Thermochemistry:

Energy Flow

and

Chemical

Reactions

• thermodynamics

• internal energy – definition, first law

• enthalpy – definition, energy diagrams, calorimetry,

theoretical calculation (heats of formation and

bond energies), stoichiometry

• hess’s law

• energy from foods

Outline

Thermochemistry is a branch of thermodynamics

that deals withthe heat involved with chemical and

physical changes.

Thermodynamics is

the study of heat and

its transformations.

Thermodynamics

Internal Energy

What is the difference of the two pictures above?

Energy state

If we go from cold water to boiling water as our reaction:

H2O (solid) H2O (gas)

energy state 1 energy state 2

Internal Energy

we define E = a variable to denote the energy of our

reacting substances

= specifically, the internal energy of the substance

undergoing reaction

∆ E = change in energy

= E (boiling water) – E (cold water)

since Energy (gas) > energy (solid)

∆ E = positive , +

For the reaction to proceed, energy has to be supplied.

But where will it come from?

A chemical system and its surroundings.

the system

the surroundings

Internal Energy

System = the reaction itself, the object of our study

Surroundings = the rest of the universe

Internal Energy

DE = Efinal - Einitial = Eproducts - Ereactants

Energy diagrams for the transfer of internal energy (E) between a

system and its surroundings.

Internal Energy

Energy changes occurs because of the exchange of

energy between the system and the surroundings!

Internal Energy

Notice that all the equations/energies refer to the system!

But we can also define it from the point of view of the

surroundings.

Case 1: surroundings supplied energy to the system∆ E = E2, surr - E1, surr = negative

We have to specify which we are referring to: system or

surroundings

State 1: State 2:

E 1, surr

E 1, sys

E 2, surr

E 2, sys

E 2, sys - E1, sys = ∆ E system = positive, +

E 2, surr - E1, surr = ∆ E surroundings = negative, -

Internal Energy

So we have a discrepancy of notations and definitions.

As per convention, whenever we have E1, E2, etc., we are

always referring to the system. Henceforth, we are always

referring to the system in all our studies of energies.

Notice, however that∆ E system + ∆ E surr = 0

since: system + surroundings = universe

∆ E universe = 0 = E2, universe - E1, universe = 0

E2, universe = E1, universe

The energy of the universe is constant!

Internal Energy

First Law of Thermodynamics:

The energy of the universe is

constant.

Thus, energy can neither be

created nor destroyed. It can

be only be converted from

one form to another.

Internal Energy

Let’s define formally E = internal energy

= total energy of the system

= summation of all the kinetic and potential

energies of the system

Experimentally: ∆ E = q + w

q = heat = energy transferred from a hotter object to a colder one

w = work = energy used to cause an object to move against a force

That means, we can change the energy of the system by

applying heat and doing work. Or,

When energy is transferred from one object to another,

it appears as work and/or as heat.

A system transferring energy as heat only.

Internal Energy

A system losing energy as work only.

Energy, E

Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq)

DE<0work done on

surroundings

Internal Energy

The Sign Conventions* for q, w and DE

q w+ = DE

+

+

-

-

-

-

+

+

+

-

depends on sizes of q and w

depends on sizes of q and w

* For q: + means system gains heat; - means system loses heat.

* For w: + means word done on system; - means work done by system.

Internal Energy

Internal Energy

DE = (sign), (number), (unit)

Sign = gives direction as to the transfer of energy

+ = E final > E initial = system gained energy from the surroundings

- = E final < E initial = system lost energy to the surroundings

Units of Energy

Joule (J)

calorie (cal)

British Thermal Unit

1 cal = 4.184 J

1 J = 1 kg*m2/s2

1 Btu = 1055 J

Internal Energy

Calorie (Cal) 1 Calorie = 1 000 calorie

Sample Problem

Hydrogen and oxygen reacts in a cylinder. As the reaction

occurs, the system loses 1150 J of heat to the

surroundings and the expanding gas product does 480 J of

work on the surroundings as it pushes against the

atmosphere. What is the change in the internal energy of

the system?

q = heat = - 1150 J

w = work = - 480 J

∆E = q + w= (-1150 J) + (- 480 J)

= - 1630 J

A system receives 425 J of heat and delivers 400 J of

work to its surroundings. What is the change in internal

energy of the system (in joules)?

