Thermochemical equations 16.3. 16.3 Thermochemical equations  Thermochemical equation = a balanced chemical equation that includes the physical states.

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Presentation transcript:

Thermochemical equations 16.3

16.3 Thermochemical equations  Thermochemical equation = a balanced chemical equation that includes the physical states of all reactants and products and the enthalpy change (∆H).  Remember that enthalpy is the heat content of a system at constant pressure.

EXOTHERMIC 4 Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) kJ 4 Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) ∆H= kJ The negative value for ∆H means that the system (reaction) is losing energy, giving off heat, is exothermic.

ENDOTHERMIC 27 kJ + NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) ∆H = 27 kJ The positive value for ∆H means that the system (reaction) is gaining energy, taking in heat, is endothermic.

Enthalpy of Combusion  ∆H comb is the enthalpy change for the complete burning of one mole of substance.  Combustion of glucose: C 6 H 12 O 6 (s) + 6O 2  6CO 2 + 6H 2 O ∆H comb = kJ The negative value means heat is given off (exothermic.) For 1 mole of glucose, 2808 kJ is given off.

Table 16-5 on page 501  ∆H comb for octane = kJ/mol  Translation: 5471 kJ is given off from the combustion of one mole of octane.  Which gives off more energy: combustion of propane or methane?  Propane (-2219 kJ/mol)  (Methane is only -891 kJ/mol)

CHANGES OF STATE  Solid to liquid; liquid to gas; solid to gas  All take in energy (endothermic)  ∆H is +  Gas to liquid; liquid to solid; gas to solid  All lose energy (exothermic)  ∆H is -

Changes of State  ∆H vap (enthalpy of vaporization) = heat required to vaporize one mole of a liquid. (liquid to gas)  ∆H fus (enthalpy of fusion) = heat required to melt one mole of a solid substance (solid to liquid)

∆H fus and ∆H vap Both are endothermic, so their values are positive. Table 16-6 on page 502 Does it take more heat to vaporize one mole of water or one mole of ammonia? Water (H vap = 40.7 kJ/mol). (Ammonia is only 23.3 kJ/mol)

STOICHIOMETRY  It’s back!!  Remember 1 mol = 6.02 x r.p. 1 mol = 22.4 L (gas at STP) 1 mol = 22.4 L (gas at STP) 1 mol = molar mass (periodic table) NOW ADD: 1 mol = __kJ (from ∆H)

How do you use it?  Write the given with UNIT.  Use unit multipliers to cancel units until you get your answer.  Look up correct ∆H on table 16-5 or 16-6, that number of kJ = 1 mol.  Round to the correct significant digits.

EXAMPLE  Calculate the heat required to melt 25.7 g of solid methanol at its melting point. (∆H= 3.22kJ/mol) 25.7 g CH 3 OH x 1 mol x 3.22 kJ = g 1 mol g 1 mol 2.58 kJ