Thermochemistry

Chapter 5Thermochemistry

John D. Bookstaver

St. Charles Community College

St. Peters, MO

2006, Prentice Hall, Inc.

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

Thermochemistry

December 16HOMEWORK

• Do questions 1,2,9,11,17,19,20

• First law of thermodynamics 23,25,29

• Enthalpy 31 to 45 odd

• Lab Notebook due by next Monday

Thermochemistry

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.

Thermochemistry

Potential Energy

Energy an object possesses by virtue of its position or chemical composition.

Thermochemistry

Electrostatic PEArises from the interaction between charged

particles.

Eel = k Q1 Q2 k = 8.99x109 Jm/C2

d Q charge of particles

d distance

If charges are same sign they repel and Eel is positive. If they have opposite sign they attract and Eel is negative.

The lower the energy of system the more stable it is. Thus the more strongly opposite charged particles attract each other.

Thermochemistry

Kinetic Energy

Energy an object possesses by virtue of its motion.

12

KE = mv2

Thermochemistry

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

Thermochemistry

Joule

• A mass of 2 kg moving at a speed of 1 m/s possesses a kinetic energy of 1 J.

calorie: the amount of heat needed to increase the temperature of 1 g of water from 14.5 0C to 15.5 0C.

• 1 Cal = 1000 cal = 1 kcal

Thermochemistry

System and Surroundings• Consider the reaction inside

the piston

• 2H2(g) +O2->2H2O(g)+heat

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). Reactants an products.

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

Thermochemistry

Work

• Energy is used to move an object over some distance.

• w = F d,

where w is work, F is the force, and d is the distance over which the force is exerted.

Thermochemistry

Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

• HEAT THE ENERGY TRANSFERRED BETWEEN A SYSTEM AND ITS SURROUNDINGS AS A RESULT OF THEIR

DIFFERENCE IN TEMPERATURE

Thermochemistry

December 17

• First law of thermodynamics

• Internal energy

• Exchange of heat – endo and exo processes

• State functions

• Enthalpy – Heat of a reaction

• Calorimetry

Thermochemistry

Transferal 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.

Thermochemistry

Transferal 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.

Thermochemistry

Transferal 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.

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.

Thermochemistry

First Law of Thermodynamics• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Use Fig. 5.5

Thermochemistry

Internal 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.

Use Fig. 5.5

Thermochemistry

Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

Use Fig. 5.5

Thermochemistry

Changes in Internal Energy

• If E > 0, Efinal > Einitial

Therefore, the system absorbed energy from the surroundings.

This energy change is called endergonic.

Thermochemistry

Changes in Internal Energy

• If E < 0, Efinal < Einitial

Therefore, the system released energy to the surroundings.

This energy change is called exergonic.

Thermochemistry

Changes in Internal Energy

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

• That is, E = q + w.

Thermochemistry

E, q, w, and Their Signs

Thermochemistry

Exchange of Heat between System and Surroundings

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

Thermochemistry

Exchange of Heat between System 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.

Thermochemistry

State Functions

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

Thermochemistry

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.

Thermochemistry

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.

Thermochemistry

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.

Thermochemistry

Work

PV work is the work associated with a change of

volume ( V) that occurs against a resisting pressure.

Thermochemistry

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

w = −PV

Thermochemistry

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 = E + PV

Thermochemistry

Enthalpy

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

H = (E + PV)

• This can be written

H = E + PV

Thermochemistry

Enthalpy

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

H = E + PV

H = (q+w) − w

H = q

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

Thermochemistry

Endothermicity and Exothermicity

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

Thermochemistry

Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Thermochemistry

Enthalpies of Reaction

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

H = Hproducts − Hreactants

Thermochemistry

Enthalpies of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Thermochemistry

The Truth about Enthalpy

1. Enthalpy is an extensive property.

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

Thermochemistry

Practice problem– Change in enthalpy – Heat of reaction

3CuCl2 + 2Al 3Cu + 2AlCl3 ΔH = -793.5 kJ

1. What is ΔH for the following reaction?

6CuCl2 + 4Al 6Cu + 4AlCl3

2. What is ΔH for the following reaction?

9 Cu + 6 AlCl3 9 CuCl2 + 6 Al

3. Given the equation at the top of the page, how much energy will be given off by the reaction of 2.07 g of Al with excess CuCl2?