Answer: ∆ E = q + w

= (+ 425) + (- 400 J)

= + 25 J

Sample Problem

Worksheet # 9-1

1. Complete combustion of 1.0 metric ton of coal

(assuming pure carbon) to gaseous carbon dioxide

releases 3.3 x 1010 J of heat. Convert this energy to (a)

kilojoules; (b) kilocalories and (c) british thermal unit.

2. A system conducts 225 cal of heat to the surroundings

while delivering 428 cal of work. What is the change in

internal energy of the system (in cal and J)?

3. What is the change in internal energy (in J) of a system

that releases 675 J of thermal energy to its

surroundings and has 525 calories of work done on it?

Worksheet # 9-1: Answers

1. Complete combustion of 1.0 metric ton of coal (assuming

pure carbon) to gaseous carbon dioxide releases 3.3 x 1010 J

of heat. Convert this energy to (a) kilojoules; (b) kilocalories

and (c) british thermal unit.

Answers: a) 3.3 x 107 kJ b) 7.9 x 106 kcal c) 3.1 x 107

Btu

2. A system conducts 225 cal of heat to the surroundings while

delivering 428 cal of work. What is the change in internal

energy of the system (in cal and J)? Answer: - 653 cal =

- 2732 J

3. What is the change in internal energy (in J) of a system that

releases 675 J of thermal energy to its surroundings and has

525 calories of work done on it? Answer: 1522 J

Some interesting

quantities of energy.

Two different paths for the energy change of a system.

Enthalpy

Pressure-volume

work.

Enthalpy

Enthalpy

There is another energy variable we call enthalpy:

H = E + PV

PV = work

∆H = ∆E + P ∆V

Formal definition: Enthalpy is the thermodynamic quantity

that is the sum of internal energy and the product of the

pressure-volume work.

The change in Enthalpy equals the heat gained or lost at

constant pressure.

Most chemical reactions occur at constant pressure, so ∆H

is more relevant than ∆E.

w = - PDV

DH = DE + PDV

qp = DE + PDV = DH

DH ≈ DE in

1. Reactions that do not

involve gases.

2. Reactions in which the

number of moles of gas does

not change.

3. Reactions in which the

number of moles of gas does

change but q is >>> PDV.

H = E + PV

where H is enthalpy

Enthalpy

Enthalpy

Let’s examine enthalpy and internal energy more closely:

We defined ∆E = E2 – E1

We can calculate ∆E from the first law of thermodynamics.

What about E2, E1 or the exact energy value of a system.

E, internal energy = summation of all the energies of the

system, including KE and PE

For a number of particles (1, 2, 3 particles) energy can be

calculated, but if we are dealing with macroscopic amounts (ex:

a mole of water, 6.02 x 1023 particles), the exact value is

incalculable and impossible!

But is it important?

Enthalpy

We do not need to calculate the energy values exactly. But

we assume that the energy of a system at a defined

condition is exact, fixed and unchanging.

Example:

1.0 mol of 2.5 L water at 90 C, 1.5 atm pressure has an

internal energy E and an enthaly H whenever this condition

is specified

Mathematically,

we call ∆E and ∆H = state functions

state function = property of a system that is determined by

specifying its conditions or its state (ex. T, P, etc.)

= the value of a state function does not depend on the

particular history of the sample, only on its present condition

Enthalpy

State function analogy: scaling a mountain

The height of the mountain is a state function. However, the

way you can climb the mountain is not.

