Lesson 6.4 Thermochemical Equations
Suggested Reading

Zumdahal Chapter 6 Sections 6.1 & 6.2.
Essential Question

How do we show the enthalpy of reaction when writing chemical
equations?
Learning Objectives


Write thermochemical equations for chemical reactions.
Apply stoichiometry to heats of reaction.
Introduction
It is convenient to write the enthalpy of reaction, ∆H, with the chemical
equation for a reaction. A thermochemical equation is the chemical
equation for a reaction (including phase labels) in which the equation is
given a molar interpretation, and the enthalpy of reaction for these molar
amounts is written directly after the equation.
Thermochemical Equations
Lets look at an example of a thermochemical equation. For the reaction of
sodium and water, you would write
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g); ∆H = -367.5 kJ
This equation says that 2 mol of sodium reacts with 2 mol of water to
produce 2 mol of sodium hydroxide and one mole of hydrogen gas, and
367.5 kJ of heat evolves. Note that thermochemical equations always
include phase labels. This is important because the enthalpy change, ∆H,
depends on the phase of the substances.
Consider the reaction of hydrogen and oxygen to produce water. If the
product is water vapor, 2 mol of H2 burn to release 483.7 kJ of heat.
2H2(g) + O2(g) → 2H2O(g); ∆H = -483.7 kJ
On the other hand, if the product is liquid water, the heat released is 571.7
kJ.
2H2(g) + O2(g) → 2H2O(l); ∆H = -571.7 kJ
In this case, additional heat is released when water vapor condenses to
liquid. (It takes 44.0 kJ of heat to vaporize 1 mol of liquid to water 25°C.)
Example: Writing a Thermochemical Equation
Aqueous sodium hydrogen carbonate solution (baking soda solution) reacts
with hydrochloric acid to produce aqueous sodium chloride, water, and
carbon dioxide gas. The reaction absorbs 11.8 kJ of heat at constant
pressure for each mole of sodium hydrogen carbonate. Write the
thermochemical equation for the reaction.
Solution
You first write the balanced chemical equation
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Because the reaction absorbs heat, the enthalpy of reaction for molar
amount of this equation is + 11.8 kJ. The thermochemical equation is
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g); ∆H + 11.8 kJ
Rules for Manipulating Thermochemical Equations
The following are two important rules for manipulating thermochemical
equations:
1. When a thermochemical equation is multiplied by any factor, the
value of ∆H for the new equation is obtained by multiplying the value
of ∆H in original equation by the same factor.
2. When a chemical equation is reversed, the value of ∆H is reversed in
sign.
Consider the thermochemical equation for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(g); ∆H = -91.8 kJ
Suppose you want the thermochemical equation to show what happens
when twice as many moles of nitrogen and hydrogen react to produce
ammonia. Because double the amount of substances are present, the
enthalpy of reaction is doubled (enthalpy is an extensive property).
Doubling of the previous equation, you obtain
2N2(g) + 6H2(g) → 4NH3(g); ∆H = -184 kJ
Suppose we reverse the first equation we wrote for the synthesis of
ammonia. Then the reaction is for the dissociation of two mol of ammonia
into its elements. The thermochemical equation is
2NH3(g) → N2(g) + 3H2(g); ∆H = +91.8 kJ
If you want to express this in terms of 1 mol of ammonia, you simply
multiply this equation by a factor of 1/2.
Lets look at another example
Manipulating Thermochemical Equations
When 2 mole of H2(g) and 1 mol of O2(g) react to give liquid water, 572 kJ
of heat evolves.
2H2(g) + O2(g) → 2H2O(l); ∆H = -572 kJ
Write the equation for 1 mol of liquid water. Give the reverse equation, in
which 1 mol of liquid water dissociates into hydrogen and oxygen.
Solution: We want to look at this equation for 1 mol of water, so we must
multiply the coefficients and ∆H by 1/2:
H2(g) + 1/2O2(g) → H2O(l); ∆H = -286 kJ
Reversing the equation give:
H2O(l) → H2(g) + 1/2O2(g); ∆H = +286 kJ
Note that the sign for ∆H is reversed.
Applying Stoichiometry to Heats of Reaction
As you have just seen, the quantity of heat obtained from a reaction
depends on the amount of reactants. Therefore, we can apply
stoichiometry to problems involving the quantity of heat. Lets look at an
example.
Consider the reaction of methane, CH4 (the principal constituent of natural
gas), burning in oxygen at constant pressure. How much heat could you
obtain from 10.0 g of methane, assuming you have an excess of oxygen?
You can answer this question if you know the enthalpy change for the
reaction of 1 mole of methane. The thermochemical equation is
CH4(g) + 2O2(g) → CO2(g) + H2O(l); ∆H = -890.3 kJ
This calculation involves the following conversions:
grams CH4 → mole CH4 → kJ of heat
That is, 10,0 g of methane burns in excess oxygen to evolve 556 kJ of
heat. If you wanted the result in kilocalories, you would have to concert
using the relationship 1 cal = 4.184 J.
Lets look at one last example.
Calculating the Heat of Reaction Using Stoichiometry
How much heat is evolved when 907 kg of ammonia is produced according
to the following equation? (Assume that the reaction occurs at constant
pressure)
N2(g) + 3H2(g) → 2NH3(g); ∆H = -890.3 kJ
The calculation involves converting grams of NH3 to moles of NH3 and then
to kilojoules of heat. The conversion sequence is
grams NH3 → mole NH3 → kJ of heat
You obtain the conversion factor for the second step from the
thermochemical equation, which says that the production of 2 mol NH3 is
accompanied by qp = -91.8 kJ.
Note that 907 kg = 9.07 x 105 g, so
Thus, 2.45 x 106 kJ of heat evolves.