THERMOCHEMISTRY

THERMOCHEMISTRYTHERMOCHEMISTRY

ThermochemistryThermochemistry

Chapter 6Chapter 6

Definitions #1Definitions #1

Energy: The capacity to do work or produce heat

Potential Energy: Energy due to position or composition

Kinetic Energy: Energy due to the motion of the object

2

2

1mvKE

Some types Energy

• Radiant energy comes from the sun and is earth’s primary energy source

• 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

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

• Which are Potential? Kinetic?

6.1

Definitions #2Definitions #2

Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between formsThe First Law of Thermodynamics: The total energy content of the universe is constant

First law of thermodynamics

Esystem + Esurroundings = 0

or

Esystem = -Esurroundings

C3H8 + 5O2 3CO2 + 4H2O

Exothermic chemical reaction!

6.3

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings

State Functions State Functions depend ONLY on the present state of the system

ENERGY IS A STATE FUNCTION

A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane!

WORK IS NOT A STATE FUNCTION

WHY NOT???

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.

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, pressure, volume, temperature

6.3

E = Efinal - Einitial

P = Pfinal - Pinitial

V = Vfinal - Vinitial

T = Tfinal - Tinitial

E = q + wE = q + wE = change in internal energy of a system

q = heat flowing into or out of the system

-q if energy is leaving to the surroundings+q if energy is entering from the surroundings

w = work done by, or on, the system

-w if work is done by the system on the surroundings

+w if work is done on the system by the surroundings

Work, Pressure, and VolumeWork, Pressure, and Volume

VPw Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increases

Work has been done on the system by the surroundings

Work Done On the System

6.3

w = F x d

w = -P V

P x V = x d3 = F x d = wFd2

V > 0

-PV < 0

wsys < 0

Work is not a state function!

w = wfinal - winitial

initial final

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands from a vacuum (0.0 atm) to a constant pressure of 3.7 atm?

w = -P V

(a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J1L•atm

= -1430 J

6.3

Enthalpy and the First Law of Thermodynamics

6.4

E = q + w

E = H - PV

H = E + PV

q = H and w = -PV

At constant pressure:

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 at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0 6.4

Energy Change in Chemical Energy Change in Chemical ProcessesProcesses

Endothermic:

Exothermic:

Reactions in which energy flows into the system as the reaction proceeds.

Reactions in which energy flows out of the system as the reaction proceeds.

+ qsystem - qsurroundings

- qsystem + qsurroundings

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)

Endothermic Endothermic ReactionsReactions

Exothermic ReactionsExothermic Reactions

Review so far…Review so far…

1)1)Energy is….Energy is….

2)2)11stst law of thermodynamics law of thermodynamics

3)3)Potential verses Kinetic EnergyPotential verses Kinetic Energy

4)4)State FunctionState Function

5)5)Enthalpy (q, heat, …)Enthalpy (q, heat, …)

6)6)Work is…Work is…

7)7) E = q + w and E = q + w and ww = - = -P P VV

8)8) EndothermicEndothermic

9)9) ExothermicExothermic

CalorimetryCalorimetry

The 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 water1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

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 = m x s

Heat (q) absorbed or released:

q = m x s x t

q = C x t

t = tfinal - tinitial

6.5

The JouleThe Joule

The unit of heat used in modern thermochemistry is the Joule

2

2111

s

mkgmeternewtonjoule

1 joule = 4.184 calories

A Bomb A Bomb CalorimeteCalorimete

rr(commonly used by (commonly used by

Oak Ridge HS)Oak Ridge HS)

A A Cheaper Cheaper

CalorimetCalorimeterer

(commonly used by (commonly used by WSHS)WSHS)

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) 4.1844.184

Ethanol (liquid) Ethanol (liquid) 2.442.44

Water (solid) Water (solid) 2.062.06

Water (vapor) Water (vapor) 1.871.87

Aluminum (solid) Aluminum (solid) 0.8970.897

Carbon (graphite,solid) Carbon (graphite,solid) 0.7090.709

Iron (solid) Iron (solid) 0.4490.449

Copper (solid) Copper (solid) 0.3850.385

Mercury (liquid) Mercury (liquid) 0.1400.140

Lead (solid)Lead (solid) 0.1290.129

Gold (solid) Gold (solid) 0.1290.129

Calculations Involving Specific Calculations Involving Specific HeatHeat

Tmsq

s = Specific Heat Capacity

q = Heat lost or gainedT = Temperature

change m = mass in grams

ORTm

qs

or q = Cp * m * T

Chemistry in Action:

Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) H = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

SubstanceSubstance HHcombustioncombustion (kJ/g)(kJ/g)

AppleApple -2-2

BeefBeef -8-8

BeerBeer -1.5-1.5

GasolineGasoline -34-34

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf6.6

Hess’s LawHess’s Law“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Hess’s Hess’s LawLaw

Hess’s Law Example Hess’s Law Example ProblemProblem

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O Reaction Ho

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

CH4 C + 2H2 +74.80 kJ




C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ

Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

C + O2 CO2 -393.50 kJ




C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

Step #3: Multiply reaction #3 by 2

2H2 + O2 2 H2O -571.66 kJ




C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

2H2 + O2 2 H2O -571.66 kJ

Step #4: Sum up reaction and H

CH4 + 2O2 CO2 + 2H2O

-890.36 kJ

Calculation of Heat of Calculation of Heat of Reaction Reaction Calculate H for the combustion of

methane, CH4: CH4 + 2O2 CO2 + 2H2OHrxn = Hf(products) - Hf(reactants)

Substance Hf

CH4 -74.80 kJ

O2 0 kJ

CO2 -393.50 kJ

H2O -285.83 kJ

Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]

Hrxn = -890.36 kJ

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

6.6

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

C (graphite) + 1/2O2 (g) CO (g)

CO (g) + 1/2O2 (g) CO2 (g)

C (graphite) + O2 (g) CO2 (g)

6.6

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn6.6

Benzene (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)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

6.6

Chemistry in Action: Bombardier Beetle Defense

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) H0 = ?

C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) H0 = 177 kJ/mol

H2O2 (aq) H2O (l) + ½O2 (g) H0 = -94.6 kJ/mol

H2 (g) + ½ O2 (g) H2O (l) H0 = -286 kJ/mol

H0 = 177 - 94.6 – 286 = -204 kJ/mol

Exothermic!

The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

Hsoln = Hsoln - Hcomponents

6.7

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?

The Solution Process for NaCl

Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol6.7

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