Thermochemistry

Specific Heat Formula

cp = Specific Heat

Q = Energy (heat) lost or gained

T = Temperature change

m = Mass

pcTmQ

Two Types of Thermal Reactions

• Exothermic: Releases Thermal Energy (heat) sign is –

• Endothermic: Absorbs Thermal Energy (heat)

sign is +

Exothermic Processes

Reactants Products + energy

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

Endothermic Processes

Reactants + energy Products

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

Enthalpy = HIn a chemical reaction, Enthalpy (H) is equal to the energy that flows as heat.

(at a constant pressure)

Example:

When 1 mol of methane gas is burned it releases 890 kJ of energy

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat

H = -890kj = exothermic reaction

Calculate H for a process in which 5.8 g of methane are burned. H = -890kJ per mol CH4

Rxn: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) +heatSolution:

Molar mass of CH4 = 16.04 g / mol

5.8 g CH4 1 mol CH4 = 0.36 mol CH4

16.04 g CH4

0.36 mol CH4 -890kJ = -320 kJ

1 mol CH4

Calculate H when 12.8 g of sulphur dioxide reacts with excess oxygen to form sulphur trioxide. H = -198.2kJ per mol SO2

Solution:

Molar mass of SO2 = 64.07 g / mol

12.8 g SO2 1 mol SO2 = 0.1998 mol SO2

64.07 g SO2

0.1998 mol SO2 -198kj = -39.6 kJ

1 mol SO2

Hydrogen peroxide decomposes according to the following thermochemical reaction:

H2O2(l) → H2O(l) + 1/2 O2(g)

ΔH = -98.2 kJ

Calculate the change in enthalpy (ΔH) when 4.00 grams of hydrogen peroxide decomposes.

Calculate the change in enthalpy (ΔH) when 36.00 grams of water are created…

The reaction that occurs in hand warmers is:

4Fe(s) + 3O2(g) 2Fe2O3(s) + heat

H of reaction = -1652kJ

How much heat is released when 2.5 grams of Fe(s) is reacted with excess oxygen?

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Law of Concert Tickets!

You can add KNOWN equations to solve UNKOWN equations

Law of Concert Tickets = Hess’s Law

FLIP (reverse) the equation – you must FLIP the sign (- +)

MULTIPY / DIVIDE equation – must Multiply / Divide the ΔH!

Two Rules:

2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ

C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = -1299.5 kJ

C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ

H2(g) + (1/2)O2(g) ---> H2O(l) ΔH° = -285.8 kJ

Example #1: Calculate the enthalpy for this reaction:

Given the following thermo chemical equations:

a) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing.

+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ

C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = -1299.5 kJ

C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ

H2(g) + (1/2)O2(g) ---> H2O(l) ΔH° = -285.8 kJ

2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ

SrO(s) + CO2(g) ---> SrCO3(s) ΔH = -234 kJ

2SrO(s) ---> 2Sr(s) + O2(g) ΔH = +1184 kJ2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g) ΔH = +2440 kJ

Example #2: Given the following data:

Find the ΔH of the following reaction:

C(s) + O2(g) ---> CO2(g)

a) first equation - flip itb) second equation - divide by two c) third equation - flip it, divide by two

+234 + (+592) + (-1220) = -394

SrO(s) + CO2(g) ---> SrCO3(s) ΔH = -234 kJ

2SrO(s) ---> 2Sr(s) + O2(g) ΔH = +1184 kJ2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g) ΔH = +2440 kJ

Example #2: Given the following data:

Find the ΔH of the following reaction:

C(s) + O2(g) ---> CO2(g)

a) first equation - flip itb) second equation - divide by two c) third equation - flip it, divide by two

+234 + (+592) + (-1220) = -394

Hess's Law

In a reaction, the change in enthalpy (ΔH)is the same - regardless if the reaction occurs in a single step or in several steps. If a series of reactions are added together, the net change in ΔH is the sum of the enthalpy changes for each step.

Rules for using Hess's Law

If the reaction is multiplied (or divided) by some factor, Δ H must also be multiplied (or divided) by that same factor.

If the reaction is reversed (flipped), the sign of Δ H must also be reversed.

Water phase changes & EnergyTemperature remains __________ during a phase change.

constant

Phase Change DiagramProcesses occur by addition of energy

Processes occur by removal of energy

Phase Diagram Represents phases as a function of temperature and pressure. Critical temperature: temperature above which the vapor can not be liquefied. Critical pressure: pressure required to liquefy AT the critical temperature. Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).

