Chapter 6:

Thermochemistry
6.1: The nature of energy

Energy: the capacity to do work or to produce heat.

This chapter we will concentrate on the heat transfer that accompanies a chemical process.

Law of conservation of energy: energy can be converted from one form to another but can be neither
created nor destroyed. (the energy of the universe is constant)

Energy can be classified as:

1) Potential energy: is energy due to position or composition.
2) Kinetic energy: is energy due to the motion of the object, KE = ½ mv2

Heat (q): Involves the transfer of energy between two objects due to a temperature difference.

Work (w): force acting over a distance.

State function or state property: a property of a system that depends only on its present state.
(does not depend on pathway). Examples of state functions:
Elevation, energy, temperature, and pressure.
Heat and work depend on pathway, so they are not state functions.
Chemical energy:

System: is the part of the universe on which we focus attention.

Surrounding: includes everything else in the universe.
Chemical system: chemical reaction.
Surrounding: the reaction container, the room, and anything else other reactants and
products.
Exothermic reaction: a reaction results in the evolution of heat.
Heat flows out of the system.
Example:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy (heat)

Energy lost by system = energy gained by surrounding

Exothermic reaction: q (negative sign)

The system’s energy decrease.
Endothermic reaction: a rection absorb energy from the surrounding.
Heat flows into the system.
Example:
N2(g) + O2(g) + energy 2NO(g)

Energy gained by system = Energy lost by surrounding

Endothermic reaction: q (positive sign)

The system’s energy increase
Internal energy (E): is the sum of the kinetic and potential energy of all the particles in the system.
The change in internal energy of a system:
ΔE = q + w where q: heat, and w: work.

Work (w) : positive sign when work is done on the system.

Work (w): negative sign when work is done by the system.

Thermodynamics: the study of energy and its interconversions.

First law of thermodynamics: the energy of the universe is constant.

Example:
Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and 1.4
kJ of work is done on the system.

q: + 15.6 kJ, w: + 1.4 kJ

ΔE = 15.6 + 1.4 = 17.0 kJ

A common type of work associated with chemical processes is work done by a gas
(expansion) or work done on a gas (compression).
w = - P ΔV
Where P: external pressure.
ΔV: volume change (V2-V1), when gas is compressed ΔV is negative (V2 < V1), and when gas
is expanded ΔV is positive (V2 > V1).

Example:
Calculate the work associated with the expansion of a gas from 46 L to 64 Lat a constant
external pressure of 15 atm.
ΔV = 64 – 46 = 18 L.
w = - 15 x 18 = -270 L.atm
Example:
A balloon is being inflated to its full by heating the air inside it. The volume of the
balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of
energy as heat. Assuming the balloon expands against a constant pressure of 1.0 atm,
calculate ΔE for the process. (1 L.atm = 101.3 J)

ΔE = q + w
q: + 1.3 x 108 J
w = - P ΔV
ΔV = 5 x 105 L
w = - 1 x 5 x 105 = - 5 x 105 L. atm

w = - 5 x 105 L. atm x (101.3 J / 1L.atm) = - 5.1 x 107 J

ΔE = (+ 1.3 x 108 J) + (- 5.1 x 107 J) = 8 x 107 J

6.2: Enthalpy and Calorimetry

Enthalpy H, which is defined as:

H = E + PV
Where E: internal energy of the system, P: pressure of the system, and V: volume of the system.

At constant pressure
ΔE = qp + w
ΔE = qp – P ΔV
Where qp is the heat at constant pressure.
qp = ΔE + P ΔV

ΔH = ΔE + Δ (PV)
P is constant so Δ (PV) = P ΔV
ΔH = ΔE + P ΔV
ΔH = qp

For chemical reaction ΔH = Hproducts - Hreactants

ΔH for endothermic reaction (+)
ΔH for exothermic reaction (-)
Example:
Calorimetry

The science of measuring heat, is based on observing the

temperature change when a body absorbs or discharge energy as
heat.

Calorimeter: a device used experimentally to determine the

heat associated with a chemical reaction.

Heat capacity (J/ °C) or (J/K): the energy required to raise the
temperature of a substance by 1°C.

Specific heat capacity (J/g . °C) or (J/g . K) : the energy

required to raise the temperature of 1 g of a substance by 1°C.