H, E = state functions

q, w = not state functions, path dependent (value depends

on how the process took place)

Enthalpy

Exothermic Reactions: ∆H = H final – H initial = negative, -

energy is released by the system

Endothermic reactions: ∆H = H final – H initial = positive, +

energy is absorbed by the system

The sign of enthalpy (like internal energy) indicates the

direction of heat transfer during a process that occurs at

constant pressure.

Enthalpy diagrams for exothermic and endothermic processes.

Enth

alp

y, H

Enth

alp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

HinitialHfinal

Hfinal

H2O(l)

H2O(g)

heat out heat inDH < 0 DH > 0

A Exothermic process B Endothermic process

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)

Enthalpy

Enthalpy

Video: endothermic reaction (endo final.wmv)

Enthalpy

It turns out experimentally that:

∆E = energy transferred at constant volume conditions

∆H = energy transferred at constant pressure conditions

In most chemical reactions, we are mostly concerned with

enthalpy. As most chemical reactions occur at constant

pressure.

Thermochemical equations

2H2 (gas)+ O2 (gas)→ 2H2O (gas) ∆H = - 483.6 kJ

The enthalpy value is given at the end of the chemical

equation.

Some Important Types of Enthalpy Change

heat of combustion (DHcomb)

heat of formation (DHf)

heat of fusion (DHfus)

heat of vaporization (DHvap)

C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)

K(s) + 1/2Br2(l) KBr(s)

NaCl(s) NaCl(l)

C6H6(l) C6H6(g)

Enthalpy

Write a balanced equation and draw an approximate

enthalpy diagram for each of the following:

a) Combustion of one mol of methane in oxygen

b) The evaporation of liquid water

Answer:

a) CH4 + 2O2 → CO2 + 2H2O, exothermic

b) H2O(l) → H2O(g), endothermic

Sample Problem

Enth

alp

y, H

CH4 + 2O2

CO2 + 2H2O Enth

alp

y, H

H2O(l)

H2O(g)

heat out heat in

Worksheet # 9-2

In each of the following cases, determine the sign of

enthalpy, state whether the reaction is exothermic or

endothermic and draw an enthalpy diagram.

1. H2 (g) + ½ O2 (g) → H2 O (l) + 285.8 kJ

2. CO2 (s) + 260 kJ → CO2 (g)

3. The heat released for the combustion of liquid ethanol

4. The vaporization of liquid ammonia

Worksheet # 9-2: AnswersE

nth

alp

y, H

Enth

alp

y, H

heat out

heat in

exothermic

endothermic

H2 (g) + ½ O2 (g)

H2 O (l) 2CO2 + 3H2O

C2H5OH + 3O2

CO2 (s)

CO2 (g)

+ 260 kJ

NH3 (l)

NH3 (g)

- 285.8 kJ

Enthalpy

Properties of enthalpy:

1. Enthalpy is an extensive property. Thus its magnitude is

directly proportional to the amount of substance reacted

or produced.

Implication: in a thermochemical equation, if the equation is multiplied by 2, the ∆H is also multiplied by 2. If the

equation is divided by 2, the ∆H is also divided by two.

Example: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ

2CH4(g) + 4O2 (g) → 2CO2 (g) + 4H2O(l) ∆H = 2(- 890) kJ

Enthalpy

2. ∆H reaction = - ∆H reverse reaction

CO2 (g) + 2H2O (l) → CH2 (g) + 2O2 (g) ∆H = 890 kJ

3. ∆H is depends on the state of the chemical species used in

the equation. The state of the substance has to be shown.

Example:

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

CO2 (g) + 2H2O (g) → CH4 (g) + 2O2 (g) ∆H = 978 kJ

Enthalpy

Enthalpy and chemical stoichometry-The calculation of the enthalpy for a particular reaction and

its relation to the chemical equation is governed property 1

of enthalpy – it is an extensive property.

