THERMOCHEMISTRY THERMOCHEMISTRY AND AND THERMODYNAMICSTHERMODYNAMICS

Energy (E)Energy (E)

The ability to do work or The ability to do work or produce heat ; the sum of produce heat ; the sum of all potential and kinetic all potential and kinetic energy in a system is energy in a system is known as the internal known as the internal energy of the systemenergy of the system

Potential energyPotential energy

In chemistry this is usually the In chemistry this is usually the energy stored in bonds (i.e., energy stored in bonds (i.e., when gasoline burns there are when gasoline burns there are differences in the attractive differences in the attractive forces between the nuclei and forces between the nuclei and the electrons in the reactants the electrons in the reactants and the products)and the products)

Kinetic energyKinetic energy

Energy of motion, usually of Energy of motion, usually of particles, proportional to particles, proportional to Kelvin temperature; kinetic Kelvin temperature; kinetic energy depends on the mass energy depends on the mass and the velocity of the object: and the velocity of the object:

KE = ½ mvKE = ½ mv22

Law of Conservation of Law of Conservation of EnergyEnergy

Energy never created nor Energy never created nor destroyeddestroyed

AKAAKA energy of the universe energy of the universe is constantis constant

AKAAKA First Law of First Law of ThermodynamicsThermodynamics

Heat (q)Heat (q)

TTransfer of energy in a process ransfer of energy in a process (flows from a warmer object to a (flows from a warmer object to a cooler one – heat transfers cooler one – heat transfers because of temperature because of temperature difference but, remember, difference but, remember, temperature is not a measure of temperature is not a measure of energy—it just reflects the energy—it just reflects the motion of particles)motion of particles)

Enthalpy (H)Enthalpy (H)Heat content at constant pressureHeat content at constant pressure

– Enthalpy of reaction (Enthalpy of reaction (HHrxnrxn)) – heat absorbed or – heat absorbed or released by a chemical reactionreleased by a chemical reaction

– Enthalpy of combustion (Enthalpy of combustion (HHcombcomb)) -- heat -- heat absorbed or released by burning (usually with Oabsorbed or released by burning (usually with O22))

– Enthalpy of formation (Enthalpy of formation (HHff)) – heat absorbed or – heat absorbed or released when released when ONE ONE molemole of compound is formed of compound is formed from elements in their standard statesfrom elements in their standard states

– Enthalpy of fusion (Enthalpy of fusion (HHfusfus)) -- heat absorbed to -- heat absorbed to melt 1 mole of solid to liquid @MPmelt 1 mole of solid to liquid @MP

– Enthalpy of vaporization (Enthalpy of vaporization (HHvapvap)) -- heat -- heat absorbed to change 1 mole liquid to gas @BPabsorbed to change 1 mole liquid to gas @BP

SystemSystem - Area of the universe we are - Area of the universe we are focusing on (i.e., the experiment)focusing on (i.e., the experiment)

SurroundingsSurroundings - everything outside of - everything outside of the systemthe system

Endothermic Endothermic - net absorption of - net absorption of energy (heat) by the system; energy is energy (heat) by the system; energy is a reactant; (i.e., baking soda and a reactant; (i.e., baking soda and vinegar when mixed get very cold to vinegar when mixed get very cold to the touch)the touch)

A Classic A Classic ENDOthermic reaction!ENDOthermic reaction!

The equation for the reaction is:

Ba(OH)2 8 H2O(s ) + 2 NH4SCN(s ) --> Ba(SCN)2(s ) + 10 H2O(l ) + 2 NH3(g )

ExothermicExothermic – – net release of net release of energy (heat) by the system; energy energy (heat) by the system; energy is a product; (i.e., burning methane is a product; (i.e., burning methane gas in the lab burner produces heat; gas in the lab burner produces heat; light sticks give off light which is also light sticks give off light which is also energy)energy)

Chemiluminesence Chemiluminesence

State FunctionState Function – A property – A property independent of past or future independent of past or future behavior; (it does not matter which behavior; (it does not matter which road brought you to school today–road brought you to school today–you started at your house and you started at your house and ended here –there are probably ended here –there are probably lots of ways for that to happen) lots of ways for that to happen)

The mother of all The mother of all EXOthermic EXOthermic Reactions!!Reactions!!

