Energetics

Thermochemistry and Thermodynamics

Dr. Pál Bauer

2009

Part I

Thermochemistry

Thermochemistry deals with the energy (heat) transfer during

chemical reactions

Key Concepts of Lecture

• Energy.

• Temperature.

• Zeroth law of thermodynamics.

• Internal energy.

• State functions.

• Transfer of energy in forms of heat and work.

• First law of thermodynamics.

• Enthalpy.

• Thermochemical equations.

.

• Calorimetry.

• Hess’s law.

• Standard enthalpy of formation.

• Problems.

Energy

• The ability to do work or transfer heat.– Work: Energy used to cause an object that

has mass to move.– Heat: Energy used to cause the temperature

of an object to rise.

E = q + w

Energy is the ability 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

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

Temperature: Temperature is a measure of average atom/molecule speeds.

For Example:

At 25 F, air molecules have an average speed ~1080 mi/hr

At 100 F, air molecules have an average speed ~1160 mi/hr

TemperatureTemperature is a measure of average atom/molecule speeds.

Temperature Measurement:

Average molecule speeds are difficult to measure directly.

Temperature is almost always measured indirectly (by one of it’s side effects).

One such side effect is thermal expansion. Most materials expand when hotter and contract when cooler.

Cool Iron bar

Temperature: Temperature is a measure of average atom/molecule speeds.

Temperature Measurement:

Average molecule speeds are difficult to measure directly.

Temperature is almost always measured indirectly (by one of it’s side effects).

One such side effect is thermal expansion. Most materials expand when hotter and contract when cooler.

Hot Iron bar

Thermal expansion is a very small effect.

For example: Steel expands 0.006% for every 10 oF

From 20 oF to 100 oF:

A 1.0 m steel bar expands 0.5 mm

From 20 oF to 100 oF:

A 500 ft steel bridge expands 3 in.

Thermal expansion is a very small effect.

The effect is greatly exaggerated in a ‘liquid in glass’ thermometer.

Thermal expansion is a very small effect.

Temperature Scales

100 oC212 oF

0 oC32 oF

- Absolute temperature T(K) = tc + 273,15

Thermal expansion is a very small effect.

Bi-metallic Strip or Spring

Upon heating: metal 2 expands more than metal 1

The zeroth law of thermodynamics is a generalization about the

thermal equilibrium among bodies, or thermodynamic systems, in contact. It

results from the definition and properties of temperature. It can be stated as:

"If A and C are each in thermal equilibrium with B, A is also in

thermal equilibrium with C."

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

Types of Energy

• potential energy- (PE) due to position or composition– ex. attractive or repulsive forces

• kinetic energy- (KE) due to motion of the object– KE = ½mv2 :depends on mass and volume

Units 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 J = 1 kg m2

s2

Transfer of Energy

Etotal = Ek + Ep + Internal energy of system (U)

• Two Ways to Transfer Energy:– Heat- (q) transfer of energy between two objects because of

a temperature difference– Work- (w) force acting over a distance

U = q + w

ΔU = Δq + Δw

Internal Energy

• (U) sum of potential and kinetic energy in system

• can be changed by work, heat, or both• U = PE + KE• ∆U = q + w

Signs• signs are very important• signs will always reflect the system’s point of

view unless otherwise stated

∆U q w

change in internal energy

heat work

exothermic- - -

endothermic + + +

First Law of Thermodynamics

• ΔU =Δq + Δw• Energy cannot be created nor

destroyed.• Therefore, the total energy of the

universe is a constant.• Energy can, however, be converted

from one form to another or transferred from a system to the surroundings or vice versa.

Pathway

• the specific conditions of energy transfer

• energy change is independent of pathway because it is a state function

• work and heat depend on pathway so are not state functions

• state function- depends only on current conditions, not past or future

Thermodynamics

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

6.7

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state.– In the system below, the water could have reached

room temperature from either direction.

State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the

system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

State Functions

• However, q and w are not state functions.

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

in the two cases.