Thermochemistry

December 20

• Enthalpy practice

• Calorimetry

Thermochemistry

Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Thermochemistry

Heat Capacity and Specific Heat• Calorimeter = apparatus that measures heat flow.• Heat capacity = the amount of energy required to raise

the temperature of an object (by one degree). (J °C-1)

• Molar heat capacity = heat capacity of 1 mol of a substance. (J mol-1 °C-1)

• Specific heat = specific heat capacity = heat capacity of 1 g of a substance. (J g-1 °C-1)

Tq substance of gramsheat specific

Thermochemistry

• Remember a change in T in K and in 0C are equal in magnitude

• T in K = T in 0C

Thermochemistry

Molar heat capacity

• The heat capacity of one mole of substance. The amount of heat required to increase the temperature of 1 mole of substance 1 K.

Thermochemistry

Constant Pressure Calorimetry• Atmospheric pressure is constant!

• HEAT LOST = HEAT GAINED

T

qq

qH P

solution of grams

solution ofheat specificsolnrxn

TmCTmC Object 1 Object 2

Thermochemistry

Constant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/mol-K), we can measure H for the reaction with this equation:

q = m s T

Thermochemistry

Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

s =q

m T

Thermochemistry

Sample problem 1

Find the specific heat of water if 209 J is required to increased the temperature of 50.0 g of water by 1.00K s=q / m T

Thermochemistry

Sample Problem 2

• How much heat is needed to warm 250g of water from 25 0C to 75 0C.

• Specific heat of water 4.18 J/gK

• Find the molar heat capacity of water.

Thermochemistry

Sample problem 3

25.0 mL of 0.100 M NaOH is mixed with 25.0 mL of 0.100 M HNO3, and the temperature of the mixture increases from 23.1°C to 49.8°C. Calculate ΔH for this reaction in kJ mol-1. Assume that the specific heat of each solution is equal to the specific heat of water, 4.184 J g-1 °C-1.

Thermochemistry

Sample problem 4

An unknown metal is to be analyzed for specific heat. A 12.84-g piece of the metal is placed in boiling

water at 101.5°C until it attained that temperature. The metal is then removed from the boiling water

and placed into a styrofoam cup containing 54.92 g of water at 23.2°C. The final temperature of the

water was 26.1°C. What is the specific heat of the metal sample, in J g-1 °C-1?

Thermochemistry

Bomb Calorimetry

A more sophisticated model is the bomb calorimeter, it has a chamber where a chemical reaction takes place and a device to start the reaction.

Thermochemistry

HEAT CAPACITY OF THE CALORIMETER

• Heat is released when combustion occurs. The heat is absorbed by the calorimeter and the T of the water rises.

• THE HEAT CAPACITY OF THE CALORIMETER HAS TO BE CALCULATED. It is done by measuring T in a reaction that releases a known amount of heat. The heat capacity of the calorimeter Ccal= amount of heat absorbed / T

Thermochemistry

TCq calrxn

• Reaction carried out under constant volume.

• Use a bomb calorimeter.• Usually study combustion.

Bomb Calorimetry (Constant Volume Calorimetry)

Thermochemistry

Bomb Calorimetry

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

• For most reactions, the difference is very small.

Thermochemistry

Sample problem 5

A 0.5269 g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.

