Chemical Energy
Skills to develop

Discuss the energy aspect of chemical reactions

Identify compounds that provides energy in reactions

Calculate the amount of energy accompanying a chemical reaction

Identify exothermic and endothermic changes

Estimate work performed due to volume change under constant pressure, including the
application of ideal gas law.

Explain the term enthalpy

Calculate the energy change using standard enthalpy of reaction and standard enthalpy of
formation.
Chemical Energy
--- energy released or absorbed in chemical
electron affinity, electronegativity, lattice energy, etc.)
reactions. (bond energy, ionization potential,
For example,
H2(g) + 1/2 O2  H2O (l),
H = -285.8 kJ/mol
The negative value means that the heat is released or exothermic.
H2O (l) = H2(g) + 1/2 O2,
H = +285.8 kJ/mol (endothermic)
Use an energy level diagram to show the relative content of
energy. The energy content of H2(g) + 0.5 O2 is 285.8 kJ
higher than a mole of water, H2O.
------------H2(g) + 1/2O2
2 H2(g) + O2 = 2 H2O (l), H = -571.6 kJ/mol
286 kJ
-286 kJ
Example 1
------------H2O
How much energy is release when a
balloon containing 0.15 mole of hydrogen is ignited in the air?
Solution
The amount released is 0.15 mol * 285.8 kJ/mol = 42.9 kJ
Discussion: The sudden release of energy causes an explosion.
Pressure-volume Work in Chemical Reactions
1 Pa 1 m3 = 1 N m-2 m3 = 1 N m (or J)
Since 1 atm = 101300 Pa, and 1 L = 0.001 m3,
1 atm L = 101300 Pa * 0.001 m3 = 101.3 J.
The P-V work under constant pressure (P) is simply the pressure times the change in volume dV.
w = - P V
w = -  d (P V)
= -  P dV -  V dP
The negative sign is retained, because the work done by the system is negative. However, the integral of P V
work depends on the path, and we will not discuss the details now.
Example 2
In the reaction to produce oxygen,
KClO3(s) = KCl(s) + 3/2 O2(g),
calculate the amount of work produced by 8.2 g of KClO3.
Solution
The molar mass of KClO3 is 123.5 g/mol and 8.2 g is 0.067 mol. Thus, the amount of oxygen produced is
0.10 (= 0.067*2/3) mol. Apply the ideal gas law (P V),
PV=nRT
w = P V
= n R T = 0.10 mol*8.312 (J / (mol K))*298 K = 248 J
Discussion: The work done is due to the formation of gas O2 which expands against the atmosphere of 1.0
atm or 101.3 kPa. The volume changes of the solids are insignificant compared to that of the gas.
Enthalpy
The enthalpy, usually represented by H is the energy released in a chemical reaction under constant pressure,
H = qP. The enthalpy is a convenient property to evaluate for reactions taking place at constant pressure.
Enthalpy differ from internal energy, E, defined in Energy as the energy input to a system at constant volume.
The energy released in a chemical reaction raises the internal energy, E, and does work under constant
pressure at the expense of energy stored in compounds. Thus,
H = qP = E + P dV
Of course, the enthalpy change (dH) of a chemical reaction depends on the amount of reactants, the
temperature, and pressure. Under normal conditions, the ideal gas law can be applied to give reasonable
results.
Like the internal energy, enthalpy is also a thermodynamic state function, depending only on the initial and
final states of the system, but not on the rate of reaction.
Standard Enthalpy of Reaction, Ho
In order to make the data useful for scientific and engineering applications, there is a general agreement to
report and tabulate enthalpy changes for a mole of reaction at a standard temperature
and pressure. Such quantities are called the standard enthalpy of reaction.
Example 3
The standard enthalpy for the combustion of methane is 890.4 kJ per
mole,
CH4(g) + 2 O2(g) - CO2(g) + 2 H2O(g), Ho = -890.4 kJ/mol
calculate the standard enthalpy change when 1.0 cubic meter of natural
gas is burned converting to gaseous products.
Solution
When 1.0 mol or 22.4 L of CH4, at 273K and 1 atm, is oxidized completely, the standard enthalpy change
is 890.4 kJ. One cubic meter is 1000 L (/22.4 = 44.6 mol). Thus, the standard enthalpy of change is,