Entropy (S)Entropy (S) – measure of disorder – measure of disorder in the system (measure of chaos) in the system (measure of chaos)

Gibb’s Free Energy (G)Gibb’s Free Energy (G)– criteria – criteria for sponteneity and amount of free for sponteneity and amount of free energy to do workenergy to do work

ThermodynamicsThermodynamics – study of – study of energy and its interconversionsenergy and its interconversions

WorkWork – force acting over distance – force acting over distance

ENERGY AND WORK (see definition of energy)

E = q(heat) + w(work)

Signs of q: +q if heat absorbed–q if heat released

w = -PVNOTE: Energy is a state function. (Work and

heat are not.)

Signs of w (commonly related to work done by or to gases)

+ w if work done on the system (i.e., compression)-w if work done by the system (i.e., expansion)

When related to gases, work is a function of pressure (pressure is force per unit of area) and volume

Exercise 1Exercise 1Internal Energy Internal Energy

Calculate ∆Calculate ∆EE for a system for a system undergoing an endothermic undergoing an endothermic process in which 15.6 kJ of process in which 15.6 kJ of heat flows and where 1.4 kJ heat flows and where 1.4 kJ of work is done on the of work is done on the system.system.

Solution:Solution:

∆∆EE = 17.0 k = 17.0 kJJ

Exercise 2Exercise 2 PV Work PV Work

Calculate the work Calculate the work associated with the associated with the expansion of a gas from 46 expansion of a gas from 46 L to 64 L at a constant L to 64 L at a constant external pressure of 15 atm. external pressure of 15 atm.

Solution:Solution:

Work = -270 LWork = -270 Latmatm

Exercise 3Exercise 3 Internal Energy, Heat, and Internal Energy, Heat, and

WorkWork A balloon is being inflated to A balloon is being inflated to

its full extent by heating the its full extent by heating the air inside it. In the final stages air inside it. In the final stages of this process, the volume of of this process, the volume of the balloon changes from 4.00 the balloon changes from 4.00 × 106 L to 4.50 × 106 L by × 106 L to 4.50 × 106 L by the addition of 1.3 × 108 J of the addition of 1.3 × 108 J of energy as heat. energy as heat.

Assuming that the balloon Assuming that the balloon expands against a constant expands against a constant pressure of 1.0 atm, calculate pressure of 1.0 atm, calculate ∆∆EE for the process. for the process.

(To convert between L (To convert between L atm atm and J, use 1 L and J, use 1 L atm = 101.3 atm = 101.3 J.)J.)

Solution:Solution:

∆∆EE = 8 × 107 J= 8 × 107 J

ENTHALPY

Measure only the change in enthalpy, H (the difference between the potential energies of the products and the reactants) H is a state function

H = q at constant pressure (i.e. atmospheric pressure)(true most of the time for us and a very handy fact!)Enthalpy can be calculated from several sources including:

StoichiometryCalorimetryFrom tables of standard valuesHess’s LawBond energies

Sample Problem ASample Problem A: :

Upon adding solid potassium Upon adding solid potassium hydroxide pellets to water the hydroxide pellets to water the following reaction takes place:following reaction takes place:

KOH (s) KOH (s) KOH(aq) + 43 kJ KOH(aq) + 43 kJ

Answer the following questions Answer the following questions regarding the addition of 14.0 regarding the addition of 14.0

g of KOH to water:g of KOH to water:

Does the beaker get warmer or colder?Does the beaker get warmer or colder?

Is the reaction endothermic or Is the reaction endothermic or exothermic?exothermic?

What is the enthalpy change for the What is the enthalpy change for the dissolution of the 14.0 grams of KOH?dissolution of the 14.0 grams of KOH?

Answers:Answers:

(a) warmer (a) warmer

(b) exothermic (b) exothermic

(c) -10.7 kJ (c) -10.7 kJ

CalorimetryCalorimetry

The process of measuring heat The process of measuring heat based on observing the based on observing the temperature change when a body temperature change when a body absorbs or discharges energy as absorbs or discharges energy as heat.heat.

Coffee-cup calorimetryCoffee-cup calorimetry

In the lab, this is In the lab, this is how we how we experiment to experiment to find energy of a find energy of a particular system. particular system. We use a We use a Styrofoam cup, Styrofoam cup, reactants that reactants that begin at the same begin at the same temperature and temperature and look for look for change change in temperaturein temperature. .

After all data is collected (mass or After all data is collected (mass or volume; initial and final volume; initial and final temperatures) we can use the temperatures) we can use the specific heat formula to find the specific heat formula to find the energy released or absorbed. We energy released or absorbed. We refer to this process as constant refer to this process as constant pressure calorimetry. pressure calorimetry.

** q = ** q = H @ these conditions.**H @ these conditions.**

Bomb calorimetryBomb calorimetry

WWeighed reactants eighed reactants are placed inside a are placed inside a steel container steel container and ignited. Often and ignited. Often referred to as referred to as constant volume. constant volume. This is used by This is used by industry to industry to determine number determine number of food calories of food calories that we consume!that we consume!

