Thermochemistry

▪ The nature of energy

▪ The first law of thermodynamics
▪ Enthalpy
▪ Enthalpies of reaction
▪ Calorimetry
▪ Hess’s law
▪ Enthalpies of formation
 The nature of energy
❖Thermochemistry is the study of heat transfer in chemical reactions.

❖Energy is the capacity to do work or transfer heat.

❖Energy can be transferred as heat e.g. heat flow from warmer to cooler objects.

❖Work is energy used to cause an object with mass to move.

❖There are two overall types of energy:

1. kinetic energy (K.E.) -the energy of motion. Any object in motion has kinetic energy. K.E = ½ MV²

2. Potential energy (P.E) refers to stored energy that can be due to an object's position. P.E. = mgh g=9.8 m/s²

h-height; m-mass; g-acceleration due to gravity

❖ Units for energy is joules. J = 1 kg.m2/s2

❖ 1 Calorie = 4.184 J
↑ More kinetic energy
 Terminologies

❖ In thermochemistry, the system refers to whatever you are interested in learning about e.g. In a chemical reaction, the

reactant and products is the system. The container is the surrounding

❖ The surroundings refer to everything the system comes into contact with e.g. if you are doing a chemical reaction in a

beaker or container, beaker or container is the surroundings.

❖ Heat refers to the flow of energy between a system and its surroundings when there is a temperature difference.

❖ Work is flow of energy between a system and its surroundings when there is no temperature difference.

❖ Energy changes occurs in form of heat or work.

W=fxd w-work, f- force, d-distance

❖ The system is usually a closed system- where energy can be transferred with its surrounding but not matter.
 Questions

In a thermodynamic study, a chemist focused on the properties of a solution in a flask

sealed with a stopper.

a. What is the system in the study?

b. What are the surrounding in the study?

c. What type of system is in the study?

 Types of systems
• Open system- There is an exchange of mass and energy. E.g. An open frying pan.

• Closed system-There is exchange of energy with its surroundings but no transfer

of matter. E.g. A cup of tea with a lid.

• Isolated system- There is no exchange of matter or energy. E.g. A thermos flask is

used to keep things either cold or hot. Thus a thermos does not allow energy

transfer.
 CALORIES

• Calories is used to describe energy in food and is equal to 1000 calories.

• Calories are also the amount of energy you burn exercising.

• 1 calorie = 4.184 J

• 1 kcal = 4184 J

• 1000 calories = 1.00 kcal

Convert: a. 1.69 J to cal

b. 196 cal to kilojoules

c. 68 cal to kcal

d. 6.89 cal to kilojoules

a. 1.69 J to cal
solution: 1.69 J x cal/4.184 J = 0.404 cal

b. 196 cal to kilojoules

Solution: 196 cal x 4.184 J/1.00 cal = 820 J
To convert J to kJ = 820 J/1000 = 0.820 J
C. 68 cal to kcal
Solution: 68 cal /1000 =0.068 kcal

d. 6.89 cal to kJ
 The first law of thermodynamics

Energy is conserved. It means it can neither be created nor destroyed. Energy lost is gained by its environment

and vice versa.

-Internal energy (∆E) is the sum of all the kinetic and potential energies of all its components making up the

system.

The internal energy is a state function i.e., the property of a system depends only on its present state which is

determined by variables, such as temperature and pressure.

The exchange of energy between the system and surroundings are of two kinds namely: Work and heat.

∆E = Efinal – Einitial

A positive ∆E indicate that Efinal is greater than Einitial meaning the system gained energy from its surrounding.
A negative ∆E indicate that Efinal is less than Einitial meaning the system lost

energy to its surrounding.

• Energy moves in and out of a system because of the difference in temperature

between the system and its surroundings.

• Work is the energy exchange that results when a force (F) moves an object

through a distance (d).

W=FXd
If a vessel containing gas with a piston. The top of the piston is
the weight and part of the surrounding. If the temperature of the
surrounding is increased and heat passes from the surrounding
to the vessel, the energy of the surrounding will decrease.