Example:

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

relationships: 1 mol CH4 is to ∆H = - 890 kJ

2 mol O2 (g) is to ∆H = - 890 kJ

1 mol CO2 is to ∆H = - 890 kJ

2 mol H2O is to ∆H = - 890 k

Sample Problem

Consider the equation:

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

1. How many moles of methane, CH4 (g) has to be burned

to generate 1500 kJ of energy?

2. How many grams of carbon dioxide is produced if the

enthalpy generated in the combustion is 1830 kJ?

Consider the following balanced thermochemical equation

for a reaction sometimes used for H2S production:

1/8 S8 + H2 → H2S ΔH = - 20.2 kJ

a) Is this an exothermic or endothermic reaction?

b) What is the ΔHrxn for the reverse reaction?

c) What is ΔH when 3.2 mol of S8 reacts?

d) What is ΔH when 20.0 g of S8 reacts?

Answers:

a) Exothermic

b) 20.2 kJ

c) -517 kJ

d) -12.6 kJ

Sample Problem

AMOUNT (mol)

of compound A

AMOUNT (mol)

of compound B

HEAT (kJ)

gained or lost

molar ratio from

balanced equation

DHrxn (kJ/mol)

Summary of the relationship

between amount (mol) of

substance and the heat (kJ)

transferred during a reaction.

Enthalpy

Worksheet # 9 - 3

1. Consider the following balanced thermochemical equation for the

decomposition of the mineral magnesite:

MgCO3 → MgO + CO2 ∆H = 117.3 kJ

(a) is heat absorbed or released in the reaction?(b) what is the ∆H for the reverse reaction?(c) what is the ∆H when 5.35 mol of CO2 reacts with excess MgO?(d) What is the ∆H when 2.36 mol of magnesite reacts?(e) what is the ∆H when 0.15 mol of MgO is produced from the reaction?

2. When 1.0 mol of NO forms from its elements, 90.29 kJ of heat is absorbed.(a) write a balanced thermochemical equation for this reaction.(b) how much heat is involved when 1.50 mol of NO decomposes to its elements?(c) how much heat is involved when 1.75 grams of NO decomposes to its elements?(d) how much heat is involved when 3.50 grams of N2 is produced?

Worksheet # 9 – 3: Answers

1. Consider the following balanced thermochemical equation for the

decomposition of the mineral magnesite:

MgCO3 → MgO + CO2 ∆H = 117.3 kJ

(a) is heat absorbed or released in the reaction? Absorbed(b) what is the ∆H for the reverse reaction? -117.3 kJ(c) what is the ∆H when 5.35 mol of CO2 reacts with excess MgO?-627.6 kJ(d) What is the ∆H when 2.36 mol of magnesite reacts? + 276.8 kJ(e) what is the ∆H when 0.15 mol of MgO is produced from the reaction? + 17.6 kJ

2. When 1.0 mol of NO forms from its elements, 90.29 kJ of heat is absorbed?(a) write a balanced thermochemical equation for this reaction?

½ N2 + ½ O2 → NO ∆H = 90.29 kJ(b) how much heat is involved when 1.50 mol of NO decomposes to its elements?(c) how much heat is involved when 1.75 grams of NO decomposes to its elements?(d) how much heat is involved when 3.50 grams of N2 is produced?

b) -135.4 kJ c) -5.3 kJ d) -22.6 kJ

Enthalpy

Calorimetry = measurement of heat flow

Calorimeter = apparatus that measures heat flow

∆H reaction can be determined experimentally using a calorimeter.

Coffee-cup calorimeter.

Enthalpy

Enthalpy

What did the

thermometer say to

the graduated

cylinder?

"You may have

graduated but I've got

many degrees."

Enthalpy

In chemical equations, energy changes can be calculated by

monitoring the temperature changes in the reaction.

q is proportional to ∆T

In equation form: q α ∆T

q = C ∆T

where C = constant of proportionality

= heat capacity

= amount of heat required to raise the temperature of

a substance by 1.0 Kelvin

The larger the heat capacity the greater the heat required to

produce a given rise in temperature.