Sorry, Chester!Sorry, Chester!

This is exactly why This is exactly why Cheetoes are more fun to Cheetoes are more fun to eat!eat!

Heat capacityHeat capacity

Energy required to raise temp. by Energy required to raise temp. by

1 degree (Joules/ 1 degree (Joules/ C) C)

Specific heat capacity Specific heat capacity (Cp)(Cp)

Same as specific heat but specific Same as specific heat but specific to to

1 gram of substance1 gram of substance

Molar heat capacityMolar heat capacity

Same as heat capacity but specific Same as heat capacity but specific to one mole of substanceto one mole of substance

(J/mol(J/mol K or J/mol K or J/mol C )C )

Energy (q)Energy (q)

Released or gained -- q = mReleased or gained -- q = mCpCpTT

q = quantity of heat ( Joules or q = quantity of heat ( Joules or

calories) calories) m = m = mass mass in gramsin grams ΔT = ΔT = TTff - T - Tii (final – initial) (final – initial)

CCpp = specific heat capacity ( J/g = specific heat capacity ( J/gC) C)

Specific heat of water Specific heat of water (liquid state)(liquid state)

= 4.184 J/g= 4.184 J/gC ( or 1.00 cal/g C ( or 1.00 cal/g C)C) Water has one of the highest specific heats known! Water has one of the highest specific heats known!

That is why the earth stays at such an even That is why the earth stays at such an even temperature all year round! Cool huh? temperature all year round! Cool huh?

Heat lost by substance = heat gained by waterHeat lost by substance = heat gained by water (if this does not happen, calculate the heat capacity (if this does not happen, calculate the heat capacity

of the substance)of the substance)

Units of EnergyUnits of Energy

– caloriecalorie--amount of heat needed to --amount of heat needed to raise the temp. of 1.00 gram of water raise the temp. of 1.00 gram of water 1.00 1.00 C C

– kilocaloriekilocalorie--duh!; the food calorie with --duh!; the food calorie with

a capital C.a capital C.

– joulejoule--SI unit of energy; 1 cal = 4.184 J--SI unit of energy; 1 cal = 4.184 J

Sample Problem B:Sample Problem B:

In a coffee cup calorimeter, In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are 100.0 mL of 1.0 M HCl are mixed. Both solutions were mixed. Both solutions were originally at 24.6originally at 24.6C. After the C. After the reaction, the final reaction, the final temperature is 31.3temperature is 31.3C. C.

Assuming that all solutions have Assuming that all solutions have a density of 1.0 g/cm3 and a a density of 1.0 g/cm3 and a specific heat capacity of 4.184 specific heat capacity of 4.184 J/gJ/gC, calculate the enthalpy C, calculate the enthalpy change for the neutralization of change for the neutralization of HCl by NaOH. HCl by NaOH.

Assume that no heat is lost to Assume that no heat is lost to the surroundings or the the surroundings or the calorimeter.calorimeter.

Solution:Solution:

∆∆HH = -5.6 kJ/mol = -5.6 kJ/mol

Exercise 4Exercise 4 Enthalpy Enthalpy

When 1 mole of methane When 1 mole of methane (CH(CH44) is burned at constant ) is burned at constant pressure, 890 kJ of energy is pressure, 890 kJ of energy is released as heat. Calculate released as heat. Calculate ∆∆HH for a process in which a for a process in which a 5.8-g sample of methane is 5.8-g sample of methane is burned at constant pressure.burned at constant pressure.

Solution:Solution:

∆∆HH = heat flow = -320 kJ = heat flow = -320 kJ

Exercise 5Exercise 5 Constant-Pressure Constant-Pressure CalorimetryCalorimetry

When 1.00 L of 1.00 When 1.00 L of 1.00 MM Ba(NOBa(NO33))22 solution at 25.0 solution at 25.0C is C is mixed with 1.00 L of 1.00 mixed with 1.00 L of 1.00 M M NaNa22SOSO44 solution at 25 solution at 25C in a C in a calorimeter, the white solid calorimeter, the white solid BaSOBaSO44 forms and the forms and the temperature of the mixture temperature of the mixture increases to 28.1increases to 28.1C.C.

