thermochemistry

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Chapter 5Chapter 5ThermochemistryThermochemistry

Thermochemistry is an aspect of thermodynamics involving the study of the relationships between chemical reactionsand energy changes involving heat.

HeatEnergy used to cause the temperature of an object to increase.

WorkEnergy used to cause an object that has mass to move.

w = F d

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Energy Energy -the capacity to do work or transfer heat.

Kinetic Energy, Ek -Energy an object possesses by virtue of its motion. 1

2Ek = ⎯ mv2

Potential Energy, Potential Energy, EEpp::Energy an object possesses by virtue of its position or chemical composition.

Thermal energy is kinetic energy in the form of random motion of the particles in any sample of matter. As temperature increases, thermal energy increases.

Chemical energy is energy possessed by atoms as a result of their state of chemical combination. The energy of 2 mol of H2 and 1 mol of O2 is different than that of 2 mol H2O.

Units of EnergyUnits of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: The calorie (cal).

1 cal = 4.184 J

• 1 kcal = 1 Calorie (food “calorie”)

1 J = 1 ⎯⎯kg m2

s2

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System and SurroundingsSystem and Surroundings

• The system includes the molecules we want to study (e.g. hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

• In thermochemistry we study the exchange of energy between the system and surroundings.

Transferal of EnergyTransferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

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Transferal of EnergyTransferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

Transferal of EnergyTransferal of Energya) The potential energy of this ball

of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

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Internal EnergyInternal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

ΔE = Efinal − Einitial

Changes in Internal EnergyChanges in Internal Energy

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

ΔE = q + w

First Law of Thermodynamics is obeyed

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ΔΔEE, , qq, , ww, and sign convention, and sign conventionΔE = q + w

(greater than zero) (less than zero)

System absorbedenergy from the

surroundings

System releasedenergy to the surroundings.

Exchange of Heat between Exchange of Heat between System and SurroundingsSystem and Surroundings

•When heat is absorbed by the system from the surroundings, the process is endothermic.

•When heat is released by the system to the surroundings, the process is exothermic.

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State FunctionsState FunctionsThe internal energy of a system is

independent of the path by which the system achieved that state.

State FunctionsState Functions

• However, q and ware not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its ΔE is the same.– But q and w are

different in the two cases.

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WorkWork

WorkofGasExpansion.MOV

EnthalpyEnthalpy• Enthalpy is a thermodynamic function equal to the internal

energy plus pressure*volume: H = E + PVWhen the system changes at constant pressure, the change in enthalpy, ΔH, is

ΔH = Δ(E + PV)This can be written

ΔH = ΔE + PΔV

Since ΔE = q + w and w = −PΔV, we can substitute these into the enthalpy expression:

ΔH = ΔE + PΔVΔH = (q+w) − wΔH = q

The enthalpy change, ΔH, is defined as the heat gained or lost by the system under constant pressure.

ΔH = qp

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Properties of EnthalpyProperties of Enthalpy

1. Enthalpy is a state function.2. Enthalpy is an extensive property.3. Enthalpy is reversible.4. ΔH for a reaction depends on the

state of the products and the stateof the reactants.

Endothermic and ExothermicEndothermic and Exothermic

• A process is endothermicwhen ΔH is positive (>0).

•A process is exothermicwhen ΔH is negative (<0).

ΔH = Hfinal – Hinitial or ΔH = Hproducts − Hreactants

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Enthalpies of ReactionEnthalpies of ReactionThis quantity, ΔH, is called the enthalpy of reaction, or the heat of reaction.

FormationofWater.MOV

Thermite.mov

A thermochemical equation is an equation for which ΔH is given:

CH4(g) + 2O2(g) CO2(g) + H2O(l) ΔH = -890 kJ

N2(g) + O2(g) 2NO(g) ΔH = +180.8 kJ

The enthalpy changes assume the coefficients are moles of the substancesEnthalpy change is a stoichiometric quantity

CalorimetryCalorimetry

• Calorimetry, the measurement of heat released or absorbed by a chemical reaction.

• A calorimeter is the device used to measure heat

• The quantity of heat transferred by the reaction causes a change in temperature of the solution.

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Heat Capacity and Specific HeatHeat Capacity and Specific Heat• The amount of energy required to raise the

temperature of a substance by 1 K (1°C) isits heat capacity (C in units of J/K).

• We define specific heat capacity (or simply specific heat; Cs or s in units of J/gK) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

Specific heat =heat transferred

mass × temperature change

Cs = s =q

m × ΔT

C =q

ΔT

Constant Pressure CalorimetryConstant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/mol-K), we can measure ΔH for the reaction with this equation:q = m × s × ΔT

Indirectly measure the heat change for the system

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A metal pellet with mass 100.0 g, originally at 88.4 °C, is dropped into 125 g of water originally at 25.1 °C. The final temperature of both the pellet and the water is 31.3 °C.

