Thermochemical Thermochemical equationsequations16.316.3

16.3 16.3 Thermochemical Thermochemical equationsequations

Thermochemical equation = a Thermochemical equation = a balanced chemical equation that balanced chemical equation that includes the physical states of all includes the physical states of all reactants and products reactants and products and the and the enthalpy change (∆H).enthalpy change (∆H).

Remember that enthalpy is the Remember that enthalpy is the heat heat contentcontent of a system at constant of a system at constant pressure.pressure.

EXOTHERMICEXOTHERMIC

4 Fe 4 Fe (s)(s) + 3O + 3O22 (g)(g) 2Fe 2Fe22OO33 (s)(s) + 1625 kJ + 1625 kJ

4 Fe 4 Fe (s)(s) + 3O + 3O22 (g)(g) 2Fe 2Fe22OO33 (s)(s) ∆H= -1625 kJ ∆H= -1625 kJ

The negative value for ∆H means that the The negative value for ∆H means that the system (reaction) is losing energy, giving off system (reaction) is losing energy, giving off heat, is exothermic.heat, is exothermic.

ENDOTHERMICENDOTHERMIC

27 kJ + NH27 kJ + NH44NONO33 (s)(s) NH NH44++

(aq)(aq) + NO + NO33--(aq)(aq)

NHNH44NONO33 (s)(s) NH NH44++

(aq)(aq) + NO + NO33--(aq)(aq) ∆H = 27 kJ ∆H = 27 kJ

The positive value for ∆H means that the The positive value for ∆H means that the system (reaction) is gaining energy, system (reaction) is gaining energy, taking in heat, is endothermic.taking in heat, is endothermic.

Enthalpy of CombusionEnthalpy of Combusion

∆∆HHcombcomb is the enthalpy change for the is the enthalpy change for the complete burning of one mole of complete burning of one mole of substance.substance.

Combustion of glucose:Combustion of glucose:

CC66HH1212OO66(s) + 6O(s) + 6O22 6CO 6CO22 + 6H + 6H22OO

∆∆HHcombcomb = - 2808 kJ = - 2808 kJ

The negative value means heat is given off The negative value means heat is given off (exothermic.)(exothermic.)

For 1 mole of glucose, 2808 kJ is given off.For 1 mole of glucose, 2808 kJ is given off.

Table 16-5 on page 501Table 16-5 on page 501

∆∆HHcombcomb for octane = -5471 kJ/mol for octane = -5471 kJ/mol

Translation: 5471 kJ is given off from the Translation: 5471 kJ is given off from the combustion of one mole of octane.combustion of one mole of octane.

Which gives off more energy: combustion of Which gives off more energy: combustion of propane or methane?propane or methane?

Propane (-2219 kJ/mol) Propane (-2219 kJ/mol) (Methane is only -891 kJ/mol) (Methane is only -891 kJ/mol)

CHANGES OF STATECHANGES OF STATE

Solid to liquid; liquid to gas; solid to gasSolid to liquid; liquid to gas; solid to gas All take in energy (endothermic)All take in energy (endothermic) ∆∆H is +H is +

Gas to liquid; liquid to solid; gas to solidGas to liquid; liquid to solid; gas to solid All lose energy (exothermic)All lose energy (exothermic) ∆∆H is -H is -

Changes of StateChanges of State

∆∆HHvapvap (enthalpy of vaporization) = (enthalpy of vaporization) =

heat required to vaporize one mole of heat required to vaporize one mole of a liquid. (liquid to gas)a liquid. (liquid to gas)

∆∆HHfus fus (enthalpy of fusion) =(enthalpy of fusion) =

heat required to melt one mole of a solid heat required to melt one mole of a solid substance (solid to liquid)substance (solid to liquid)

∆∆HHfus fus and and ∆∆HHvapvap

Both are endothermic, so their values are Both are endothermic, so their values are positive.positive.

Table 16-6 on page 502Table 16-6 on page 502

Does it take more heat to vaporize one Does it take more heat to vaporize one mole of water or one mole of ammonia?mole of water or one mole of ammonia?

Water (HWater (Hvapvap = 40.7 kJ/mol). = 40.7 kJ/mol).

(Ammonia is only 23.3 kJ/mol) (Ammonia is only 23.3 kJ/mol)

STOICHIOMETRYSTOICHIOMETRY

It’s back!!It’s back!! Remember Remember

1 mol = 6.02 x 101 mol = 6.02 x 102323 r.p. r.p.

1 mol = 22.4 L (gas at STP)1 mol = 22.4 L (gas at STP)

1 mol = molar mass (periodic table)1 mol = molar mass (periodic table)

NOW ADD:NOW ADD:

1 mol = __kJ (from 1 mol = __kJ (from ∆∆H) H)

How do you use it?How do you use it?

Write the given with UNIT.Write the given with UNIT. Use unit multipliers to cancel units until Use unit multipliers to cancel units until

you get your answer.you get your answer. Look up correct Look up correct ∆∆H on table 16-5 or 16-6, H on table 16-5 or 16-6,

that number of kJ = 1 mol.that number of kJ = 1 mol. Round to the correct significant digits. Round to the correct significant digits.

EXAMPLEEXAMPLE

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

25.7 g CH25.7 g CH33OH x OH x 1 mol1 mol x x 3.22 kJ 3.22 kJ ==

32.05 g 1 mol32.05 g 1 mol

2.58 kJ2.58 kJ