Standard Enthalpies of Formation

Learning Goal:Learning Goal:

•You will be able to write formation equations, find the You will be able to write formation equations, find the enthalpies of formation and use them & Hess’ Law to enthalpies of formation and use them & Hess’ Law to find the enthalpy of a given reaction.find the enthalpy of a given reaction.

Standard Enthalpy of Formation the quantity of energy associated with the formation of the quantity of energy associated with the formation of

one mole of a substance from its elements in their one mole of a substance from its elements in their standard states (at SATP)standard states (at SATP)

Symbol: Symbol: ΔΔHH°°ff Units: kJ/molUnits: kJ/mol

For example:For example:

C(s) + ½ OC(s) + ½ O2 (g)2 (g) + 2 H + 2 H2 (g)2 (g) CH CH33OHOH (l) (l) Δ Δ HH°°ff = -239.1 = -239.1 kJ/molkJ/mol

2 Al2 Al (s) (s) + 3/2 O + 3/2 O2 (g)2 (g) Al Al22OO3 (s)3 (s) Δ Δ HH°°ff = -1675.7 kJ/mol = -1675.7 kJ/mol

All of the elements on the left sides of the equations are All of the elements on the left sides of the equations are in their standard states (SATP)in their standard states (SATP)

Formation Equations Are equations that show the production of a Are equations that show the production of a

compound from its elements.compound from its elements.

How to Write Formation Equations:How to Write Formation Equations:

Step 1Step 1: : Write one mole of the product in the state that has Write one mole of the product in the state that has been specified.been specified.

Example:Example: FeSO FeSO4 (s)4 (s)

Step 2Step 2:: Write the reactant elements in their standard states Write the reactant elements in their standard states ((HOFBrINClHOFBrINCl should be diatomic – all others are monatomic) should be diatomic – all others are monatomic) (reference periodic table. Pg 838).(reference periodic table. Pg 838).

FeFe (s) (s) + S + S (s) (s) + O + O2 (g)2 (g) FeSO FeSO4 (s)4 (s)

Step 3Step 3: : Write coefficients for the reactants to give a balanced Write coefficients for the reactants to give a balanced equation yielding one mole of product – you may have to use equation yielding one mole of product – you may have to use fractionsfractions!!

Fe (s) + S (s) + 2 O2 (g) FeSO4 (s)

Let’s Practice: Write the formation equation for solid sodium chloride.Write the formation equation for solid sodium chloride.

Write the formation equation for solid magnesium Write the formation equation for solid magnesium carbonatecarbonate

Calculating ΔH A third way of calculating ΔH is possible using standard A third way of calculating ΔH is possible using standard

enthalpies of formation. enthalpies of formation.

Table C.6 pg 799 gives standard enthalpies of formation Table C.6 pg 799 gives standard enthalpies of formation for common molecules.for common molecules.

According to HessAccording to Hess’’s Law, the enthalpies of known s Law, the enthalpies of known equations may be used to calculate the enthalpy of an equations may be used to calculate the enthalpy of an unknown reaction.unknown reaction.

For exampleFor example: The formation of hydrogen and carbon : The formation of hydrogen and carbon monoxide gas from water and graphite:monoxide gas from water and graphite:

HH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g) ΔH = ?ΔH = ?

Use the formation equations for each of the products and Use the formation equations for each of the products and reactants to create the target equation.reactants to create the target equation.

Calculating ΔH0, continuedHH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g)

ReactantsReactants

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) O(g) HH00f H2Of H2O = - 241.8 kJ/mol = - 241.8 kJ/mol

CC (s) (s) – graphite in standard state – graphite in standard state HH00ff = 0 kJ/mol = 0 kJ/mol

ProductsProducts

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g) (g) --> CO(g) HH00f COf CO = - 110.5 kJ/mol = - 110.5 kJ/mol

HH2 (g)2 (g) – hydrogen in standard state – hydrogen in standard state HH00ff = 0 kJ/mol = 0 kJ/mol

HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) (g) HH00 = 241.8 kJ/mol = 241.8 kJ/mol

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g) (g) --> CO(g) HH00 = - 110.5 kJ/mol = - 110.5 kJ/mol

HH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g) HH00 = 131.3 kJ/mol = 131.3 kJ/mol

Using Using ΔΔHH°°ff to calculate enthalpy change: to calculate enthalpy change:

where n = the amount in moles of each product or reactantwhere n = the amount in moles of each product or reactant

Example: Calculate the standard change in enthalpy for Example: Calculate the standard change in enthalpy for the thermite reaction:the thermite reaction:

Al(s) + FeAl(s) + Fe22OO3 (s)3 (s) Al Al22OO3 (s)3 (s) + Fe + Fe(s)(s)

)(reactants Hn - (products) Hn H ffrxn

Sample Problem

Calculate the heat of combustion of methanol.

CH3OH (l) + 3/2 O2 (g) --> CO2 (g) + 2 H2O (g) H0comb = ?

Multistep Energy Calculations

Several energy calculations may be required to solve multistep problems

–Heat flow q = mcΔT

–Enthalpy changes ΔH = n ΔHx

–Hess’s Law ΔHtarget = ∑ΔHknown

–Enthalpies of Formation

)(reactants Hn - (products) Hn H ffrxn

Sample Problem, pg 336

Sample problem cont’d

Sample problem cont’d