Thermochemistry Notes

Thermochemistry Notes

I. Thermochemistry deals with the changes in energy that accompany a chemical reaction. Energy is measured in a quantity called enthalpy, represented as H. The change in energy that accompanies a chemical reaction is represented as H. Page 519

a. The energy absorbed or released as heat in a chemical or physical change is measured in a calorimeter. In one kind of calorimeter, known quantities of reactants are sealed in a reaction chamber, which is immersed in a known quantity of water in an insulated vessel. Therefore, the energy given off (or absorbed) during the reaction is equal to the energy absorbed (or given off) by the known quantity of water. The amount of energy is determined from the temperature change of the known mass of surrounding water.

Calorimeter

Using the change in temperature, T, determined from calorimeter one can use the following equation to determine the quantity of energy gained or lost during the reaction or for a physical change:

q = (cp)(m)(T)T in kelvinq represents the energy lost or gained (in

J)m is the mass of the sample (in g)cp is the specific heat of a substance at a

given temperature pg513

• Practice Problem: 1. How much heat energy is needed to raise the temperature of a 33.0 gram sample of aluminum from 24.0C to 100.C?

• 2. Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293K to 313K.

• 3. During a chemical reaction carried out in a calorimeter the temperature of water within the calorimeter raised from 24.0C to 125C. If 250. grams of water were present calculate the amount of heat energy the water gained.

b. In thermochemical equations the quantity of energy released or absorbed as heat during a reaction is written and is represented by H.

Example: 2H2(g) + O2(g) 2H2O(l) H= - 571.6 kJ/mol

c. H can be used to determine if the reaction is exothermic or endothermic. If the H value of an equation is negative that represents an exothermic reaction. (Meaning energy is released, therefore the energy of the products would be less.)

If the H value of an equation is positive that represents an endothermic reaction. Example:

2H2(g) + O2(g) 2H2O(g)

H= - 483.6 kJ/mol H reactants = 1450.8 kJ/mol H products = 967.2 kJ/mol

Type of Reaction: Exothermic

2H2O(g) 2H2(g) + O2(g)

H= + 483.6 kJ/mol H reactants = 967.2 kJ/mol

H products = 1450.8 kJ/molType of Reaction: Endothermic

d. Hess’s law provides a method for calculating the H of a reaction from tabulated data. This law states that if two or more chemical equations are added, the H of the individual equations may also be added to find the H of the final equation. As an example of how this law operates, look at the three reactions below.

(1) 2H2(g) + O2(g) 2H2O(l)H = 571.6 kJ/mol

(2)2H2O2(l) 2H2(g) + 2O2(g)H = +375.6 kJ/mol

(3)2H2O2(l) 2H2O(l) + O2(g)H = ? kJ/mol

When adding equations 1 and 2, the 2 mol of H2(g) will cancel each other out, while only 1 mol of O2(g) will cancel.

(3)2H2O2(l) 2H2O(l) + O2(g)H = ? kJ/mol

Warm-up• What type of energy and energy transfer do you see in this picture•Thank you Ms. Bouwman (from McNeil High School) for this wonderful power point!

Basic Thermochemistry

Courtesy of lab-initio.com

Energy is the capacity to do work

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Electrical energy is the energy associated with the flow of electrons

• Potential energy is the energy available by virtue of an object’s position

6.1

Heat (Enthalpy) Change, ΔHDefinition:Definition: The amount of heat energy released or The amount of heat energy released or absorbed during a process.absorbed during a process.

EnergyEnergyEnergy is the capacity to do workEnergy is the capacity to do work, and can take many forms

Potential energy is stored energy or the energy of position Kinetic energy is the energy of motion Thermal energy (heat) is an outward manifestation of movement at the atomic level

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy

900C400C

greater thermal energy6.2

Heat

The flow of thermal energy from one object to another.

Heat always flows from warmer to cooler objects.