If gas is enclosed in the vessel and the temperature of the gas is

increased and the gas pressure increase, the gas will expand and
by lifting the weight, work is done. The system does work.
∆E can relate with work

∆E = q + w q-heat added or released from the system

w-work done on the system

When heat is added to a system or work is done on a system, the internal energy increases.

q is positive means system gained energy

q is negative means systems lost energy

w is positive means work done on the system

w is negative means work done by the system

1. Two gases A and B are placed in container that has a piston. A and B react to
form product C. As the reaction occurs, the system loses 2000 J heat to the
surrounding. The piston moves downward, the volume of the gas decreases and
the surrounding do 480 J work on the system. Calculate the change in the
internal energy.
∆E = q + w

Energy is lost therefore q = -2000 J

Work is done on the system, w = +480 J

∆E = (-2000 J) + (480 J) = -1520 J

∆E is negative indicating that energy is transferred from the system to the surrounding
2. Calculate the change in internal energy of a system that absorb 140 J of heat and

does 85 J of work on the surroundings.

∆E = q + w

q = +140 J because it absorbs energy

w = -85 J because the work is done by the system on the surrounding

∆E =140 J + (-85 J) = 55 J

∆E = + 55 J indicates energy is gained by the system

A gas expands and does P-V work on the surroundings equal to 325 J.
At the same time, it absorbs 127 J of heat. Calculate the change in
energy of the gas.
E=Q+W
Q = +127 J
W = -325 J
E = +127 J – (325 J) = - 198 J
1. A system absorbs 85kJ of heat from its surrounding while doing 29 kJ of work on the

surroundings. Calculate the change in internal energy.

2. A chemical reaction releases 8.65 kJ of heat and does no work on the surroundings.

Calculate the change in internal energy.

2. A chemical reaction releases 8.65 kJ of heat and does no work on the
surroundings. Calculate the change in internal energy.
ΔE = q + w
Q = -8.65 kJ
W=0
ΔE = -8.65 kJ
 Exothermic and Endothermic processes

• A process in which a system absorb heat is known as endothermic. E.g. melting of

ice- heat flows into system from the surrounding. (The glass feels cold)

• Exothermic reaction is a process in a system in which heat is lost. E.g.

combustion of gasoline- heat flows from the system to the surrounding.

• The conditions that affects internal energy are temperature and pressure.

• The total internal energy = total quantity of matter (because energy is an

extensive property)
 Enthalpy

• Enthalpy is a thermodynamic function that accounts for heat flow in processes occurring at

constant pressure when no form of work is performed in the system other than P-V work.

• It is an extensive property of a substance and useful to determine the heat absorbed or evolved

from a chemical reaction. It depends on the amount of substance.

• An extensive property depends on the size of the system under study. It is directly proportional to

the mass. Examples of extensive properties are volume, internal energy, mass, enthalpy, entropy,

• It is a state function and depends only on its present state which is determined by variables such

as the temperature and pressure.

• The change in enthalpy of a reaction at a given temperature and pressure is known as the
enthalpy of reaction.

• ΔH =H(products)-H(reactants)

• H is a state function and it is independent on the details of the reaction. It depends on the
initial state of the reactant and the final state of the product. ΔH =H(Final)-H(Initial)

• The change in enthalpy of a reaction is equal to the heat of reaction at constant pressure.

ΔH = qp

• Example N2 + 3H2 2NH3 qp = ΔH = -91.8 kJ

 Enthalpy and Internal Energy
• H = E +PV E-Internal energy

• ΔH = H final –H initial = (E final + PV final) – (E initial + PV initial)

• ΔH = ΔE + PΔV

• ΔE = ΔH - PΔV

• - PΔV is the energy required by a system to change the volume against the constant pressure of the
atmosphere.

• Reaction can occur at constant volume or constant pressure.

• Pressure-volume work (P-V work) is work involved in the expansion and compression of gases.

When pressure is constant w = -P ∆V

• P- pressure and ∆V is the change in volume of the system (∆Vfinal - ∆Vinitial)

• When a gas expand, ∆V is a positive quantity and w is a negative quantity.