For 1.0 mol substance , C = C (mol)

Lets look at another experiment:

1.0 L water and 10.0 ml water are heated separately such that each would

have its temperature increase by 10.0 C. What would be the result.

The 10.0 ml water sample will have a greater increase in energy as some of

its molecules can turn into gas.

Implication: mass is also a factor in measuring enthalpies.

q = C ∆T

As there is no mass in the equation, it can be in the heat capacity factor.

C = m c

where m = mass of the substance

c = specific heat of the substance

Thus, q = m c ∆T

Enthalpy

Heat Capacity C = q / ΔT

Specific heat capacity (c) = q / (mass x ΔT)

unit: J/K

unit: J/gK

Molar heat capacity (C mol) = q / (moles x ΔT)

unit: J/mol K

Enthalpy

Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

Specific Heat

Capacity (J/g*K)

SubstanceSpecific Heat

Capacity (J/g*K)

Substance

Compounds

water, H2O(l)

ethyl alcohol, C2H5OH(l)

ethylene glycol, (CH2OH)2(l)

carbon tetrachloride, CCl4(l)

4.184

2.46

2.42

0.864

Elements

aluminum, Al

graphite,C

iron, Fe

copper, Cu

gold, Au

0.900

0.711

0.450

0.387

0.129

wood

cement

glass

granite

steel

Materials

1.76

0.88

0.84

0.79

0.45

Enthalpy

Relation of enthalpy and q (from calorimetry experiments)

q = m c ∆T = q solution

q solution = - q reaction = ∆H reaction

∆H reaction = - m c ∆T

A 295 g aluminum engine part at an initial temperature of

3.0 C absorbs 85.0 kJ of heat. What is the final

temperature of the part (specific heat of aluminum = 0.900

J/g K)?

Answer: Tf = 323 C

Sample Problem

q = m c ∆T85 000 J = 295 g Al (0.9 J/gK) (Tf – 3.0 C)

When 165 ml of water at 22 C is mixed with 85 mL of water

at 82 C, what is the final temperature? Assume that no

heat is lost to the surroundings (density of water is 1.0

g/mL).

(m c ∆T) water 1 = (m c ∆T) water 2

165 ml ( Tf – 22 C ) = 85 ml (Tf – 82 C)

Answer: Tf = 42 C

Sample Problem

Worksheet # 9 - 4

1. Calculate q when 12.0 grams of water is heated from 20

degrees to 100 degrees? ( c water = 4.184 J/g C)

2. The specific heat capacity of silver is 0.24 J/g C.

a) Calculate the energy required to raise the temperature of

150.0 grams Ag from 273 K to 298 K.

b) Calculate the energy required to raise the temperature of

1.0 mol Ag by 1.0 C

c) It takes 1.25 kJ of energy to heat a sample of pure silver

from 12.0 C to 15.2 C. Calculate the mass of the sample

of silver.

3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at

330 K. Calculate the final temperature of the mixture

assuming no heat loss to the surroundings.

Worksheet # 9 – 4: Answers

1. Calculate q when 12.0 grams of water is heated from 20

degrees to 100 degrees? Ans = 4016.6 J

2. The specific heat capacity of silver is 0.24 J/g C.

a) Calculate the energy required to raise the temperature of

150.0 grams Ag from 273 K to 298 K. Ans = 900 J

b) Calculate the energy required to raise the temperature of

1.0 mol Ag by 1.0 C. Ans = 25.9 J

c) It takes 1.25 kJ of energy to heat a sample of pure silver

from 12.0 C to 15.2 C. Calculate the mass of the sample

of silver. Ans = 1627.6 g

3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at

330 K. Calculate the final temperature of the mixture

assuming no heat loss to the surroundings. Ans = 311 K