Calculate the heat capacity C (in J/°C) and specific heat capacity (in J/g°C) of the pellet. The specific heat of water is 4.184 J/g°C.

Example CalculationExample Calculation

Calorimetry CalculationsCalorimetry Calculations

• The calorimeter and its contents are the surroundings, so qsurr is found from the mass, heat capacity, and temperature change (q = msΔT)

• The chemical reaction is the system, and qsys is ΔH

• The law of conservation of energy requires that

qsurr + ΔH = 0 or ΔH = -qsurr

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Example CalculationExample Calculation

• When 200. g of a AgNO3 solution mixes with 150. g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34oC. Assume 350. g of solution and a heat capacity (Cs) of 4.18 J/g•oC. Calculate ΔH for the equation

Ag+(aq) + I-(aq) AgI(s)

Example CalculationExample Calculation

• When 200. g of a AgNO3 solution mixes with 150. g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34oC. Assume 350. g of solution and a heat capacity of 4.18 J/g•oC. Calculate ΔH for the equation

Ag+(aq) + I-(aq) AgI(s)

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Bomb CalorimetryBomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ΔE, not ΔH.

• For most reactions, the difference is very small.

Standard Enthalpy ValuesStandard Enthalpy ValuesΔH is well known for many reactions, and it is inconvenient to measure ΔH for every reaction in which we are interested.

• However, we can estimate ΔH using ΔHvalues that are published and the properties of enthalpy.

• Most ΔH values are labeled ΔHo

• Measured under standard conditions– P = 1 atm (but for gases P = 1 bar)– T = usually 298.15 K (25 oC)– (Concentration = 1 mol/L)

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HessHess’’s Laws Law

Hess’s law states when two or more thermochemical equations are added, the enthalpy change of the resulting equation is the sum of those for the added equations

C(s) + O2(g) CO2(g) ΔH = -393.5 kJCO2(g) CO(g) + ½O2(g) ΔH = +283.0 kJ

C(s) + ½O2(g) CO(g) ΔH = -110.5 kJ

HessHess’’s Laws Law

Hess’s law states that “If a reaction is carried out in a series of steps, ΔHfor the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

ΔH is a state function.

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• Given the thermochemical equations2WO2(s) + O2(g) 2WO3(s) ΔH = -506 kJ2W(s) + 3O2(g) 2WO3(s) ΔH = -1686 kJ

calculate the enthalpy change for2W(s) + 2O2(g) 2WO2(s)

HessHess’’s Law Examples Law Example

Enthalpies of FormationEnthalpies of Formation• An enthalpy of formation, ΔHf, is defined as the

enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Standard Enthalpy of FormationStandard Enthalpy of Formation

• Only one enthalpy value is needed for each substance, called the standard enthalpy of formation

• The standard enthalpy of formation is the enthalpy change when one mole of a substance in its standard state is formed from the most stable form of the elements in their standard states

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Standard StateStandard State

• Enthalpy changes depend on the temperature and pressure at which they are measured–When applying Hess’s law, all values must

refer to the same conditions of pressure and temperature

• The standard state of a substance at a specified temperature is the pure form at 1 atm pressure–Tabulated values for enthalpy refer to the

standard state, usually at a temperature of 25oC

Standard Enthalpies of FormationStandard Enthalpies of Formation

Standard enthalpies of formation, ΔHf, are measured under standard conditions (25°C and 1.00 atm pressure).

°

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Calculation of Calculation of ΔΔHHWe can calculate ΔH in this way:

ΔH = Σ n ΔHf(products) - Σ m ΔHf(reactants)

where n and m are the stoichiometric coefficients.

°

Example Calculation:Example Calculation:Enthalpy of reactionEnthalpy of reaction

• Use standard enthalpies of formation to calculate the enthalpy change for the reactionP4O10(s) + 6H2O(g) 4H3PO4(s)Substance ΔHf

o (kJ/mol)P4O10(s) -2940H2O(g) - 242H3PO4(s) -1279

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Standard Enthalpy of FormationStandard Enthalpy of Formation

• The symbol used for standard enthalpy of formation is ΔHf

o, where the superscript o designates standard state– Some examples of standard enthalpies of

formation are:

C(graphite) + O2(g) CO2(g) ΔHfo[CO2(g)]

H2(g) + ½O2(g) H2O(l) ΔHfo[H2O(l)]

2Na(s) + Se(s) + 2O2(g) Na2SeO4(s) ΔHfo[Na2SeO4(s)]

H2(g) H2(g) ΔHfo[H2(g)] = 0

Calculation of Calculation of ΔΔHH

•Imagine this as occurringin 3 steps:

C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)

C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g)

4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)

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Calculation of Calculation of ΔΔHH

•Imagine this as occurringin 3 steps:

C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)

C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g)

4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)

Calculation of Calculation of ΔΔHH

•Imagine this as occurringin 3 steps:

C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)

C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g)

4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)

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C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)

C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g)

4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)

C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)

Calculation of Calculation of ΔΔHH

• The sum of these equations is:

C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)Calculation of Calculation of ΔΔHH

ΔH = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) - (-103.85 kJ)= -2219.9 kJ

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Enthalpy of reaction

Example 1:

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

ΔHfo(CO2) = -393.5 kJ

The enthalpy of reaction is -393.5 kJ.