Molar heat capacity (J/ mol . °C) or (J/mol . K): the energy
required to raise the temperature of 1 mole of a substance by
1°C.
 At constant pressure:
ΔH = qp
Heat (q) = mass of substance x specific heat x change in temperature
q = m x s x ΔT
Example:
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a
calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that
the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.184
J/g . °C, and that the density of the final solution is 1 g/mL, calculate the enthalpy change per mole of BaSO4
formed.
Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2NaNO3(aq)
ΔH = qp = q
heat evolved by reaction = heat absorbed by solution = mass of solution x specific heat x change in temperature
Mass of solution = total solution volume x density of solution = 2.00 L x 1 g/mL x 1000mL/1L = 2.0 x 103 g
Change in temperature = 28.1°C – 25°C = 3.1°C
q = 2 x 103 g x 4.184 J/g . °C x 3.1°C = 2.6 x 104 J
ΔH = qp = q = -2.6 x 104 J, since the reaction is exothermic.
Since 1 mole Ba(NO3)2 and 1 mole Na2SO4 are mixed 1 mole solid is formed.
ΔH = -2.6 x 104 J/mol = - 26 kJ/mol
Example:

A sample weighing 45.2 g of a metal at 85.1°C was dropped into a coffee-cup calorimeter
containing 55.6 mL of water at 25°C and as a result, the temperature of water increased to
27.2°C. What is the specific heat of this metal? (specific heat of water is 4.184 J/g . °C)

Heat released by metal = heat gained by water

Heat gained by water = mass of water x specific heat of water x change in water temperature
q = 55.6 g x 4.184 J/g . °C x (27.2°C - 25°C) = 511.8 J

Heat released by metal = mass of metal x specific heat of metal x change of metal temperature
q = 511.8 J = 45.2 g x s x ( 85.1°C - 27.2°C)
s = 0.196 J/g . °C
Calorimetry at constant volume:

H = E + PV
ΔH = ΔE + Δ (PV)
at constant volume: ΔV = 0
ΔH = ΔE
ΔE = q + w = q – PΔV
ΔE = qv
Where qv: heat at constant volume

qv = change in temperature x heat capacity of calorimeter

qv = ΔT x heat capacity
Example:
Example:
6.3: Hess’s law

The change in enthalpy in going from a given set of reactants to a given set of products
is the same whether the process takes place in one step or a series of steps.

ΔH = Hproducts - Hreactants

Example:
The oxidation of nitrogen to produce nitrogen dioxide:
In one step reaction:
N2(g) + 2O2(g) 2NO2(g) ΔH = 68 kJ

This reaction can be carried out in 2 steps:

N2(g) + O2(g) 2NO(g) ΔH1 = 180 kJ
2NO(g) + O2(g) 2NO2(g) ΔH2 = - 112 kJ

Net reaction: N2(g) + 2O2(g) 2NO2(g) ΔH = 68 kJ

Characteristics of Enthalpy changes:
1. If the rection is reversed, the sign of ΔH is also reversed.
2. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is
multiplied by the same integer.

Example:
Xe(g) + 2F2(g) XeF4(s) ΔH = - 251 kJ
1) What is ΔH for the decomposition of XeF4 into the elements?

XeF4(s) Xe(g) + 2F2(g) ΔH = + 251 kJ

2) What is ΔH for this reaction: 2XeF4(s) 2Xe(g) + 4F2(g)

ΔH = 2 x (+ 251 kJ) = + 501 kJ

Example:
Two forms of carbon are graphite and diamond, using the enthalpies of combustion for
graphite (-394 kJ/mol) and diamond ( -396 kJ/mol), calculate ΔH for the conversion of
graphite to diamond.
Cgraphite(s) Cdiamond(s)

The combustion reactions are:

Cgraphite(s) + O2(g) CO2(g) ΔH = - 394 kJ
Cdiamond(s) + O2(g) CO2(g) ΔH = - 396 kJ

Solution:
Cgraphite(s) + O2(g) CO2(g) ΔH = - 394 kJ
CO2(g) Cdiamond(s) + O2(g) ΔH = - ( -396 kJ)

Cgraphite(s) Cdiamond(s) ΔH = 2 kJ
Example:
Calculate ΔH for the synthesis of diborane from its elements according to this equation

Solution:
1) Reaction (a) will be used as it is.
2) Rection (b) must be reversed and the sign of ΔH also must be reversed.
3) Rection (c) must be multiplied by 3, and ΔH must be multiplied by 3.
4) Reaction (d) must be multiplied by 3, and ΔH must be multiplied by 3.

2B(s) + 3/2O2(g) B2O3(s) ΔH = - 1273 kJ

B2O3(s) + 3H2O(g) 3O2(g) + B2H6(g) ΔH = + 2035 kJ
3H2(g) + 3/2O2(g) 3H2O(l) ΔH = 3 (- 286 kJ)
3H2O(l) 3H2O(g) ΔH = 3 (44 kJ)
2B(s) + 3H2(g) B2H6(g) ΔH = + 36 kJ
6.4: Standard Enthalpies of Formation (ΔH°f):
The change in enthalpy that accompanies the formation of one mole of a compound from
its elements with all substances in their standard state.

Examples:
Example:
Calculate ΔH° for this reaction: 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

ΔH°reaction = [4(34 kJ)] + [6(-286 kJ)] – [(4(-46 kJ)] + [7(0 kJ)]

ΔH°reaction = - 1396 kJ
Example:
Example:

This value of ΔH° is for 2 mole of methanol (64g), since the molar mass of
methanol is 32 g/mol
This value of ΔH° is for 2 mole of octane (228.4 g), since the molar mass of
octane is 114.2 g/mol