Assuming that the Assuming that the calorimeter absorbs only a calorimeter absorbs only a negligible quantity of heat, negligible quantity of heat, that the specific heat capacity that the specific heat capacity of the solution is 4.18 J/of the solution is 4.18 J/C C g, g, and that the density of the and that the density of the final solution is 1.0 g/mL,final solution is 1.0 g/mL,

calculate the enthalpy change calculate the enthalpy change per mole of BaSOper mole of BaSO44 formed. formed.

Solution:Solution:

Enthalpy change per Enthalpy change per

mole of BaSOmole of BaSO44 = =

-26 kJ/mol-26 kJ/mol

Exercise 6Exercise 6 Constant-Volume Constant-Volume CalorimetryCalorimetry

It has been suggested that It has been suggested that hydrogen gas obtained by the hydrogen gas obtained by the decomposition of water might be decomposition of water might be a substitute for natural gas a substitute for natural gas (principally methane). To (principally methane). To compare the energies of compare the energies of combustion of these fuels, the combustion of these fuels, the following experiment was carried following experiment was carried out using a bomb calorimeter out using a bomb calorimeter with a heat capacity of 11.3 kJ/with a heat capacity of 11.3 kJ/C.C.

When a 1.50-g sample of When a 1.50-g sample of methane gas was burned with methane gas was burned with excess oxygen in the excess oxygen in the calorimeter, the temperature calorimeter, the temperature increased by 7.3increased by 7.3C. C.

When a 1.15-g sample of When a 1.15-g sample of hydrogen gas was burned with hydrogen gas was burned with excess oxygen, the temperature excess oxygen, the temperature increase was 14.3increase was 14.3C. C.

Calculate the energy of Calculate the energy of combustion (per gram) combustion (per gram) for hydrogen and for hydrogen and methane.methane.

Solution:Solution:

Methane = 55kJ/gMethane = 55kJ/g

Hydrogen = 141 kJ/gHydrogen = 141 kJ/g

Sample Problem CSample Problem C: :

Camphor (CCamphor (C1010HH1616O) has a heat of O) has a heat of combustion of 5903.6 kJ/mol. When combustion of 5903.6 kJ/mol. When a sample of camphor with mass of a sample of camphor with mass of 0.1204 g is burned in a bomb 0.1204 g is burned in a bomb calorimeter, the temperature calorimeter, the temperature increases by 2.28increases by 2.28C. C.

Calculate the heat capacity of the Calculate the heat capacity of the calorimeter.calorimeter.

Solution:Solution:

Heat capacity of the Heat capacity of the calorimeter = calorimeter =

2.05 kJ/2.05 kJ/C C

TablesTables

HHff = enthalpy of formation. = enthalpy of formation.

= Production of ONE mole of = Production of ONE mole of

compound FROM ITS ELEMENTS compound FROM ITS ELEMENTS

in their standard states (in their standard states ()) = 0 for ELEMENTS in standard states= 0 for ELEMENTS in standard states

Standard States: 25Standard States: 25C (298 K), 1 atm, 1MC (298 K), 1 atm, 1M

Big Mamma Big Mamma EquationEquation

HHrxnrxn = = HHff (products) – (products) – HHff (reactants)(reactants)

Sample Problem DSample Problem D: :

Calculate the Calculate the HHrxnrxn for the following: for the following:

3 Al(s) + 3 NH3 Al(s) + 3 NH44ClOClO44(s) (s)

AlAl22OO33(s) + AlCl(s) + AlCl33(s) + 3 NO(g) + 6 H(s) + 3 NO(g) + 6 H22O(g)O(g)

Given the following values:Given the following values:

SubstanceSubstance HHff (kJ/mol) (kJ/mol)

NHNH44ClOClO44(s) (s) -295 -295

AlAl22OO33(s)(s) -1676 -1676

AlClAlCl33(s) (s) -704 -704

NO(g) NO(g) 90.0 90.0

HH22O(g)O(g) -242 -242

Solution:Solution:

HHrxnrxn = -2680 kJ (exo) = -2680 kJ (exo)

Sample Problem ESample Problem E: :

Sometimes all values are not Sometimes all values are not found in the table of found in the table of thermodynamic data. For most thermodynamic data. For most substances it is impossible to go substances it is impossible to go into a lab and directly synthesize into a lab and directly synthesize a compound from its free a compound from its free elements. The heat of formation elements. The heat of formation for the substance must be found for the substance must be found by working backwards from its by working backwards from its heat of combustion.heat of combustion.