Meaning work is done by the system on the surrounding

• When gas is compressed, ∆V is a negative quantity and w is a positive quantity.

Meaning work is done on the system by the surrounding

 ∆H is positive means system has gained heat from the surroundings-

endothermic process.

∆H is negative means system has lost heat to the surrounding-exothermic

process.
• In a chemical reactor a mixture of gases reacts. The system releases 1705 kJ
of heat into its surroundings. A work of 3780 kJ is done by the system on
the surroundings. a) Calculate the change of enthalpy for this process. b)
Calculate the change of internal energy for this process.
• A. The change of enthalpy for the process ∆H = -1705 kJ heat release
• B Calculate the change of internal energy for this process
ΔE = ΔH – PΔV
W = work is done by the system on the surrounding,
ΔE = -1705 kJ – 3780 kJ = -5485 kJ
• In a reaction chamber hydrogen reacts with oxygen. The system
releases 358 kJ of heat while doing 737 kJ of work on the
surroundings. Determine the change of internal energy for this
process.
ΔE = ΔH – PΔV
ΔE = -358 kJ – 737 kJ = -1095 kJ
ΔE = ΔH – PΔV

ΔE = -358 kJ – 737 kJ = -1095 kJ

• Calculate the change in internal energy for the reaction below carried
out at constant pressure of 40 atm at a volume of -1.12 dm³.

• N2 +3H2 2 NH3 ΔH = -92.2 kJ

Note the unit of P is Pascal and ΔV = m³ PΔV = Joules

ΔE = ΔH – PΔV
ΔE = ΔH – PΔV
ΔH = -92.2 kJ
P = 40 atm x 1.013 x 10⁵ Pa/ 1 atm = 40.52 x 10⁵ = 4.052 x 10⁶ Pa
ΔV= -1.12 x 10ˉ³m³
ΔE = ΔH – PΔV
ΔE= -92.2 kJ – (4.052 x 10⁶ Pa)(-1.12 x 10ˉ³m³ )
= -92.2 kJ – (-4.538 x10³ J)
= -92.2 kJ + 4.538 kJ = -87.662 kJ
 Enthalpies of reaction

∆H = Hproducts – Hreactants

Enthalpy change that accompanies a reaction is known as enthalpy of reaction ∆Hrxn

1. Enthalpy change is an extensive property- The magnitude of ∆H is directly proportional

to the amount of reactants consumed in the process.

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

It means 1 mole of CH4 and 2 moles of O2 releases -890 kJ of heat.

2. The enthalpy change of a reaction is equal in magnitude but opposite in sign to ∆H for the

reverse reaction.

e.g. For example: CH4 (g) + 2O2 → CO2 (g) + 2H2O (l) ∆H = -890 kJ

Therefore the reverse CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 ∆H = +890 kJ
• A system that releases heat to the surroundings is an exothermic reaction with a negative ΔH because the

enthalpy of the products is lower than the enthalpy of the reactants of the system.

• For endothermic reactions, heat is absorbed. This means that the reactants absorb heat in order to form the

products. This results in the products having a higher enthalpy than the reactants.

Activation

energy is the

minimum energy

required to cause

a process (such

as a chemical

reaction) to occur.
 Class activity

1. At temperature and pressure, 2H2 (g) + O2 (g) → 2H2O (g) ∆H = -242 kJ

a. Under these conditions which has a higher enthalpy: 2H2 (g) or 2H2O (g)

b. Is the enthalpy change, ∆H, an intensive or extensive quantity?

c. What is the value of ∆H for the process, 2H2O (g) → 2H2 (g) + O2 (g)
• A. 2H2
• B. Extensive quantity
• C. ∆H = +242 kJ
 Calorimetry

• The measurement of heat flow is calorimetry and a device known as calorimeter is used to

measure the heat flow.

• Heat capacity (C) of an object is the amount of heat required to raise the temperature by 1

K or 1oC. The greater the heat capacity, the greater the heat required to produce an increase

in temperature.

