Thermochemistry-Part 1Chapter 7

Brad Collins

Thermochemistry• Thermodynamics: The study of energy

• Thermochemistry: The study of energy in chemical reactions

• Energy: The capacity to do work

• Work = force x distance

• Force = mass ÷ area

7.1

Systems and SurroundingsThe system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

7.1

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

An 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)

energy + H2O (s) H2O (l)

7.1

Exothermic Endothermic

Exo vs. EndothermicAn Illustration

7.1

Types of Energy• Potential energy is the energy of position, or stored

energy.

• 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 an atom.

• Elastic energy is the energy stored in a compressed gas or coiled spring (sometimes called mechanical energy)

7.1

Types of Energy• Kinetic energy is the energy of motion

• Radiant (electromagnetic) radiation comes from the sun and is the earth's primary energy source

• Thermal energy is the energy from the random motion of atoms and molecules

• Sound energy is the transmitted vibrational energy of atoms and molecules in bulk.

7.1

Energy can be converted from one form to another, but cannot be created or destroyed.

First Law of Thermodynamics

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

More Thermochemistry Definitions

Temperature is a measure of the thermal energy.

900C 400Cgreater thermal energy

Temperature = Thermal EnergyX

7.1

Units of Heat• Heat is measured in joules or calories

• A joule (J) is defined as the energy expended in applying a force of 1 newton over a distance of 1 meter

• 1 J = N x m2 = Pa x m3 = (kg x m2 )/s2

• A calorie (cal) is defined as the energy required to raise 1 gram of water 1ºC.

• A kilocalorie (kcal, or Cal) is defined as the energy required to raise 1 kilogram of water 1ºC. Used for foods.

• 1 calorie = 4.184 joules

7.1

Chemistry in Action:Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

1 cal = 4.184 J1 Cal = 1000 cal = 4184 J

heat of reaction = –2801 kJ/mol

7.1

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.

C = m·sHeat (q) absorbed or released:

q = m·s·Δt

q = C·ΔtΔt = tfinal - tinitial

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.

Specific Heat

7.1

Practice ProblemHow much heat is required to raise the temperature of 12.5 g of copper, 57ºC?

q = m·s·∆Tspecific heat (s) of copper = 0.365 J/g·ºC

7.1

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

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

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

6.5

Calorimetry• The measurement of heat (q)

transferred between a system and its surroundings

• Uses a calorimeter

• Types of calorimeters:

• Constant volume

• Constant pressure

7.1

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qbomb + qrxn

qsys = 0 (an isolated system)

qrxn = – (qwater + qbomb)qwater = msΔtqbomb = CbombΔt

Reaction at Constant V

Bomb Calorimeter

Cbomb is the heat capacity of the calorimeter

7.1

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = msΔt

qcal = CcalΔt

Ccal is the heat capacity of the calorimeter

Reaction at constant pressure

7.1

Practice ProblemA lead pellet having a mass of 26.47 g was heated to a temperature of 89.98ºC and placed in a constant-pressure calorimeter containing 100.0 g of water at 22.50ºC. If the calorimeter has a heat capacity of approximately zero and the final temperature of the water plus lead pellet is 23.17ºC, what is the specific heat of the lead pellet?

7.1

SolutionHeat lost by the pellet is gained by the water in the calorimeter. Because the calorimeter has a heat capacity of zero, no heat is lost to the calorimeter, so:

qpellet + qwater = 0

Heat gained by the water is:

qpellet = – qwater

qwater = m·s·∆T, where ∆T = 23.17ºC – 22.50ºCqwater = 100.0 g·4.184 J/g·ºC·0.67ºC = 280.328 J

Heat lost by the pellet is:qpellet = – qwater = –280.328 J

Specific heat of lead is:

slead = qlead/m·∆T = –280.328 J/(26.47 g·(–66.81ºC)) = 0.158 J/g·ºC∆T = 23.17ºC – 89.98ºC

∆T = Tf(ºC) – Ti(ºC)

7.1

Enthalpy (H)• Enthalpy (H) is used to measure the heat flow into or out of a system for a

process that occurs at constant pressure.

• Enthalpy change (∆H) is the difference between the enthalpy under initial conditions and enthalpy under final conditions

• ∆H = Hfinal – Hinitial

• H and ∆H are both state functions

• Enthalpy is a molar quantity, meaning it is expressed in J/mol

• Under normal lab conditions ∆H = q

• Negative values for ∆H indicate exothermic processes.

• Positive values for ∆H indicate endothermic processes.

7.2

Enthalpy Change (∆H)

ΔH = H (products) – H (reactants)

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

Hproducts < Hreactants

ΔH < 0Hproducts > Hreactants

ΔH > 0 7.2

Enthalpy Change Examples

H2O (s) H2O (l) ΔH = 6.01 kJ

Is ΔH negative or positive?