Phase Changes

Effect of Pressure on Boiling PointBoiling Point of Water at Various Locations

Location Feet above sea level

Patm (kPa) Boiling Point (C)

Top of Mt. Everest, Tibet 29,028 32 70

Top of Mt. Denali, Alaska 20,320 45.3 79

Top of Mt. Whitney, California 14,494 57.3 85

Leadville, Colorado 10,150 68 89

Top of Mt. Washington, N.H. 6,293 78.6 93

Boulder, Colorado 5,430 81.3 94

Madison, Wisconsin 900 97.3 99

New York City, New York 10 101.3 100

Death Valley, California -282 102.6 100.3

Phase Diagram Represents phases as a function of temperature and pressure. Critical temperature: temperature above which the vapor can not be liquefied. Critical pressure: pressure required to liquefy AT the critical temperature. Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).

Phase changes by Name

Water

Carbon dioxidePhase

Diagram for

Carbondioxide

CarbonPhase

Diagram for

Carbon

Phase Diagram for Sulfur

Reaction Pathway

• Shows the change in energy during a chemical reaction

Exothermic Reaction• reaction that

releases energy

• products have lower PE than reactants

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

energyreleased

Endothermic Reaction• reaction that

absorbs energy

• reactants have lower PE than products

2Al2O3 + energy 4Al + 3O2

energyabsorbed

Latent Heat of Phase Change

Molar Heat of Fusion

The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.

The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

Molar Heat of Solidification

Latent Heat of Phase Change #2Molar Heat of Vaporization

The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.

The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

Molar Heat of Condensation

Latent Heat – Sample ProblemProblem: The molar heat of fusion of water is6.009 kJ/mol. How much energy is needed to convert 60

grams of ice at 0C to liquid water at 0C?

Massof ice

MolarMass ofwater

Heatof

fusion

kiloJoulesOHmol

kJ

OHg

OHmolOHg20

1

009.6

02.18

160

22

22

Heat of Solution

Substance

Heat of Solution (kJ/mol)

NaOH -44.51

NH4NO3 +25.69

KNO3 +34.89

HCl -74.84

The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

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

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

6.2

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)

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

DH = H (products) – H (reactants)

DH = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

DH < 0Hproducts > Hreactants

DH > 0 6.3

Thermochemical Equations

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

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

6.3

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DH = -890.4 kJ

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

6.3

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

• The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

• If you reverse a reaction, the sign of DH changes

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

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

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

6.3

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

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

Thermochemical Equations

6.3

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

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

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

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

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 = msDt

q = CDt

Dt = tfinal - tinitial

6.4

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

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

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

6.4

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qbomb + qrxn

qsys = 0

qrxn = - (qwater + qbomb)

qwater = msDt

qbomb = CbombDt

6.4

Reaction at Constant V

DH ~ qrxn

DH = qrxn

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = msDt

qcal = CcalDt

6.4

Reaction at Constant PDH = qrxn

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 49 of 50

Terminology

• Energy, U– The capacity to do work.

• Work– Force acting through a distance.

• Kinetic Energy– The energy of motion.

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 50 of 50

Energy

• Kinetic Energy

ek =

12 mv2 [ek ]

=

kg m2

s2 = J

w = Fd [w ] =

kg ms2 = Jm

• Work

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 51 of 50

Energy• Potential Energy

– Energy due to condition, position, or composition.

– Associated with forces of attraction or repulsion between objects.

• Energy can change from potential to kinetic.

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 52 of 50

Energy and Temperature

• Thermal Energy– Kinetic energy associated with random molecular

motion.– In general proportional to temperature.– An intensive property.

• Heat and Work– q and w.– Energy changes.

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 53 of 50

Heat

Energy transferred between a system and its surroundings as a result of a temperature difference.

• Heat flows from hotter to colder.– Temperature may change.– Phase may change (an isothermal process).

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 54 of 50

Units of Heat

• Calorie (cal)– The quantity of heat required to change the

temperature of one gram of water by one degree Celsius.

• Joule (J)– SI unit for heat

1 cal = 4.184 J

Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 55 of 50

Heat Capacity

• The quantity of heat required to change the temperature of a system by one degree.

– Molar heat capacity.• System is one mole of substance.

– Specific heat capacity, c.• System is one gram of substance

– Heat capacity• Mass specific heat.

q = mcT

q = CT