Ice gets warmer while

hand gets cooler

Cup gets cooler while hand gets

warmer

3 Types of Heat Transfer• Radiation- the transfer of energy by

electromagnetic waves. • Convection – Transfer of energy by currents• Conduction – Transfer of energy by touching

objects

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

6.2

energy + H2O (s) H2O (l)

Exothermic ProcessesExothermic Processes

Reactants Reactants Products + energy Products + energy

Processes in which energy is released as it proceeds, and Processes in which energy is released as it proceeds, and surroundings become warmersurroundings become warmer

Endothermic ProcessesEndothermic Processes

Reactants + energy Reactants + energy Products Products

Processes in which energy is absorbed as it proceeds, Processes in which energy is absorbed as it proceeds, and surroundings become colderand surroundings become colder

Water phase changesTemperature remains __________ during a phase change.constant

Phase Change DiagramPhase Change DiagramProcesses occur by addition of energy Processes occur by addition of energy Processes occur by removal of energyProcesses occur by removal of energy

Thermochemical Calculations

Units for Measuring HeatThe JouleJoule is the SI system unit for measuring heat:

The caloriecalorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree

2

2111smkgmeternewtonJoule

Joulescalorie 18.41

Specific HeatSpecific Heat

The amount of heat The amount of heat required to raise the required to raise the temperature of one temperature of one gram of substance by gram of substance by one degree Celsius.one degree Celsius.

1

2

3

45 6

7

8

9

1 102

3

45 6

7

8

9

11

Specific Heat (cp, sometimes s, but usually c)

Things heat up or cool down at different rates.

Land heats up and cools down faster than water, and aren’t we lucky for that!?

Specific heat is the amount of heat required to raise the temperature of 1 kg (but in Chem we use g) of a material by one degree (C or K, they’re the same size).

Cp water = 4184 J / kg C (“holds” its heat)

Cp sand = 664 J / kg C (less E to change)

This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

Calculations Involving Specific Heat

ccpp = Specific Heat

QQ = Heat lost or gained

TT = Temperature change

OROR

mm = Mass

TmQcp

pcTmQ

Specific HeatSpecific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

SubstanceSubstance Specific Heat (J/g·K)Specific Heat (J/g·K)Water (liquid) Water (liquid) 4.184.18Ethanol (liquid) Ethanol (liquid) 2.442.44Water (solid) Water (solid) 2.062.06Water (vapor) Water (vapor) 1.871.87Aluminum (solid) Aluminum (solid) 0.8970.897Carbon (graphite,solid) Carbon (graphite,solid) 0.7090.709Iron (solid) Iron (solid) 0.4490.449Copper (solid) Copper (solid) 0.3850.385Mercury (liquid) Mercury (liquid) 0.1400.140Lead (solid)Lead (solid) 0.1290.129Gold (solid) Gold (solid) 0.1290.129

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = ms

Heat (q) absorbed or released:

q = mst

q = Ct

t = tfinal - tinitial

6.4

Specific Heat CapacitySpecific Heat Capacity

If 25.0 g of Al cool from 310 If 25.0 g of Al cool from 310 ooC to 37 C to 37 ooC, how many joules of heat C, how many joules of heat energy are lost by the Al?energy are lost by the Al?

heat gain/lose = q = (c)(mass)(∆T)

where ∆T = Twhere ∆T = Tfinalfinal - T - Tinitialinitial

q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = - 6120 Jq = - 6120 J

Notice that the negative sign on q signals heat “lost by” or transferred OUT Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.of Al.

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g • 0C

t = tfinal – tinitial = 50C – 940C = -890C

q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

6.4

CALORIMETRY

CalorimetryThe amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimetercalorimeter.

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qbomb + qrxn

qsys = 0

qrxn = - (qwater + qbomb)

qwater = mst

qbomb = Cbombt

6.4

Reaction at Constant V

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = mst

qcal = Ccalt

6.4

Reaction at Constant P

First, some heat from reaction warms the water, which we know the mass of and “c” for…

qwater = (c)(water mass)(∆T)THEN, some heat from reaction warms “bomb,” which has a known specific heat for the entire apparatus (typically), so we don’t need the mass…qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

BOOM! Combustible material ignited BOOM! Combustible material ignited at constant volume! This heats up the at constant volume! This heats up the “bomb”, which heats up the water “bomb”, which heats up the water surrounding it…surrounding it…

PracticeA sample of iron metal is added to 75.00 grams of water originally at 35.0°C in a calorimeter. The final temperature of the metal and water in the calorimeter is measured to be 95.0°C.

a) Describe the transfer of heat energy that occurs in the calorimeter.

(b) Assuming no heat is lost to the outside, how many joules of heat energy are transferred?

