + (2 mol)(-241.8) –

(3/2 mol)(0.0 kJ) + ∆H of [CH3OH(l)]

∆H of [CH3OH(l)] = 638.5 kJ + (-393.5 kJ) + (-483.6)

∆H of [CH3OH(l)] = -238.6 kJ