Thermochemistry

Sample problem 6

• After 50 mL of 1.0 M HCl and 50mL of 1.0M NaOH are mixed in a coffe cup calorimeter, the temperature of the solution increases form 21.0 0C to 27.5 0C. Calculate the enthalpy change for the rx in kJ/mol of HCl assuming that the calorimeter loses only a negligible quantity of heat, total volume of solution is 100mL, that its density is 1.0 g/mL and the specific heat is 4.18J/gk

Thermochemistry

December 21

• MORE HESS LAW

• ENTHALPIES OF FORMATION

Thermochemistry

Hess’s Law

H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.

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

Thermochemistry

Hess’s Law

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

Thermochemistry

Hess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Thermochemistry

Example

The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water:

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

Thermochemistry

C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ

H2 (g) + O2 (g) H2O2 (aq) ΔH = -191.2 kJ

H2 (g) + ½ O2 (g) H2O (g) ΔH = -241.8 kJ

H2O (g) H2O (l) ΔH = -43.8 kJ

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

ΔH = ???

Thermochemistry

Enthalpies of Formation

An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Thermochemistry

Hof .

• Standard conditions (standard state): 1 atm and 25 oC (298 K).

• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.

• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

• If there is more than one state for a substance under standard conditions, the more stable one is used.

• Standard enthalpy of formation of the most stable form of an element is zero.

Thermochemistry

Enthalpies of FormationEnthalpies of Formation

Thermochemistry

CONVENTIONAL DEFINITIONS OF STANDARD STATES FOR

ELEMENTS

• Is the form in which the element exist under conditions of 1 atm and 25 C.

• For O2 is gas.

• For Na is solid

• For Hg is liquid

• For C is graphite

• By definition the H of formation of the most stable form of an element is zero.

Thermochemistry

Examples – write formation reactions for each:

1. H2SO4

2. C2H5OH

3. NH4NO3

Thermochemistry

Examples

Use the tables in the back of your book to calculate ΔH for the following reactions:

1. 4 CuO (s) 2 Cu2O (s) + O2 (g)

2. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3. NH3 (g) + HCl (g) NH4Cl (s)

Thermochemistry

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.

Thermochemistry

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• For a reaction

reactantsproductsrxn ff HmHnH

Thermochemistry

Calculation of H

We can use Hess’s law in this way:

H = nHf(products) - mHf(reactants)

where n and m are the stoichiometric coefficients.

Thermochemistry

Calculation of H

• 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)

Thermochemistry

Calculation of H

• 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)

Thermochemistry

Calculation of H

• 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)

Thermochemistry

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 H

• The sum of these equations is:

Thermochemistry

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

Calculation of H

H = [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

Thermochemistry

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

Thermochemistry

Fuels

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

Thermochemistry

HW

• PRELAB FOR HESS LAW LAB DUE TOMORROW (Lab 6 – page 25)

• Remember test on monday

• Hw for Thursday up to 5.77 odd

Thermochemistry

Example

Calculate the total energy produced from the combustion of one gram of gasoline (assume it is pure octane, C8H18). Compare that to the energy produced from the combustion of one gram of ethanol (a renewable fuel – C2H5OH)

Thermochemistry

• Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.

• For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ

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

Thermochemistry

Note that: H1 = H2 + H3

Thermochemistry

JANUARY 8

• Prelab

• Foods and fuels. Read from txbook

• Hess law problems

Thermochemistry

Foods• Fuel value = energy released when 1 g of substance is

burned.• 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.• Energy in our bodies comes from carbohydrates and fats

(mostly).• Intestines: carbohydrates converted into glucose:

C6H12O6 + 6O2 6CO2 + 6H2O, H = -2816 kJ

• Fats break down as follows:2C57H110O6 + 163O2 114CO2 + 110H2O, H = -75,520 kJ

• Fats: contain more energy; are not water soluble, so are good for energy storage.

Thermochemistry

Fuels• In 2000 the United States consumed 1.03 1017 kJ of

fuel.• Most from petroleum and natural gas.• Remainder from coal, nuclear, and hydroelectric.• Fossil fuels are not renewable.

Thermochemistry

Sources of Sources of energy energy

consumed in consumed in USAUSA