ENTHALPY (H)

Phase Change DiagramPhase Change Diagram

D

A

C

B

E

Changing Phase• Heat of Fusion - heat change for freezing and

melting• Heat of Vaporization – heat change for condensation

or evaporationFor Water: Heat fusion = 340 J/g Heat vaporization = 2,300 J/g

Heat = (mass)(heat of fusion or vaporization)

How many joules of heat are necessary to melt 500g of ice at its freezing point?= 500g * 340J/g= 170,000 J or 170KJ

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction

Hproducts < Hreactants

H < 0

Hproducts > Hreactants

H > 0 6.3

∆∆HHffoo, standard molar enthalpy of , standard molar enthalpy of

formationformation∆∆HHff

o o = Enthalpy change when 1 mol of compound is formed = Enthalpy change when 1 mol of compound is formed from the corresponding elements under standard from the corresponding elements under standard conditionsconditions

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g)

∆∆HHffoo (H (H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol

Decomposition

• ∆∆HHffo o may also be used to calculate the may also be used to calculate the

decomposition of somethingdecomposition of something

• If …If …HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) ∆HO(g) ∆Hff˚ = - 242 kJ/mol˚ = - 242 kJ/mol

Then…Then…HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) ∆H(g) ∆Hff˚ = + 242 kJ/mol˚ = + 242 kJ/mol

Enthalpy ValuesEnthalpy Values

• Depend on Depend on how the reaction is writtenhow the reaction is written and on phases of reactants and on phases of reactants and products…and products…

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g)

∆∆H˚ = -242 kJH˚ = -242 kJ

2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g)O(g)

∆∆H˚ = -484 kJH˚ = -484 kJ

HH22O(g) ---> HO(g) ---> H22(g) + 1/2 O(g) + 1/2 O22(g) (g)

∆∆H˚ = +242 kJH˚ = +242 kJHH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liquid)O(liquid)

∆∆H˚ = -286 kJH˚ = -286 kJ

Huh? So what’s that mean?

To convert 1 mol of water to 1 mol each of To convert 1 mol of water to 1 mol each of HH22 and CO and CO requiresrequires 131 kJ of energy. 131 kJ of energy.

Since delta H is positive, the “water gas” reaction is Since delta H is positive, the “water gas” reaction is

ENDOthermicENDOthermic

6.4

6.5

A problem… Using Standard A problem… Using Standard Enthalpy ValuesEnthalpy Values

Calculate the heat of combustion of methanol, i.e., Calculate the heat of combustion of methanol, i.e., ∆H∆Hoo

rxnrxn for for

∆ ∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHffoo

CHCH33OH = -238.6 KJ/molOH = -238.6 KJ/mol ∆H ∆Hffoo

COCO22 = -393.5 = -393.5

∆∆HHffoo

OO22 = 0 KJ/mol = 0 KJ/mol ∆H∆Hffoo

HH22O = -285.8O = -285.8

= -238.6 – (-393.5 + -285.8)= -238.6 – (-393.5 + -285.8)= 442.7 KJ/mol= 442.7 KJ/mol

PracticeGiven the following heats of formation:

NaCl(s): DH(f)= -400 kJ

H2SO4(l): DH(f)= -800 kJ

Na2SO4(s): DH(f)= -1400 kJ

HCl(g): DH(f)= -90 kJ

a) Find the heat of reaction of the following chemical change:2NaCl(s) + H2SO4 --> Na2SO4(s) + 2HCl(g)

b) Is this reaction exothermic or endothermic?

Energy Stoichiometry

• Remember…1. In an equation coefficients mean moles!2. To go from grams to moles divide by molar mass3. To go from moles to grams multiply my molar

mass4. Make sure you use the right energy units KJ vs. J5. Double check the sign of your energy, is it

endothermic or exothermic

Practice

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x 3013 kJ1 mol P4

x = 6470 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

PracticeThe burning of magnesium is a highly exothermic reaction.2Mg(s) + O2(g) --> 2MgO(s) + 1500 kJ

Use the thermochemical equation to calculate the energy change in kilojoules when 0.5 mol of Mg burn in an excess of O2.

You try

• Calculate the change in H when 32g of NO decomposes?– ½ N2 (g) + ½ O2 (g) NO (g) ∆H∆Hff˚ = ˚ = 90.2 kJ

Using Standard Enthalpy Using Standard Enthalpy ValuesValuesUse ∆H˚’s to calculate Use ∆H˚’s to calculate enthalpy changeenthalpy change for for

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

• When calculating ∆H information from several ∆H information from several equations is neededequations is needed

And then…

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

From either givens From either givens within the problem or reference within the problem or reference

books and tables we can find…books and tables we can find…

• HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) ∆HO(g) ∆Hff˚ = - 242 kJ/mol˚ = - 242 kJ/mol

• C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H ∆Hff˚ = - 111 kJ/mol˚ = - 111 kJ/mol

And and then… we add ‘em.HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) ∆H(g) ∆Hoo = +242 kJ = +242 kJ

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H∆Hoo = -111 kJ = -111 kJ -----------------------------------------------------------------

-------HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

∆∆HHoonetnet = +131 kJ = +131 kJ

• Positive delta H means?