System absorbs heat

Endothermic

ΔH > 0

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

7.2

Enthalpy Change Examples

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

Is ΔH negative or positive?

System gives off heat

Exothermic

ΔH < 0

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

7.2

H2O (s) H2O (l) ΔH = 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 ΔH changes

H2O (l) H2O (s) ΔH = -6.01 kJ

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

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

• Chemical equations that include the enthalpy change as part of the balanced chemical reaction

7.2

H2O (s) H2O (l) ΔH = 6.01 kJ

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

Thermochemical Equations

H2O (l) H2O (g) ΔH = 44.0 kJ

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

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

266 g P4

1 mol P4

123.9 g P4x 3013 kJ

1 mol P4x = 6470 kJ

7.2

Enthalpy and Phase Changes

7.2

Enthalpy and Phase Changes

• As heat is added to a substance it warms, and undergoes phase changes.

• The heat required to warm a substance is given by the specific heat for the phase (s, l, g) of the substance times its mass (q=m·s·∆T)

• The heat required to melt a substance is the enthalpy of fusion (∆Hfus)

#

• The heat required to vaporize a substance is the enthalpy of vaporization (∆Hvap)

H2O (s) H2O (l) ΔHfus = 6.01 kJ

H2O (l) H2O (g) ΔHvap = 44.0 kJ

7.2

Practice ProblemHow much heat is required to raise the temperature of 10.0 g of water from –10ºC to +50ºC?

7.2

Hess's Law• Germaine Henri Hess (Russian chemist)

• Thermochemical equations are additive

7.3

+

C(s) + ½O2(g) —> CO(g) ∆H = –110.5 kJ

CO(g) + ½O2(g) —> CO2(g) ∆H = –283.0 kJ

C(s) + O2(g) —> CO2(g) ∆H = –393.5 kJ

• Hess’s Law works because ∆H is a state function

State Functions

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy (E), pressure, volume, temperature

ΔE = Efinal - Einitial

ΔP = Pfinal - Pinitial

ΔV = Vfinal - Vinitial

ΔT = Tfinal - Tinitial

enthalpy (H),

ΔH = Hfinal - Hinitial

7.3

Hess’s Law PracticeCalculate the enthalpy change of:

C2H2(g) + 2Cl2(g) —> C2H2Cl4(l)Using the following thermochemical equations:

2C(s) + H2(g) —> C2H2(g) ∆H = 227 kJ2C(s) + 2Cl2(g) —> C2H2Cl4(l) ∆H = 130 kJ

7.3

Enthalpy of FormationDefinition: The enthalpy change required to form one mole of a substance from each of its elements in their standard states.

• Standard state of an element is its state in pure form at 1 atm.

• Standard states of some common elements

• Metals: are monatomic solids,except Hg(l)

• Diatomic elements: are gases, except Br(l) and I(s)

• Carbon: graphite, written as C(graphite)

• Phosphorus: white phosphorus, P(white)

• Sulfur: rhombic sulfur, S(rhombic)

• Enthalpy of formation (∆Hf) is determined experimentally

C(graphite) + O2(g) —> CO2(g) ∆Hf = –393.5 kJ

7.4

7.4

C6H12O6(s) + 6 O2(g) —> 6 CO2(g) + 6H2O(l)

∆H0rxn = [6·∆Hºf{CO2(g)} + 6·∆Hºf{H2O(l)}] – [∆Hºf{C6H12O6(s)} + 6·∆Hºf{O2(g)}]

Calculate using ∑ n·∆Hºf{products} – ∑ m·∆Hºf{reactants}

From Table, p. 253∆Hºf{CO2(g)} = –393.5 kJ/mol

∆Hºf{H2O(l)} = –241.8 kJ/mol

∆Hºf{C6H12O6(s)} = –1274.5 kJ/mol

∆Hºf{O2(g)} = 0 kJ/mol

∆H0rxn = [6·(–393.5) + 6·(–241.8)] – [–1274.5 + 6·(0)] = –2537.3 kJ

Try this one: CaCO3(s) —> CaO(s) + CO2(g)

Using ∆Hf to Calculate ∆HrxnUse ∆Hº*f of the reactants and products

*Indicates ∆Hf was measured at 25ºC and 1 atm

6.5

∆Hºrxn Practice ProblemBenzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6(l) + 15O2(g) —> 12CO2(g) + 6H2O(l)

∆Hºrxn = ∑n·∆Hºf(products) – ∑m·∆Hºf(reactants)

∆Hºrxn = [12·∆Hºf(CO2) + 6·∆Hºf(H2O)] – ∑2·∆Hºf(C6H6)

∆Hºrxn = [12·(–393.5 kJ) + 6·(–285.8 kJ)] – 2·(49.04 kJ)

–6535 kJ2 mol = –3267 kJ/mol C6H6