• The heat capacity of one mole of a substance is called molar heat capacity (Cmolar)
• The heat capacity of one gram of a substance is known as the specific
heat capacity (s). s = q/m x Δt

Δt- Change in temperature; m-mass; q-quantity of heat transferred.

Question:

How much of heat is needed to warm 250 g of water from 22 °C near to

its boiling point of 98 °C? The specific heat of capacity is 4.18 J/g-k.
What is the molar heat capacity of water?
 s = q/m x Δt
q =s x m x Δt
q = 4.18 J/g-k x 250 g x (98-22) K
q = 4.18 J/g-k x 250 g x 76 K = 7.9 X10⁴ J amount of heat is needed

Molar heat capacity = Specific heat capacity x molar mass of water

= 4.18 J/g-K x 18 g/mol = 75.2 J/mol-K

Note: ΔT in K = ΔT in °C
A degree Celsius is exactly the same as a Kelvin, so the heat capacities can be expresses equally well, and perhaps a bit
more correctly in SI, as joules per Kelvin, J/K

Since the scaling for Kelvin (K) and degrees Celsius (°C) are exactly the same, the DIFFERENCE (ΔT = Tfinal − Tinitial) is the
same is either is used for temperature calculations, but make sure not to mix these two temperature units for
BOTH Tfinal and Tinitial.
 Start here Constant pressure calorimetry

• E.g. coffee cup calorimeter- The calorimeter is not sealed and reaction occurs under constant pressure of the

atmosphere. It is used to measure the temperature change that accompanies a reaction. Heat measured is

equal to the enthalpy change (ΔH)

• Note: Be careful with the sign on ΔH. Heat out of the reaction is heat in to the calorimeter and vice versa. So

an increase in temperature means the reaction was "exothermic" and ΔH is negative. Conversely, if the

temperature drops during the reaction, then the reaction must be endothermic and ΔH is positive.

Question

1. A student mixed 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature of

the resultant solution increased from 21.0 oC to 27.5 oC. Calculate the enthalpy change for the reaction in

kJ/mol HCl, assuming the calorimeter loses only a negligible quantity of heat, that the total volume is 100 mL,

density is 1.0 g/mL and specific heat is 4.18J/g-K.

• Total volume = 50 mL + 50 mL = 100 mL
• Density = 1g/mL Density = mass/volume
• 1g/mL = mass/100 mL
• mass= 100 g
• Δt = 27.5-21 = 6.5 K
• s = q/m x Δt q = s x m x Δt
• q = - (4.18 J/g-K X 100 g x 6.5 K) =-2.7 X 10³ J = -2.7 kJ
• kJ/mol Hcl = -2.7 kJ/mol Hcl
• Mol Hcl = C X V = 1.0 M X 0.05 L = 0.05 mol
• kJ/mol Hcl = -2.7 kJ/0.05 mol = -54 kJ/mol
2. When 50 mL of 0.100 M AgNO3 and 50 mL of 0.100 M HCl are mixed in a constant-

pressure calorimeter, the temperature of the mixture increases from 22.2 oC to 23.11 oC. The

temperature increase is caused by the following reaction:

AgNO3(aq) + HCl (aq) → AgCl(S) + HNO3 (aq)

Calculate the enthalpy change for this reaction in kJ/mol AgNO3 assuming the combined

mass is 100 g and specific heat is 4.18 J/g- oC.

• s =q/m x Δt q=s x m x Δt

• m = 100 g; s = 4.18 J/g-°C

• Δt = 23.11-22.20 = 0.91 °C

• q = -(4.18 J/g-°C x 100 g X 0.91 °C) = -380.38 J = -0.38 kJ

• kJ/mol AgNO3 = -0.38 kJ /mol AgNO3

• mol AgNO3 = C X V = 0.100 M X 0.05 L = 0.005 mol

• kJ/mol AgNO3 = 0.38 kJ /mol AgNO3 = -0.38 kJ / 0.005 mol = -76 kJ/mol
 Bomb Calorimetry (Constant Volume Calorimetry)

• The common type of reaction in the calorimeter is combustion

reaction.

