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2NaHCO3 + 85 kJ Na2CO3 + H2O + CO2

Is this an endothermic or exothermic reaction?

Calculate the amount of heat transferred when 36 g of NaHCO3 is decomposed.

Thermochemical Equations

Enthalpy (H)

• Heat content of a system at constant pressure

ΔHrxn = Hfinal – Hinitial

Changes of State

• Many processes other than chemical reactions absorb or release heat

Changes of StateΔHvap

• Molar heat of vaporization : the heat required to vaporize one mole of liquid

ΔHfus• Molar heat of

fusion: the heat required to melt one mole of a solid substance

What happens in the reverse process?

ΔHvap = -ΔHcond

ΔHfus= -ΔHsolid

Molar enthalpy of condensation

Molar enthalpy of solidification

Changes of State

H2O(l) H2O(s) Δhsolid=-6.01 kJ

H2O(s) H2O(l) ΔHfus=6.01 kJ

Example:

• How many grams of ice at 0°C will melt if 2.25 kJ of heat at added?

Changes of State

H2O(l) H2O(g) ΔHvap=40.7 kJ

H2O(g) H2O(l) ΔHcond=-40.7 kJ

Example:

How much heat is absorbed when 24.8 g of H2O (l) and 101.3 kPa is converted to H2O(g) at 100°C?

Changes of State

H2O(l) H2O(g) ΔHvap=40.7 kJ

H2O(g) H2O(l) ΔHcond=-40.7 kJ

H2O(s) H2O(l) ΔHfus=6.01 kJ

H2O(l) H2O(s) ΔHsolid=-6.01 kJ

Calculate the heat required to melt 25.7 g of solid methanol at its melting point. ΔHfus= 3.22 kJ/mol

Hess’s Law

If you add 2 or more thermochemical equations to give a final equation, the you can also add the heats of reaction to give the final heat of reaction.

What is the enthalpy change in kJ for the following reaction?

2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

Use the enthalpy changes for the combustion of aluminum and iron: 2Al(s) + 3/2O2(s) Al2O3(s); ΔH = -1676.0 kJ

2Fe(s) +3/2O2(s) Fe2O3(s); ΔH = -822.1 kJ

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Determine the heat of reaction for the reaction:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Using the following sets of reactions:

N2(g) + O2(g) 2NO(g) H = 180.6 kJ

N2(g) + 3H2(g) 2NH3(g) H = -91.8 kJ

2H2(g) + O2(g) 2H2O(g) H = -483.7 kJ

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4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Using the following sets of reactions:

a. N2(g) + O2(g) 2NO(g) H = 180.6 kJ

b. N2(g) + 3H2(g) 2NH3(g) H = -91.8 kJc. 2H2(g) + O2(g) 2H2O(g) H = -483.7 kJ

Goal:

NH3:O2 :

NO:H2O:

Reverse and x 2 4NH3 2N2 + 6H2 H = +183.6 kJ Found in more than one place, SKIP IT (its hard).

x2 2N2 + 2O2 4NO H = 361.2 kJx3 6H2 + 3O2 6H2O H = -1451.1 kJ

Add it up!! Hess StyLe

b. 4NH3 2N2 + 6H2

H = +183.6 kJa. 2N2 + 2O2 4NO H = 361.2 kJ

c. 6H2 + 3O2 6H2O H = -1451.1 kJ

4NH3 + 2N2 + 6H2 + 5O2 2N2 + 6H2 + 4NO + 6H2O

ΔH = -906.3 kJ4NH3 + 5O2 4NO + 6H2O ΔH = -906.3 kJ

Is this reaction exothermic or endothermic?

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Determine the heat of reaction for the reaction:

C2H4(g) + H2(g) C2H6(g)

Use the following reactions:

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ

H2(g) + 1/2O2(g) H2O(l) H = -286 kJConsult your neighbor if necessary.

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Determine the heat of reaction for the reaction:

Goal: C2H4(g) + H2(g) C2H6(g) H = ?

Use the following reactions:

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ

H2(g) + 1/2O2(g) H2O(l) H = -286 kJ

C2H4(g) :use 1 as is C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g) H2O(l) H = -286 kJC2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ

C2H4(g) + H2(g) C2H6(g) H = -137 kJ

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