6.8 - Enthalpy of Formation

Overview

Nobody can list

\Delta H

for every possible reaction, but chemists can list one number for each compound: its standard enthalpy of formation,

\Delta H_f^\circ

. With a table of these values, you can calculate

\Delta H^\circ

for almost any reaction in one line, using an equation that's on the AP reference sheet. It's the most accurate of the Unit 6 methods for finding

\Delta H

from tables, and AP questions use it often.

What a standard enthalpy of formation is

The standard enthalpy of formation,

\Delta H_f^\circ

, of a compound is the enthalpy change when one mole of the compound forms from its elements in their standard states. The reaction it describes is called a formation reaction:

\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} \qquad \Delta H_f^\circ = -393.5\ \mathrm{kJ/mol}

\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)} \qquad \Delta H_f^\circ = -285.8\ \mathrm{kJ/mol}

\mathrm{\tfrac{1}{2}N_2(g) + \tfrac{3}{2}H_2(g) \rightarrow NH_3(g)} \qquad \Delta H_f^\circ = -46.1\ \mathrm{kJ/mol}

A formation reaction has to follow three rules:

Exactly 1 mol of the compound is made. Fractions on the reactant side are fine, which is why you see $\tfrac{1}{2}\mathrm{O_2}$ .

Only elements are reactants.

Each element is in its standard state: its most stable form at 1 atm and 25 °C. For example $\mathrm{O_2}(g)$ , $\mathrm{H_2}(g)$ , $\mathrm{N_2}(g)$ , $\mathrm{C}(s)$ as graphite, $\mathrm{Fe}(s)$ , $\mathrm{Br_2}(l)$ , and $\mathrm{Hg}(l)$ .

Check yourself:

Equation	Formation reaction?	Why
$\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}$	No	It makes 2 mol of water, not 1
$\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}$	No	$\mathrm{CO}$ is a compound, not an element
$\mathrm{C(s,\ diamond) + O_2(g) \rightarrow CO_2(g)}$	No	Graphite, not diamond, is carbon's standard state
$\mathrm{Na(s) + \tfrac{1}{2}Cl_2(g) \rightarrow NaCl(s)}$	Yes	1 mol of product, made from elements in their standard states

What the degree sign means

The ° in

\Delta H_f^\circ

means standard conditions: gases at 1 atm, solutions at 1 M, and each substance in its stated phase, usually at 25 °C (298 K). It doesn't mean 0 °C.

Elements in their standard states have a value of zero

Forming

\mathrm{O_2}(g)

from

\mathrm{O_2}(g)

isn't a change at all, so

\Delta H_f^\circ = 0

. The same is true for every element in its standard state, which is why tables either leave them out or list 0. Only the standard state gets zero: ozone (

\mathrm{O_3}

), bromine gas, and diamond all have nonzero values.

Phase matters

Water has two different values:

-285.8

kJ/mol for

\mathrm{H_2O}(l)

and

-241.8

kJ/mol for

\mathrm{H_2O}(g)

. The 44.0 kJ/mol difference is the energy needed to vaporize a mole of water at 25 °C. (That's a bit more than the 40.7 kJ/mol in Topic 6.5, which is for water boiling at 100 °C.) Always use the value that matches the phase in the equation.

Using the table

These are the values used on this page (standard tables, rounded to one decimal; other books may differ by a few tenths). On the AP exam, the question always gives you the values you need.

Substance	$\Delta H_f^\circ$ (kJ/mol)	Substance	$\Delta H_f^\circ$ (kJ/mol)
$\mathrm{CH_4}(g)$	$-74.8$	$\mathrm{H_2O}(l)$	$-285.8$
$\mathrm{C_3H_8}(g)$	$-103.8$	$\mathrm{H_2O}(g)$	$-241.8$
$\mathrm{C_2H_5OH}(l)$	$-277.7$	$\mathrm{Fe_2O_3}(s)$	$-824.2$
$\mathrm{CO_2}(g)$	$-393.5$	$\mathrm{Al_2O_3}(s)$	$-1675.7$
$\mathrm{O_2}(g)$ , $\mathrm{Al}(s)$ , $\mathrm{Fe}(s)$	0

Calculating the enthalpy of reaction

This equation is on the AP reference sheet:

\Delta H^\circ_{\mathrm{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

Write the balanced equation with phases.

Look up $\Delta H_f^\circ$ for each substance. Elements in their standard states are 0.

Multiply each value by its coefficient. $\Delta H_f^\circ$ is per mole of that compound.

Subtract: products minus reactants. Watch the signs, since subtracting a negative number means adding.

Why it works

Imagine running the reaction in two steps. First, take the reactants apart into their elements: that's the reverse of their formation reactions, so it has

\Delta H = -\sum \Delta H_f^\circ(\text{reactants})

. Second, build the products from those same elements:

\Delta H = +\sum \Delta H_f^\circ(\text{products})

. Adding the two steps gives the formula. (Topic 6.9 explains why adding steps is allowed.)

The line for the elements acts as a shared zero. Every

\Delta H_f^\circ

is a height measured from that line, so

\Delta H^\circ_{\mathrm{rxn}}

is just the height of the products minus the height of the reactants.