• It is used to determine the enthalpy of combustion.

• Since combustion reactions are usually exothermic (give off

heat), the enthalpy of combustion is typically negative.

• Heat evolved from the reaction qrxn = -Ccal X ΔT

• -Ccal is the total heat of capacity of the calorimeter.

Question:

1. Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of

methylhydrazine with oxygen produces nitrogen, carbon dioxide gases and water. When 4 g

of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter

increases from 25.00 to 39.50 oC. In a separate experiment the heat capacity of the

calorimeter is measured to be 7.794 kJ/ oC. What is the heat of reaction for the combustion of

a mole of methylhydrazine in this calorimeter?

2CH6N2(l) + 5O2 (g) → 2N2(g) + 2CO2(g) + 6H2O (l)

• qrxn = -Ccal X ΔT

• qrxn = -(7.794 kJ/°C X 14.50 °C) = -113.013 kJ

• Mole of methylhydrazine = 4g/46 g/mol = 0.087 mol

• The heat of reaction for combustion of a mole of methylhydrazine in this

calorimeter = qrxn/mole of methylhydrazine

• = -113.013 kJ/0.087 mol = -1.299 kJ/mol = -1.3 x10³ kJ/mol

2. 0.5869 g sample of lactic acid (HC3H5O3) is burned in calorimeter whose heat capacity is

4.812 kJ/ oC. The temperature increases from 23.10 to 24.95 oC. Calculate the heat of

combustion of lactic acid.

a. per gram

b. per mole
• qrxn = -Ccal X ΔT = -(4.812 kJ/°C x 1.85 °C ) = -8.902 kJ

• (a) Heat of combustion of lactic acid per gram = qrxn/mass

• = -8.902 kJ/0.5869 g

• = -15.17 kJ/g

• (b) Heat of combustion of lactic acid per mole

• Mole of lactic acid = 0.5869 g/90 g/mol = 0.0065 mol

• Heat of combustion of lactic acid per mole = qrxn/mol

• = -8.902 kJ/0.0065 mol = -1.4 x 10³ kJ/mol

 Hess’s Law

• Hess law states that if a reaction is carried out in a series of steps, ΔH

for the overall reaction will equal the sum of the enthalpy changes

for the individual steps.

1. Calculate ΔH for the reaction: C2H4 (g) + H2 (g) → C2H6 (g), from the following data.

a. C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ΔH = -1411. kJ

b. C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l) ΔH = -1560. kJ

c. H2 (g) + ½ O2 (g) → H2O (l) ΔH = -285.8 kJ

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ΔH = -1411 kJ (Intact)

2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3½ O2 (g) ΔH = +1560 kJ (Reverse)

H2 (g) + ½ O2 (g) → H2O (l) ΔH = -285.8 kJ (Intact)

C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -1411+1560-285.8 = -136.8 kJ

C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -136.8 kJ

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ΔH = -1411 kJ (Intact)

2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3½ O2 (g) ΔH = +1560 kJ (Reverse)

H2 (g) + ½ O2 (g) → H2O (l) ΔH = -285.8 kJ (Intact)

C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -136.8 kJ

2. 2-Methylpropan-1-ol can also be converted to produce diesel and jet fuel. The first step in the process is

the production of 2-methylpropene.

C4H10O(l) → C4H8(g) + H2O(g)

Using the data below, calculate the enthalpy change, in kJ mol-1, for the production of 2-methylpropene

from 2-methylpropan-1-ol.