Transfer of Energy

• exothermic- – energy is produced in reaction– flows out of system– container feels hot to the touch

• endothermic-– energy is consumed by the reaction– flows into the system– container feels cold to the touch

Transfer of Energy

Combustion of Methane Gas is exothermic

Transfer of Energy

Reaction between nitrogen and oxygen is endothermic

Transfer of Energy

• the energy comes from the potential energy difference between the reactants and products

• energy produced (or absorbed) by reaction must equal the energy absorbed (or produced) by surroundings

• usually the molecules with higher potential energy have weaker bonds than molecules with lower potential energy

Work

When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

Work

• common types of work

– expansion- work done by gas

– compression- work done on a gas

A

FP

hAPhFw

VPw

expansion +∆V -w

compression -∆V +w

P is external P is external pressure – not pressure – not internal like we internal like we normally refer tonormally refer to

Example 1

• Find the ∆E for endothermic process where 15.6 kJ of heat flows in the system and 1.4 kJ of work is done on system– Since it is endothermic, q is + and w is +

kJ

kJkJ

wqE

0.17

4.16.15

Example 2

• Calculate the work of expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm.– Since it is an expansion, ∆V is + and w is -

atmL

LLatm

VPw

270

)4664)(15(

Example 3• A balloon was inflated from 4.00 x 106 L to

4.50 x 106 L by the addition of 1.3 x 108 J of heat. Assuming the pressure is 1.0 atm, find the ∆E in Joules.

(1 L∙atm=101.3 J)– Since it is an expansion, ∆V is + and w is -

VPqwqE

)1000.4105.4)(0.1(103.1 668 LLatmJE

JJJE 778 100.8)101.5(103.1

Enthalpy• If a process takes place at constant

pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume:

H = U + PV

Enthalpy

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (U + PV)

• This can be written

H = U + PV + VΔP

Enthalpy

• Since U = q + w and w = −PV, we can substitute these into the enthalpy expression:

H = U + PV

H = (q+w) − w

H = qp

• So, at constant pressure the change in enthalpy is the heat gained or lost.

Endothermicity and Exothermicity

• A process is endothermic, then, when H is positive.

Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Enthalpy of phase changesThe phase diagram of water

Phase transitions

Enthalpies of Reactions

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

H2O (s) H2O (l) H = 6.01 kJ/mol 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 H changes

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

• 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 mol x 6.01 kJ/mol = 12.0 kJ

6.4

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

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

Thermochemical Equations

6.4

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

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

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

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

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/molf

6.6

Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Heat Capacity

• (C) how much heat it takes to raise a substance’s T by one °C or K

• the amount of energy depends on the amount of substance

K)or C(in Tin increase

J)(in absorbedheat Ccapacityheat

Heat Capacity

• specific heat capacity– (s) heat capacity per gram– in J/°C*g or J/K*g

• molar heat capacity– heat capacity per mole– in J/°C*mol or J/K*mol

Constant-Pressure Calorimetry

• uses simplest calorimeter (like coffee-cup calorimeter) since it is open to air

• used to find changes in enthalpy (heats of reaction) for reactions occurring in a solution since qP = ∆H

• heat of reaction is an extensive property, so we usually write them per mole so they are easier to use

Constant-Pressure Calorimetry

• when 2 reactants are mixed and T increases, the chemical reaction must be releasing heat so is exothermic

• the released energy from the reaction increases the motion of molecules, which in turn increases the T

Constant-Pressure Calorimetry

• If we assume that the calorimeter did not leak energy or absorb any itself (that all the energy was used to increase the T), we can find the energy released by the reaction:

E released by rxn = E absorbed by soln

∆H = qP = sP x m x ∆T

Constant-Volume Calorimetry

• uses a bomb calorimeter

• weighed reactants are placed inside the rigid, steel container and ignited

• water surrounds the reactant container so the T of it and other parts are measured before and after reaction

Constant-Volume Calorimetry

• Here, the ∆V = 0 so -P∆V = w = 0

∆U = q + w = qV for constant volume

U released by rxn = ∆T x Ccalorimeter

Example 1

• When 1 mol of CH4 is burned at constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CH4 at constant P.

• 890 kJ is released per mole of CH4

kJmolCH

kJ

gCH

moleCHgCH 320

1

890

04276.16

18.5

44

44

Example 2

• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a coffee-cup calorimeter, solid BaSO4 forms and the T increases to 28.1°C. The specific heat capacity of the solution is 4.18 J/g*°C and the density is 1.0 g/mL. Find the enthalpy change per mole of BaSO4 formed.