Find the Find the HHff of C of C66HH1212OO66(s) (s) from the following from the following information:information:

CC66HH1212OO66(s) + 6 O(s) + 6 O22(g) (g)

6 CO6 CO22(g) + 6 H(g) + 6 H22O(l) + 2800 kJO(l) + 2800 kJ

SubstanceSubstance HHff (kJ/mol) (kJ/mol)

COCO22(g)(g) -393.5 -393.5

HH22O(l)O(l) -285.8 -285.8

Solution:Solution:

HHff = =

-1276 kJ/mol for glucose-1276 kJ/mol for glucose

Exercise 10 Exercise 10

Using enthalpies of formation, calculate Using enthalpies of formation, calculate the standard change in enthalpy for the standard change in enthalpy for the thermite reaction: the thermite reaction:

2A1(s)+Fe2A1(s)+Fe22OO33(s) (s) A A1212OO33(s)+2Fe(s)(s)+2Fe(s)

This reaction occurs when a mixture of This reaction occurs when a mixture of powdered aluminum and iron (III) oxide powdered aluminum and iron (III) oxide is ignited with a magnesium fuse.is ignited with a magnesium fuse.

Solution:Solution:

Standard change in Enthalpy =-850. kJ

Hess’s LawHess’s Law

Enthalpy is not dependent on the Enthalpy is not dependent on the reaction pathway. If you can find reaction pathway. If you can find a combination of chemical a combination of chemical equations that add up to give you equations that add up to give you the desired overall equation, you the desired overall equation, you can also sum up the can also sum up the H’s for the H’s for the individual reactions to get the individual reactions to get the overall overall HHrxnrxn..

Remember this:Remember this: First decide how to rearrange equations so First decide how to rearrange equations so

reactants and products are on appropriate reactants and products are on appropriate sides of the arrows.sides of the arrows.

If equations had to be reversed, reverse the If equations had to be reversed, reverse the sign of sign of HH

If equations had be multiplied to get a correct If equations had be multiplied to get a correct coefficient, multiply the coefficient, multiply the H by this coefficient H by this coefficient since since H’s are in kJ/H’s are in kJ/MOLE MOLE (division applies (division applies similarly) similarly)

Check to ensure that everything cancels out to Check to ensure that everything cancels out to give you the exact equation you want.give you the exact equation you want.

Hint**Hint** It is often helpful to begin your work It is often helpful to begin your work backwards from the answer that you want! backwards from the answer that you want!

Sample Problem F:Sample Problem F:

Given the following equations:Given the following equations:

HH33BOBO33(aq) (aq) HBO HBO22(aq) + H(aq) + H22O(l)O(l)HHrxnrxn = -0.02 kJ = -0.02 kJ

HH22BB44OO77(aq) + H(aq) + H22O(l) O(l) 4 HBO 4 HBO22(aq)(aq)HHrxn rxn = -11.3 kJ= -11.3 kJ

HH22BB44OO77(aq) (aq) 2 B 2 B22OO33(s) + H(s) + H22O(l)O(l)HHrxnrxn = 17.5 kJ = 17.5 kJ

Find the Find the H for this overall H for this overall reactionreaction

2 H2 H33BOBO33(aq) (aq)

BB22OO33(s) + 3 (s) + 3 HH22O(l)O(l)

Solution:Solution:

H = 14.4 kJ H = 14.4 kJ endothermic endothermic

Bond EnergiesBond Energies

Energy must be added/absorbed to Energy must be added/absorbed to BREAK bonds (endothermic). Energy is BREAK bonds (endothermic). Energy is released when bonds are FORMED released when bonds are FORMED (exothermic).(exothermic).

H = sum of the energies required to H = sum of the energies required to break old bonds (positive signs) plus the break old bonds (positive signs) plus the sum of the energies released in the sum of the energies released in the formation of new bonds (negative signs).formation of new bonds (negative signs).

H = bonds broken – bonds formedH = bonds broken – bonds formed

Sample Problem G:Sample Problem G:

Using bond energies, calculate the Using bond energies, calculate the change in energy that accompanies change in energy that accompanies the following reaction:the following reaction:

HH22(g) + F(g) + F22(g) (g) 2 HF(g) 2 HF(g)

Bond TypeBond Type Bond EnergyBond Energy

H-HH-H 432 kJ/mol 432 kJ/mol

F-FF-F 154 kJ/mol 154 kJ/mol

H-FH-F 565 kJ/mol 565 kJ/mol

Solution:Solution:

Change in energy = -544 Change in energy = -544 kJ kJ

SUMMARY FOR ENTHALPYSUMMARY FOR ENTHALPY What does it really tell you about an equation?What does it really tell you about an equation?

H = + H = + reaction is endothermicreaction is endothermic H = - H = - reaction is exothermic reaction is exothermic

(favored – nature tends towards (favored – nature tends towards lower energy) lower energy)