4C(s) + 5H2(g) + ½O2(g) → C4H10O(l) ΔH = -335 kJ mol-1

4C(s) + 4H2(g) → C4H8(g) ΔH = -17 kJ mol-1

H2(g) + ½O2(g) → H2O(g) ΔH = -242 kJ mol-1

C4H10O(l) → 4C(s) + 5H2(g) + ½O2(g) ΔH = +335 kJ mol-1

4C(s) + 4H2(g) → C4H8(g) ΔH = -17 kJ mol-1

H2(g) + ½O2(g) → H2O(g) ΔH = -242 kJ mol-1

C4H10O(l) → C4H8(g) + H2O(g)

ΔH = +76 kJ mol-1
3. Self-heating cans may be used to warm drinks such as coffee. When the button on the can is pushed, a seal is
broken, allowing water and calcium oxide to mix and react. The reaction produces solid calcium hydroxide and
releases heat. If more water is used the calcium hydroxide is produced as a solution instead of as a solid. The
equation for the reaction is:

CaO(s) + H2O(ℓ) → Ca(OH)2(aq)

Using the following data, calculate the enthalpy change, in kJ mol−1, for this reaction.

Ca(s) + ½O2(g) → CaO(s) ΔH = −635 kJ mol−1

H2(g) + ½O2(g) → H2 O(ℓ) ΔH = −286 kJ mol−1

Ca(s) + O2(g) + H2(g) → Ca(OH)2(s) ΔH = −986 kJ mol−1

Ca(OH)2(s) → Ca(OH)2(aq) ΔH = −82 kJ mol−1

 CaO(s) + H2O(ℓ) → Ca(OH)2(aq)

Using the following data, calculate the enthalpy change, in kJ mol−1, for this reaction.

CaO(s) → Ca(s) + ½O2(g) ΔH = +635 kJ mol−1

H2O(ℓ) → H2(g) + ½O2(g) ΔH = +286 kJ mol−1

Ca(s) + O2(g) + H2(g) → Ca(OH)2(s) ΔH = −986 kJ mol−1

Ca(OH)2(s) → Ca(OH)2(aq) ΔH = −82 kJ mol−1

ΔH = −147 kJ mol−1
 Standard enthalpies of formation
)

• Standard enthalpy of formation (ΔH°f) is the change in enthalpy that is associated with the formation of

one mole of a compound from its element with all substances in their standard states.

Using the standard enthalpies of formation, calculate the standard enthalpy change (ΔH°) for the following

reactions:

1. 2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(g)

2. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

3. 2Na(s) + CO2 (g) Na2O(s) + CO(g)

4. SiCl4(l) + 2H2O(g) SiO2(s) + 4HCl (aq)

ΔH°f (kJ/mol) values
CH3OH(l) -238.7
O2(g) 0
CO2(g) -393.5
H2O (l) -285.8
NH3(g) -46.2
NO(g) +33.2
H2O(g) -241.8
Na(s) 0
Na2O(s) -416
CO(g) -110.5
SiCl4(l) -687.0
SiO2(s) -910.7
HCl (aq) -92.3
1. 2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l)

ΔH° = ∑ΔH°f Products – ∑ΔH°f Reactants

= (2 x ΔH°f CO2 + 4 x ΔH°f H2O) - (2 x ΔH°f CH3OH + 3 x ΔH°f O2)

= (2 x -393.5 kJ/mol + 4 x -285.8 kJ/mol) – ( 2 x -238.7 + 3 x 0)

= (-787 kJ/mol + -1143.2 kJ/mol ) – (-477.4 kJ/mol)

= -1930.2 kJ/mol + 477.4 kJ/mol

= -1452.8 kJ/mol = -1.45 x103 kJ/mol

Using the standard enthalpies of formation, calculate the standard enthalpy change (ΔH°)
for the reaction shown below:-
2CO (g) + O2 (g) → 2CO2 (g) ΔH° = -566.0 kJ/mol
4NH3 (g) +5O2 (g) → 4NO(g) +6H2O (g) ΔH° = -1133.2 kJ/mol
• ΔH°f (kJ/mol) values
• CH3OH(l) -238.7
• O2(g) 0
• CO2(g) -393.5
• H2O (l) -285.8
• NH3(g) -46.2
• NO(g) +33.2
• H2O(g) -241.8
• Na(s) 0
• Na2O(s) -416
• CO(g) -110.5
• SiCl4(l) -687.0
• SiO2(s) -910.7
• HCl (aq) -92.3