Example 2

• Write the net ionic equation for the reaction:

Ba2+ (aq) + SO42- (aq) BaSO4(s)

• Is the energy released or absorbed? What does that mean about ∆H and q?

exothermic: -∆H and –qP

• How can we calculate ∆H or heat?

heat = q = sP x m x ∆T

• How can we find the m?

use density and volume

Example 2• Find the mass:

• Find the change in T:

• Calculate the heat created:

gmLmLg

VDmV

mD

33 100.2)100.2)(/0.1(

CT 1.30.251.28

JCggC

Jq 43 106.2)1.3)(100.2)(18.4(

Example 2

• since it is a one-to-one ratio and the moles of reactants are the same, there is no limiting reactant

• 1.0 mol of solid BaSO4 is made so

∆H= -2.6x104 J/mol = -26 kJ/mol

Example 3

• Compare the energy released in the combustion of H2 and CH4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

Example 3• methane: CH4

• hydrogen: H2

• The energy released by H2 is about 2.5 times the energy released by CH4

gkJgkJH

kJCC

kJH

TCH rcalorimete

/555.1/83

83)3.7()3.11(

gkJgkJH

kJCC

kJH

TCH rcalorimete

/14115.1/162

162)3.14()3.11(

Hess’s Law

Hess’s Law

• since H is a state function, the change in H is independent of pathway

• Hess’ Law- when going from a set of reactants to a set of products, the ∆H is the same whether it happens in one step or a series of steps

Example 1

Example 1

• N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ

OR

• N2(g) + O2(g) 2NO(g) ∆H = 180 kJ

2NO(g) + O2(g) 2NO2(g) ∆H = -112 kJ

N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ

Rules

1. If a reaction is reversed, the sign of ∆H must be reversed as well.

– because the sign tells us the direction of heat flow as constant P

2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction.

If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way, because ∆H is an extensive property

Example 2

• Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond.

Cgraphite (s) Cdiamond (s) ∆H=?

Example 2

(1) Cgraphite(s) + O2(g) CO2(g) ∆H=-394kJ/mol

(2) Cdiamond(s) + O2(g) CO2(g) ∆H=-396kJ/mol

• to get the desired equation, we must reverse 2nd equation:

(1) Cgraphite(s) + O2(g) CO2(g) ∆H=-394kJ/mol

-(2) CO2(g) Cdiamond(s) + O2(g) ∆H=396kJ/mol

Cgraphite (s) Cdiamond (s)∆H=-394 + 396

∆H=2 kJ/mol

Standard Enthalpies of Formation

Standard Enthalpy of Formation

• ∆Hf°

• change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states

• ° means that the process happened under standard conditions so we can compare more easily

●● ∆∆HfHfoo = 0 for elements in their standard states. = 0 for elements in their standard states.

Standard States• For a COMPOUND:

– for gas: P = 1 atm– pure liquid or solid state– in solution: concentration is 1 M

• For an ELEMENT:– form that it exists in at 1 atm and 25°C

O: O2(g) K: K(s) Br: Br2(l)

Using Standard Enthalpies of Formation

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

where – n = number of moles of products/reactants– ∑ means “sum of”– ∆Hf° is the standard enthalpy of formation for

reactants or products

• ∆Hf° for any element in standard state is zero so elements are not included in the summation

Using Standard Enthalpies of Formation

• since ∆H is a state function, we can use any pathway to calculate it

• one convenient pathway is to break reactants into elements and then recombine them into products

Using Standard Enthalpies of Formation

0)()()( 422

CHfOHfCOf HHHH

Using Standard Enthalpies of Formation

Example 1

• Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

• break them apart into elements and then recombine them into products

Example 1

Example 1

• can be solved using Hess’ Law:

(1) 4NH3(g) 2N2(g) + 6H2(g) -4∆Hf°NH3

(2) 7O2(g) 7O2(g) 0

(3) 2N2(g) + 4O2(g) 4NO2(g) 4 ∆Hf°NO2

(4) 6H2(g) + 3O2(g) 6H2O(l) 6 ∆Hf°H2O

)286

(6)34

(40)46

(4mol

kJmol

mol

kJmol

mol

kJmolH

kJH 1396

Example 1

• can also be solved using enthalpy of formation equation:

kJH 1396

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

]0)46(4[)]286(6)34(4[ΔHreaction

ExExaample 2mple 2ExExaample 2mple 2

∆∆HHoorxnrxn = ∆H = ∆Hff

oo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O) O)

- {3/2 ∆H- {3/2 ∆Hffoo

(O(O22) + ∆H) + ∆Hffoo

(CH(CH33OH)} OH)}

= (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ)

- {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}

∆∆HHoorxnrxn = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol

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

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

Example 3

• Calculate the standard enthalpy change for the following reaction:

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

kJ.850ΔH

)]826(1)0(2[)]0(2)1676(1[ΔH

reaction

reaction