H = 44.6 mol * 890.4 kJ/mol = 39712 kJ or 39 million joules.
Further examples of standard enthalpy of reactions:
2 H(g) - H2(g)
Ho = -436 kJ/mol
2 O(g) - O2(g)
Ho = -249 kJ/mol
H2O(l) - H2O(g)
Ho = 44 kJ/mol at 298 K
H2O(l) - H2O(g)
H = 41 kJ/mol at 373 K, non-standard condition
Mg(s) + S(s) - MgS(s)
Ho = -598 kJ/mol
2 H(g) + O(g) - H2O(g)
Ho = -847 kJ/mol
Cu(s) + 1/2O2(g) - CuO(s)
Ho = -157 kJ/mol
1
/2N2(g) + O2(g) - NO2(g)
Ho = 34 kJ/mol
Mg(s) + 1/2O2(g) - MgO(s)
Ho = -602 kJ/mol
2 P(s) + 3 Cl2(g) - 2 PCl3(s)
Ho = -640 kJ/mol
2 P(s) + 5 Cl2(g) - 2 PCl5(s)
Ho = -880 kJ/mol
C(graphite) + 2 O(g) - CO2(g)
Ho = -643 kJ/mol
C(graphite) + O2(g) - CO2(g)
Ho = -394 kJ/mol
C(graphite) + 2 H2(g) - CH4(g) Ho = -75 kJ/mol
2 Al(s) + Fe2O3(s) - Al2O3(s) + 2Fe(s) Ho = -850 kJ/mol
As we shall see, the application of Hess Law will make these data very useful. For example, applying Hess
law using a few of these reactions enable us to calculate the heat of combustion of methane to form liquid
water (as opposed to gaseous water) and carbon dioxide,
CH4 + 2 O2 - 2 H2O(l) + CO2(g)
H = -980 kJ/mol.
Enthalpy is an important topic in thermodynamics. Various methods have been devised for the accurate
measurement of heat of reaction under constant pressure or under constant volume. This link gives a more
advanced treatment on enthalpy.
Standard Enthalpy of Formation, Hf
When the standard enthalpy is for a reaction that forms a compound from its basic elements also at the
standard state, the standard enthalpy of reaction is called the standard enthalpy of formation, represented by
Hof. Unless specified, the temperature is assumed to be at 298 K.
Some more examples:
Mg(s) + S(s)  MgS(s)
Hof = -598 kJ/mol
P(s) + 3/2 Cl2(g)  PCl3(g)
Hof = -320 kJ/mol
P(s) + 5/2 Cl2(g)  PCl5(g)
Hof = - 440 kJ/mol
H2(g) + 1/2 O2(g)  H2O(g)
Hof = -286 kJ/mol
1
/2N2(g) + O2(g)  NO2(g)
Hof = + 34 kJ/mol
Cu(s) + 1/2 O2(g)  CuO(s)
Hof = -157 kJ/mol
Mg(s) + 1/2 O2(g)  MgO(s)
Hof = -602 kJ/mol
C(graphite) + O2(g)  CO2(g)
Hof = -394 kJ/mol
C(graphite) + 4 H2(g)  CH4(g)
Hof = -75 kJ/mol
In all the above equations of reaction, the right hand side has only one product and that its coefficient is 1. A
general rule is to consider standard enthalpy of formation of all elements at the standard condition to be zero.
Then, there is no need to write the complete equation for the tabulation of the standard enthalpy of formation.
The above list can be simplified to give:
Compound Hof
MgS
CuO
PCl3
PCl5
H 2O
NO2
MgO
CO2
CH4
-598 kJ/mol
-157
-320
-440
-286
+ 34
-602
-394
-75
Example 4
For NH3, Hf = -46.1 kJ/mol. Estimate energy released when 10 g of N2
react with excess of H2 to form ammonia.
Solution
Ten grams of nitrogen is less than 1 mol, and we carry out the calculation in the following manner:
1 mol N2
- 46.1 kJ
10 g N2 ---------- ---------- = - 32.9 kJ
28.1 g N2 0.5 mol N2
Thus, 32.9 kJ is released when 10 g of N2 is consumed.
The principle of conservation of energy leads to the formulation of the Hess law. It's application makes the
enthalpy of reaction and standard enthalpy of formation very useful.
Confidence Building Questions
Which one of the following will you buy it for its (potential) chemical energy: gasoline, beer,
diamond, dry ice, a music CD, or a chemistry text book?
Skill:
Identify compounds that provide energy.
A bottle of beer also provides a bit of energy, but the energy aspect is insignificant.
Which of the following processes is endothermic?
a. condensing steam into water
b. burning a candle
c. melting ice cream
d. cooling hot coffee
e. formation of snow flakes
Skill:
Identify process as endothermic or exothermic reaction.
Which one of the following compounds when formed from its elements at the standard condition
is endothermic?
a. CuO
b. PCl3
c. PCl5
d. H2O
e. NO2
Discussion:
The common sense tells us that the formation of a though d are exothermic. Check the standard
enthalpy of formation. A lot of energy is required to break the triple bond in 0.5 N2 + O2 - NO2