Warm-Up

• Calculate the change in H when 64g of NO is formed?– ½ N2 (g) + ½ O2 (g) NO (g) ∆H∆Hff˚ = ˚ = 90.2 kJ

Enthalpy Day 2

H2O (s) H2O (l) H = 6.01 kJ

1. The coefficients always refer to the number of moles of a substance

Things to remember about Thermochemical Equations

2. If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

3. If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ

6.3

H2O (s) H2O (l) H = 6.01 kJ

4. The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

6.3

H2O (l) H2O (g) H = 44.0 kJ

What if you have limited information?Calculate the enthalpy of the reaction: 2 NO(g) + O2(g) --> 2 NO2(g)

Given the following reactions and enthalpies of formation:½ N2(g) + O2(g) --> NO2(g) ∆∆ H = 33.2 kJ½ N2(g) + ½ O2(g) --> NO(g) ∆∆ H = 90.2 kJ

2 NO(g) + O2(g) --> 2 NO2(g)2/2 N2(g) + 2 O2(g) --> 2 NO2(g) ∆∆ H = 33.2 kJ *2

2 NO(g) --> 2/2 N2(g) + 2/2 O2(g) ∆∆ H = - 90.2 kJ * 2

-114 kJ

+

You try…Calculate the enthalpy of the reaction: 4B(s)

+3O2(g) --> 2B2O3(s)

Given the following information:B2O3(s) + 3H2O(g) --> 3O2(g) + B2H6(g) ∆ ∆ H = +2035 kJ2B(s) + 3H2(g) --> B2H6(g) ∆ ∆ H = +36 kJH2(g) + ½ O2(g) --> H2O(l) ∆∆ H = -285 kJH2O(l) --> H2O(g) ∆∆ H = +44 kJ

Equilibrium and

Le Chatelier’s Principle

Chemical Equilibrium

Reversible ReactionsReversible Reactions: : A chemical reaction in which the A chemical reaction in which the

products can react to re-form the products can react to re-form the reactantsreactants

Chemical EquilibriumChemical Equilibrium: : When the rate of the forward reactionWhen the rate of the forward reactionequals the rate of the reverse reactionequals the rate of the reverse reactionand the concentration of products andand the concentration of products andreactants remains unchangedreactants remains unchanged

2HgO(s) 2Hg(l) + O2(g) Arrows going both directions ( ) indicates equilibrium in a chemical equation

LeChatelier’s PrincipleWhen a system atequilibrium is placed understress, the system willundergo a change in sucha way as to relieve thatstress.

Henry Le Chatelier

When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away.

Le Chatelier Translated:

When you add something to a system at When you add something to a system at equilibrium, the system equilibrium, the system shiftsshifts in such a way as toin such a way as to use up what you’ve added.use up what you’ve added.

LeChatelier Example #1A closed container of ice and water at equilibrium. The temperature is raised.

Ice + Energy Ice + Energy Water Water

The equilibrium of the system shifts to the _______ to use up the added energy.

rightright

LeChatelier Example #2A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the container.

NN22OO4 4 (g) + Energy (g) + Energy 2 NO 2 NO22 (g) (g)

The equilibrium of the system shifts to the _______ to use up the added NO2.leftleft

LeChatelier Example #3A closed container of water and its vapor at equilibrium. Vapor is removed from the system.

water + Energy water + Energy vapor vapor

The equilibrium of the system shifts to the _______ to replace the vapor.rightright

LeChatelier Example #4A closed container of N2O4 and NO2 at equilibrium. The pressure is increased.

NN22OO4 4 (g) + Energy (g) + Energy 2 NO 2 NO22 (g) (g)

The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation.

leftleft

What about this?

3 H2(g) + N2(g) <--> 2 NH3(g)

What happens when…a)More N2 is added right/left/no change

b)Pressure is increased right/left/no changec) Temp. is deceased right/left/no change

Or this?

• The system depicted here is maintained at a temperature of 30 °C. If the temperature of the system is doubled, the system will achieve equilibrium by which of the following responses?

a) The temperature of the liquid water will exceed the temperature of the vapor.

b) The temperature of the vapor will exceed the temperature of the liquid water.

c) A higher percentage of the water vapor in the container will condense to liquid.

d) A higher percentage of the water will move into the vapor phase.