Lesson 2 of 5Energy Changes and Rates of Reaction

Unit 3 · Lesson 2 of 5

Hess's Law and Enthalpies of Formation

Finding enthalpy changes you cannot measure directly by combining known equations (Hess's law), by using standard enthalpies of formation, and by estimating from average bond energies.

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Key things to remember

1Hess's law: $Δ H$ depends only on the initial and final states, so $Δ H_{target} = \sum Δ H_{steps}$ .
2Reverse an equation: change the sign of $Δ H$ . Multiply an equation by a factor: multiply $Δ H$ by the same factor.
3 $Δ H_{r x n}^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum n Δ H_{f}^{\circ} (reactants)$
4 $Δ H_{f}^{\circ} = 0$ for any element in its standard state, e.g. $O_{2} (g)$ or $C (s, graphite)$ .
5Bond energies: $Δ H \approx \sum (broken) - \sum (formed)$ ; breaking bonds absorbs energy, forming bonds releases it.

Some enthalpy changes are impossible to measure in a calorimeter. Try to burn carbon to make only carbon monoxide and you will always get some carbon dioxide too; try to make methane directly from carbon and hydrogen and nothing happens at a useful rate. Chemists get around this with one powerful idea: enthalpy is a state function. Like your altitude on a mountain, the change depends only on where you start and where you finish, not on the path you take. So if you can find any route from reactants to products made of steps you do know, you can add them up.

Hess's law

The enthalpy change for a reaction is the same whether it happens in one step or in a series of steps. If a target equation is the sum of several equations, its

Δ H

is the sum of their

Δ H

values:

Δ H_{target} = Δ H_{1} + Δ H_{2} + \dots

Fig 1A Hess's law cycle: the direct route and the two-step route start and end at the same place, so their total enthalpy changes are equal.

To use Hess's law, you rearrange the given equations so that, when you add them, everything cancels except the target equation. Only two moves are allowed, and each changes $Δ H$ in a predictable way.

Look at each substance in the target equation and find a given equation that contains it (ideally one that contains it only once).
Reverse an equation if that substance is on the wrong side. Change the sign of its $Δ H$ .
Multiply an equation (including by fractions) so the substance has the right coefficient. Multiply its $Δ H$ by the same factor.
Add the adjusted equations, cancel anything that appears equally on both sides, and check that exactly the target equation is left.
Add the adjusted $Δ H$ values.

Standard Enthalpies of Formation

Rather than memorizing thousands of reaction enthalpies, chemists tabulate one number per compound. The standard enthalpy of formation, $Δ H_{f}^{\circ}$ , is the enthalpy change when 1 mol of a compound forms from its elements in their standard states (their most stable form at $25^{\circ} C$ and $100 kPa$ ). For example, the formation equation for liquid water is $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ , $Δ H_{f}^{\circ} = - 285.8 kJ/mol$ . Fractions are fine here, because the product must have a coefficient of exactly 1.

Elements are the zero point

An element in its standard state, such as

O_{2} (g)

H_{2} (g)

N_{2} (g)

Fe (s)

C (s, graphite)

, has

Δ H_{f}^{\circ} = 0

because forming it from itself involves no change. Other forms of an element, like ozone

O_{3} (g)

or diamond, are not the standard state and have non-zero values.

Δ H_{r x n}^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum n Δ H_{f}^{\circ} (reactants)

Multiply each formation enthalpy by its coefficient n, then take products minus reactants.

This formula is Hess's law in disguise. Imagine breaking every reactant down into its elements (the reverse of formation, so the signs flip — that is the minus sign), then building the products from those same elements. Both steps use formation enthalpies, and the elements in the middle contribute zero. Watch the states: $H_{2} O (l)$ and $H_{2} O (g)$ have different formation enthalpies, and the $44.0 kJ/mol$ difference is the energy needed to vaporize water at $25^{\circ} C$ , so always use the value that matches the state in the equation.

Substance	Standard enthalpy of formation (kJ/mol)
$CO_{2} (g)$	$- 393.5$
$CO (g)$	$- 110.5$
$H_{2} O (l)$	$- 285.8$
$H_{2} O (g)$	$- 241.8$
$CH_{4} (g)$	$- 74.8$
$C_{3} H_{8} (g)$	$- 103.8$
$C_{2} H_{5} OH (l)$	$- 277.7$
$NH_{3} (g)$	$- 46.1$
$NO (g)$	$+ 90.3$
$NO_{2} (g)$	$+ 33.2$
$CaCO_{3} (s)$	$- 1206.9$
$CaO (s)$	$- 635.1$
$Fe_{2} O_{3} (s)$	$- 824.2$
$Al_{2} O_{3} (s)$	$- 1675.7$

Common standard enthalpies of formation at 25 °C. Values in other tables may differ slightly in the last digit.

Estimating with Average Bond Energies

A reaction rearranges atoms, so it can be pictured as two stages: break all the bonds in the reactants, then form all the bonds in the products. Breaking a bond always absorbs energy (endothermic), because you must pull bonded atoms apart against their attraction. Forming a bond always releases energy (exothermic). The bond energy is the energy needed to break 1 mol of a particular bond in the gas phase. If the bonds formed are stronger overall than the bonds broken, more energy is released than absorbed and the reaction is exothermic.

