Thermochemistry Thermochemistry Standard Enthalpies of FormationStandard Enthalpies of Formation

Standard Enthalpy of FormationStandard Enthalpy of Formation

∆∆HHff°°

change in enthalpy that accompanies the change in enthalpy that accompanies the formation of one mole of a compound from formation of one mole of a compound from its elements in standard statesits elements in standard states

° means that the process happened under ° means that the process happened under standard conditions so we can compare standard conditions so we can compare more easily more easily

Standard StatesStandard States For a COMPOUND:For a COMPOUND:

for gas: P = 1 atmfor gas: P = 1 atm pure liquid or solid statepure liquid or solid state in solution: concentration is 1 Min solution: concentration is 1 M

For an ELEMENT:For an ELEMENT: form that it exists in at 1 atm and 25°Cform that it exists in at 1 atm and 25°C

O: OO: O22(g)(g) K: K(s)K: K(s) Br: BrBr: Br22(l)(l)

Writing Formation EquationsWriting Formation Equations

always write equation where 1 mole of always write equation where 1 mole of compound is formed (even if you must use compound is formed (even if you must use non-integer coefficients)non-integer coefficients)

NONO22(g):(g):

½N½N22(g) + O(g) + O22(g) (g) NO NO22(g)(g)

∆∆HHff°= 34 kJ/mol°= 34 kJ/mol

CHCH33OH(l):OH(l):

C(s) + 2HC(s) + 2H22(g) + ½O(g) + ½O22(g) (g) CHCH33OH(l) OH(l)

∆∆HHff°= -239 kJ/mol°= -239 kJ/mol

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

reaction p f(products) r f(reactants)ΔH Σn ΔH Σn ΔH

where where n = number of moles of products/reactantsn = number of moles of products/reactants ∑ ∑ means “sum of”means “sum of” ∆∆HHff° is the standard enthalpy of formation for ° is the standard enthalpy of formation for

reactants or productsreactants or products

∆∆HHff° for any element in standard state is zero so ° for any element in standard state is zero so

elements are not included in the summationelements are not included in the summation

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

since ∆H is a state function, we can use since ∆H is a state function, we can use any pathway to calculate itany pathway to calculate it

one convenient pathway is to break one convenient pathway is to break reactants into elements and then reactants into elements and then recombine them into productsrecombine them into products

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

0)()()( 422

CHfOHfCOf HHHH

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

Example 1Example 1

Calculate the standard enthalpy change Calculate the standard enthalpy change for the reaction that occurs when ammonia for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide is burned in air to make nitrogen dioxide and waterand water

4NH4NH33(g) + 7O(g) + 7O22(g) (g) 4NO 4NO22(g) + 6H(g) + 6H22O(l)O(l)

break them apart into elements and then break them apart into elements and then recombine them into productsrecombine them into products

Example 1Example 1

Example 1Example 1

can be solved using Hess’ Law:can be solved using Hess’ Law:

(1)(1) 4NH4NH33(g) (g) 2N 2N22(g) + 6H(g) + 6H22(g)(g) -4 ∆H -4 ∆Hff°°NH3NH3

(2)(2) 7O7O22(g) (g) 7O 7O22(g)(g) 0 0

(3)(3) 2N2N22(g) + 4O(g) + 4O22(g) (g) 4NO 4NO22(g)(g) 4 ∆H 4 ∆Hff°°NO2NO2

(4)(4) 6H6H22(g) + 3O(g) + 3O22(g) (g) 6H 6H22O(l)O(l) 6 ∆H 6 ∆Hff°°H2OH2O

)286

(6)34

(40)46

(4mol

kJmol

mol

kJmol

mol

kJmolH

kJH 1396

4NH4NH33(g) + 7O(g) + 7O22(g) (g) 4NO 4NO22(g) + 6H(g) + 6H22O(l)O(l)

Look up:

-46 kJ/mol

34 kJ/mol

-286 kJ/mol

values are in Appendix 3: p. A8-A12values are in Appendix 3: p. A8-A12

Example 1Example 1

can also be solved using enthalpy of formation can also be solved using enthalpy of formation equation:equation:

kJH 1396

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

]0)46(4[)]286(6)34(4[ΔHreaction

Example 2Example 2 Calculate the standard enthalpy change for the Calculate the standard enthalpy change for the

following reaction:following reaction:

2Al(s) + Fe2Al(s) + Fe22OO33(s) (s) Al Al22OO33(s) + 2Fe(s)(s) + 2Fe(s)

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

reactionΔH [1( 1676) 2(0)] [2(0) 1( 826)]

= 0 kJ/mol -826 kJ/mol -1676 kJ/mol 0 kJ/mol0f

H

reactionΔH 850.kJ

Example 3Example 3

Compare the standard enthalpy of Compare the standard enthalpy of combustion per gram of methanol with per combustion per gram of methanol with per gram of gasoline (it is Cgram of gasoline (it is C88HH1818).).

Write equations:Write equations:

2CH2CH33OH(l) + 3OOH(l) + 3O22(g) (g) 2CO 2CO22(g) + 4H(g) + 4H22O(l)O(l)

2C2C88HH1818(l) + 25O(l) + 25O22(g)(g)16CO16CO22(g) + 18H(g) + 18H22O(l)O(l)

Example 3Example 3

Calculate the enthalpy of combustion per mole:Calculate the enthalpy of combustion per mole:

]2[]42[ )()()(3 322

OHCHfOHfCOfOHCH HHHH

OHCH moles 2for kJ 1454ΔH 3OHCH3

]2[]1816[ )()()( 18822188

HCfOHfCOfHC HHHH

1884

HC HC moles 2for kJ 1009.1ΔH188

Example 3Example 3

Convert to per gram using molar mass:Convert to per gram using molar mass:

g

kJ22.7

g 32.0

OHCH mol 1

OHCH mol 2

kJ 1454 3

3

g

kJ.874

g 114.2

HC mol 1

HC mol 2

kJ 101.09 188

188

4

so octane is about 2x more effectiveso octane is about 2x more effective