Example 1: Burning methane, again

(a) Use the table to calculate

\Delta H^\circ

for:

\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}

\sum \Delta H_f^\circ(\text{products}) = (1)(-393.5) + (2)(-241.8) = -393.5 - 483.6 = -877.1\ \mathrm{kJ}

\sum \Delta H_f^\circ(\text{reactants}) = (1)(-74.8) + (2)(0) = -74.8\ \mathrm{kJ}

\Delta H^\circ = -877.1 - (-74.8) = -802.3\ \mathrm{kJ/mol_{rxn}}

Answer:

-802.3\ \mathrm{kJ/mol_{rxn}}

. In Topic 6.7, bond enthalpies estimated

-808\ \mathrm{kJ/mol_{rxn}}

for the same reaction. They're close, and the value from enthalpies of formation is the more accurate one.

(b) What is

\Delta H^\circ

if the water forms as a liquid instead? Replace

-241.8

with

-285.8

\Delta H^\circ = [(-393.5) + 2(-285.8)] - (-74.8) = -965.1 + 74.8 = -890.3\ \mathrm{kJ/mol_{rxn}}

Answer:

-890.3\ \mathrm{kJ/mol_{rxn}}

, which is 88.0 kJ more negative. That's the extra heat released when 2 mol of water vapor condenses at 25 °C:

2(44.0) = 88.0

kJ. Liquid water has lower enthalpy than water vapor, so making liquid releases more heat.

Example 2: The thermite reaction

The thermite reaction gets hot enough to melt iron, and it's used to weld railroad rails:

\mathrm{2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)}

(a) Calculate

\Delta H^\circ

. Both metals are elements in their standard states, so their values are 0.

\Delta H^\circ = [(-1675.7) + 2(0)] - [2(0) + (-824.2)] = -1675.7 + 824.2 = -851.5\ \mathrm{kJ/mol_{rxn}}

(b) How much heat is released when 10.0 g of aluminum reacts? (This is the Topic 6.6 skill.)

n_{\mathrm{Al}} = \frac{10.0\ \mathrm{g}}{26.98\ \mathrm{g/mol}} = 0.3706\ \mathrm{mol\ Al}

q = 0.3706\ \mathrm{mol\ Al} \times \frac{-851.5\ \mathrm{kJ}}{2\ \mathrm{mol\ Al}} = -158\ \mathrm{kJ}

Answer: 158 kJ of heat is released.

Example 3: Working backward (AP-style)

Ethanol is added to gasoline as a fuel. Its combustion is:

\mathrm{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)} \qquad \Delta H^\circ = -1366.7\ \mathrm{kJ/mol_{rxn}}

(a) Use this value and the table values for $\mathrm{CO_2}(g)$ and $\mathrm{H_2O}(l)$ to calculate $\Delta H_f^\circ$ of $\mathrm{C_2H_5OH}(l)$ . Call the unknown

x

and set up the same equation:

-1366.7 = [2(-393.5) + 3(-285.8)] - [x + 3(0)]

-1366.7 = -1644.4 - x \quad\Rightarrow\quad x = -1644.4 + 1366.7 = -277.7\ \mathrm{kJ/mol}

Answer:

\Delta H_f^\circ = -277.7

kJ/mol, which matches the table.

(b) Check the propane value used in Topic 6.6,

\Delta H^\circ = -2219.9\ \mathrm{kJ/mol_{rxn}}

for

\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}

\Delta H^\circ = [3(-393.5) + 4(-285.8)] - [(-103.8) + 5(0)] = -1180.5 - 1143.2 + 103.8 = -2219.9\ \mathrm{kJ/mol_{rxn}}

(c) Which fuel releases more heat per gram? Divide each

\Delta H^\circ

by the molar mass of the fuel (1 mol of fuel per mole of reaction):

\text{ethanol: } \frac{1366.7\ \mathrm{kJ}}{46.07\ \mathrm{g}} = 29.7\ \mathrm{kJ/g} \qquad \text{propane: } \frac{2219.9\ \mathrm{kJ}}{44.09\ \mathrm{g}} = 50.3\ \mathrm{kJ/g}

Answer: propane releases more heat per gram. One reason: ethanol already has an oxygen atom bonded to carbon, so it's partly oxidized before it burns, and there's less energy left to release.

Bond enthalpies vs. enthalpies of formation

	Bond enthalpies (Topic 6.7)	Enthalpies of formation (Topic 6.8)
Formula	$\Delta H \approx \sum \text{broken} - \sum \text{formed}$	$\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$
Order	Reactants minus products	Products minus reactants
On the reference sheet?	No	Yes
What you need	Lewis diagrams to count bonds	A balanced equation with phases
Accuracy	An estimate (average values, gases only)	Accurate for the stated phases

Common mistakes

Forgetting the coefficients. In Example 1, water's value counts twice.

Subtracting in the wrong order. It's products minus reactants.

Sign errors with negatives. $-877.1 - (-74.8) = -802.3$ , not $-951.9$ .

Giving elements a nonzero value, or giving zero to a form that isn't the standard state (like $\mathrm{O_3}$ or diamond).

Using the wrong phase. $\mathrm{H_2O}(l)$ and $\mathrm{H_2O}(g)$ have different values.

Writing a formation equation that makes 2 mol. A formation reaction always makes exactly 1 mol of the compound, even if that needs fractions.

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