Δ H \approx \sum (bond energies of bonds broken) - \sum (bond energies of bonds formed)

Bonds broken minus bonds formed. Note the order is the reverse of the formation-enthalpy formula.

Bond	Average bond energy (kJ/mol)	Bond	Average bond energy (kJ/mol)
H–H	436	C–H	413
O=O	498	O–H	463
N≡N	945	N–H	391
Cl–Cl	242	H–Cl	432
C–C	347	C=C	614
C=O (in CO₂)	799	C–O	358

Typical average bond energies. Tables differ slightly; use the values given in each problem.

Why bond energies only estimate

A C–H bond in methane is not exactly as strong as a C–H bond in ethanol, so tables list averages over many molecules. Bond energies also apply to gases only. That is why a bond-energy estimate usually lands close to, but not exactly on, the value from formation enthalpies.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Use Hess's law to find

Δ H

for

C (s) + \frac{1}{2} O_{2} (g) \to CO (g)

, given (1)

C (s) + O_{2} (g) \to CO_{2} (g)

Δ H_{1} = - 393.5 kJ

and (2)

CO (g) + \frac{1}{2} O_{2} (g) \to CO_{2} (g)

Δ H_{2} = - 283.0 kJ

Example 2

Find

Δ H

for

C (s) + 2 H_{2} (g) \to CH_{4} (g)

from these combustion data: (1)

C (s) + O_{2} (g) \to CO_{2} (g)

- 393.5 kJ

; (2)

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

- 285.8 kJ

; (3)

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l),

- 890.3 kJ

Example 3

Calculate

Δ H^{\circ}

for the combustion of propane,

C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (l),

using

Δ H_{f}^{\circ}

C_{3} H_{8} (g) = - 103.8

CO_{2} (g) = - 393.5

H_{2} O (l) = - 285.8 kJ/mol

Example 4

Estimate

Δ H

for

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (g)

from average bond energies: C–H 413, O=O 498, C=O (in

CO_{2}

) 799, O–H

463 kJ/mol

Where marks get lost

Reversing an equation but forgetting to change the sign of its $Δ H$ , or multiplying the equation but not its $Δ H$ .
Using reactants minus products with formation enthalpies. It is always products minus reactants for $Δ H_{f}^{\circ}$ (but bonds broken minus bonds formed for bond energies).
Forgetting to multiply each $Δ H_{f}^{\circ}$ by its coefficient in the balanced equation.
Giving elements such as $O_{2} (g)$ a non-zero formation enthalpy, or using the value for $H_{2} O (g)$ when the equation shows $H_{2} O (l)$ .
Counting bonds from the formula instead of the structure — for example, treating $CO_{2}$ as having C–O single bonds instead of two C=O double bonds, or missing that each water molecule has two O–H bonds.

Flashcards 8

Hess's law

What happens to Delta H when you reverse an equation?

Standard enthalpy of formation

Formation-enthalpy formula

+ 4 more

Practice

Given

N_{2} (g) + O_{2} (g) \to 2 NO (g)

Δ H = + 180.6 kJ

and

2 NO (g) + O_{2} (g) \to 2 NO_{2} (g)

Δ H = - 114.2 kJ

, find

Δ H

for

N_{2} (g) + 2 O_{2} (g) \to 2 NO_{2} (g)

in kJ (one decimal place).

An equation has

Δ H = - 92.2 kJ

. What is

Δ H

if the equation is reversed and every coefficient is doubled?

Calculate

Δ H^{\circ}

in kJ for

C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (l)

using

Δ H_{f}^{\circ}

C_{2} H_{5} OH (l) = - 277.7

CO_{2} (g) = - 393.5

H_{2} O (l) = - 285.8 kJ/mol

. Give one decimal place.

Lime is made by heating limestone:

CaCO_{3} (s) \to CaO (s) + CO_{2} (g)

. Using

Δ H_{f}^{\circ}

CaCO_{3} (s) = - 1206.9

CaO (s) = - 635.1

CO_{2} (g) = - 393.5 kJ/mol

, find

Δ H^{\circ}

in kJ (one decimal place).

The thermite reaction is

2 Al (s) + Fe_{2} O_{3} (s) \to Al_{2} O_{3} (s) + 2 Fe (s) .

Using

Δ H_{f}^{\circ}

Fe_{2} O_{3} (s) = - 824.2

and

Al_{2} O_{3} (s) = - 1675.7 kJ/mol

, find

Δ H^{\circ}

in kJ (one decimal place).

Estimate

Δ H

in kJ for

H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)

using bond energies H–H 436, Cl–Cl 242 and H–Cl

432 kJ/mol

Estimate

Δ H

in kJ for

N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)

using bond energies N≡N 945, H–H 436 and N–H

391 kJ/mol

Which substance has a standard enthalpy of formation of exactly zero?

Which equation is the correct formation equation for carbon dioxide?

Bond energies from a table are averages taken over many different molecules. What kind of answer does a bond-energy calculation therefore give for

Δ H

Match each change made to a thermochemical equation with its effect on Delta H.

Reverse the equation

Multiply all coefficients by 3

Divide all coefficients by 2

Add it to a second equation

Put the steps for estimating Delta H from bond energies in order.

1Count each type of bond broken in the reactants and formed in the products
2Subtract: bonds broken minus bonds formed
3Add the bond energies for bonds broken and for bonds formed
4Draw or picture the structure of every reactant and product

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