Example 2:

2C(s) + O2(g) 2CO(g)

ΔHfo(CO) = -110.5 kJ

The enthalpy of reaction is 2*(-110.5 kJ) = -221.0 kJ.

ΔH = Σ n ΔHf(products) - Σ m ΔHf(reactants)

Hess’ Law example

2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ

C3H4(g) + 4O2(g) 3CO2(g) + 2H2O(l) ΔH = -1937 kJ

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔH = -2220 kJ

Use Hess’ Law to calculate the enthalpy of the following hydrogenation reaction:

C3H4(g) + 2H2(g) C3H8(g)

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Calculate the standard enthalpy of formation of gaseous diborane(B2H6) using the following thermochemical information:

4B(s) + 3O2(g) 2B2O3 (s) ΔH° = -2509.1 kJ

2H2(g) + O2(g) 2H2O(l) ΔH° = -571.7 kJ

B2H6(g) + 3O2(g) B2O3(s) + 3H2O(l) ΔH° = -2147.5 kJ

HessHess’’s Laws Law

• Calculate ΔH for3 C2H2(g) C6H6(l)

• Given the thermochemical equations:ΔH(kJ)

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) -16922C6H6(l) + 15O2(g) 12CO2(g) +6H2O(l) -6339

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Enthalpy Question

Consider the reaction:

2Mg(s) + O2(g) 2MgO(s) ΔH = -1204 kJ

a) Exothermic or endothermic?

b) Calculate amount of heat transferred when 2.4 g of Mg(s) reacts at constant pressure.

c) How many grams of MgO are produced during an enthalpy change of -96.0 kJ?

d) How many kJ of heat are absorbed when 7.50 g of MgO(s) is decomposed into Mg(s) and O2(g) at constant pressure?

Energy ResourcesEnergy Resources

• Modern society requires large quantities of energy that is generated from the earth’s natural resources. The three principal sources of this energy are–Fossil fuels–Nuclear reactors–Hydroelectric plants

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Fossil FuelsFossil Fuels

• Fossil fuels originated from the decay of living organisms millions of years ago, and account for about 80% of the energy generated in the U.S.

• The fossil fuels used in energy generation are– Natural gas, which is 70 - 80% methane (CH4)– Liquid hydrocarbons obtained from the distillation

of petroleum– Coal - a solid mixture of large molecules with a

H/C ratio of about 1

Problems with Fossil FuelsProblems with Fossil Fuels• Fossil fuels are nonrenewable

resources–at current consumption rates, natural

gas and petroleum will be practically depleted at some point.

• Impurities in fossil fuels are a major source of pollution

• Burning fossil fuels produces large amounts of CO2

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Air PollutionAir Pollution• A large fraction of the fossil fuels (especially

coal) contains sulfur, which is converted into SO2 when the fuel is burned– Atmospheric SO2 is slowly converted into SO3 that

dissolves in water and forms sulfuric acid, a major contributor to acid rain

• Chemical scrubbers can be used to treat stack gases and remove the SO2

• Research is in progress to develop methods of removing sulfur from fuels before they are burned

Greenhouse EffectGreenhouse Effect

• The earth receives energy from the sun, and loses it by radiation

• The greenhouse effect results in the warming of the environment, if more energy is received than is radiated

• An increase in the carbon dioxide in the atmosphere from burning fossil fuels reduces radiation losses and contributes to warming by the greenhouse effect

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Nuclear EnergyNuclear Energy• Since the end of World War II nuclear power

plants have become the second most important source of energy– about 21% of electricity in the U.S. come from nuclear

power plants– In France more than 80% of the electricity is generated

using nuclear energy

Pros and Cons of Nuclear EnergyPros and Cons of Nuclear Energy• Relatively low fuel costs• Nuclear plants do not

contribute to air pollution• Since no CO2 is generated,

nuclear plants do not contribute to the greenhouse effect

• Waste from nuclear reactors remain dangerously radioactive for several decades

• High cost of decommissioning obsolete reactors

• Nuclear accidents, while rare, have enormous potential for damage

Solar EnergySolar Energy• Only a very small fraction (< 0.1%) of

the energy provided by the sun, is sufficient to meet the energy needs of the world

• Currently, no practical way is available to convert this energy into useful form for most of the energy needs of mankind

• In remote areas, when power consumption is modest, current solar energy technology is practical

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FuelsFuels

The vast majority of the energy consumed in this country comes from fossil fuels.

Energy in FoodsEnergy in FoodsMost of the fuel in the food we eat comes from carbohydrates and fats.