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Unit 7: The Energy in Chemical Reactions—Thermodynamics and Enthalpy

Section 10: Standard Enthalpies of Formation

When a chemical reaction describes the formation of one mole of a compound from its constituent elements in their standard states, the enthalpy change of the reaction is called the "standard enthalpy of formation" (ΔHf°) for that compound. Standard state is the most stable state of that substance at a pressure of 1 atm and a temperature of 25°C. For liquid water, the equation would be:

H2(g) + ½O2(g) → H2O(l)    ΔHf° = -285.8 kJ/mol

The equation fits all the criteria: One mole of water is formed. The hydrogen and oxygen are in standard states (they are in the gas phase, and they are diatomic).

The following equations do not have all the species in standard states...

N(g) + ½O2(l) → NO(g)

...because nitrogen should be diatomic, and oxygen should be a gas, not a liquid, at 1 atm and 25°C.

C(diamond) + O2(g) → CO2(g)

...because at 1 atm and 25°C, the most stable form of carbon is graphite, not diamond.

Important fact: By definition, the ΔHf° of an element in its standard state is zero.

Thousands of ΔHf° values have been determined and recorded. Why do we care about them? These ΔHf° values are very useful because they provide yet another way to determine the change in enthalpy of a reaction without doing it experimentally. It still uses Hess's Law, but it uses it on tables containing very specific enthalpy values for only formation reactions. Here's how it works.

Take the equation for the combustion of methane:

CH4(g) + 2O2(g) → H2O(l) + CO2(g)

If we write the equations for the formation of everything in this equation and look up the ΔHf° values on a table, we get:

CH4(g) C(s) + 2H2(g) → CH4(g) -103.85 kJ/mol
O2(g)O2(g) → O2(g)0 kJ/mol
H2O(l)H2(g) + ½O2(g) → H2O(l)-285.8 kJ/mol
CO2(g)C(s) + O2(g) → CO2(g)-393.5 kJ/mol

Once we have these equations and values, we will always be able to use Hess's Law to find the ΔH of the overall reaction.

CH4(g) → C(s) + 2H2(g)

O2(g) → O2(g)

2H2(g) + O2(g) → 2H2O(l)

C(s) + O2(g) → CO2(g)

or

CH4(g) + O2(g) + 2H2(g) + O2(g) + C(s) + O2(g)C(s) + 2H2(g) + O2(g) + 2H2O(l) + CO2(g)

leaving us with

CH4(g) + 2O2(g) → 2H2O(l) + CO2(g)

Notice that:

  • The first equation is the ΔHf° of CH4(g), only reversed.
  • The third equation is double the ΔHf° of H2O(l).
  • The fourth equation is the ΔHf° of CO2(g).

And so, the ΔH of the reaction is:

ΔH = [(ΔHf° of CO2(g)) + 2(ΔHf° of H2O(l))] – [(ΔHf° of CH4(g)) + 2(ΔHf° of O2(g))]

= [-393.5 + 2(-285)] - [(-103.85) + 2(0)] = -859.65 kJ/mol

Because all the ΔHf° values can be looked up in a table, we can determine the enthalpy change of the combustion reaction without burning any CH4. This technique always works, and is generalized by the formula below where n represents the stoichiometric coefficients of the substances in the balanced reaction:

ΔHrxn = ΣnΔHf°(products) - ΣnΔHf°(reactants)

Put simply: To figure out the change in enthalpy for any equation, add up the ΔHf° values of the products and multiply them by their coefficients; then subtract all the ΔHf° values of the reactants multiplied by their coefficients.

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