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Enthalpies of Formation and Reaction. Definitions: Standard state A gas at 1 atm An aqueous solution with a concentration of 1 M at a pressure of 1 atm Pure liquids and solids The most stable form of elements at 1 atm and 25°C
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Enthalpies of Formation and Reaction Definitions: • Standard state • A gas at 1 atm • An aqueous solution with a concentration of 1 M at a pressure of 1 atm • Pure liquids and solids • The most stable form of elements at 1 atm and 25°C • allotrope: one or 2 or more forms of an element in the same physical state (diamond and graphite are allotropes of carbon; O2(g) and ozone, O3(g) are allotropes of oxygen).
Enthalpies of Formation and Reaction Definitions (Cont’d) • standard enthalpy of formation (ΔHf°) • enthalpy change for formation of 1 mole of an element or compound from its elements in their standard states (i.e. naturally occurring form at 1 atm and 25˚C) Note: ΔHf° = 0 for any element in its naturally occurring (most stable) form
ΔHf° Values Examples of ΔHf° are as follows: Ag(s)+ Cl2(g)AgCl(s)ΔH = –127.1 kJ so ΔHf°(AgCl, s) = -127.1 kJ N2(g) + O2(g)NO2(g)ΔH = +33.2 kJ so ΔHf° (NO2, g) = +33.2 kJ
Which of these are standard enthalpy of formation reactions? • N2(g) + 3 H2(g) 2 NH3(g) • Cgraphite(s) + 2 H2(g) CH4(g) • Ca(s) + Br2(g) CaBr2(s)
Example When 50.0 g of nitrogen react with excess hydrogen to form ammonia gas, 164.5 kJ of heat are liberated (released) at standard state conditions. Calculate the standard enthalpy of formation (in kJ/mol) for ammonia gas.
Direct Method for Calculating ΔH° Calculation of ΔH°(superscript ° denotes standard state conditions, at 1 atm and 25°C) For the reaction:aA + bBcC + dD where a,b,c,d = stoichiometric coefficients ΔH° = Σn ΔHf°(products) – Σm ΔHf°(reactants) = [c (C) + d (D)] – [a (A) + b (B)] = (total energy of products) – (total energy of reactants)
Example Use the table above to calculate ΔH° for the reaction below: Cgraphite(s) + O2(g) CO2(g) ΔH°= (1mol)(CO2, g) – [(1mol)(Cgraphite, s) + (1mol) (O2, g)] ΔH°=
Example Consider the reaction for the combustion of glucose: C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)DH°= –2537 kJ Use the standard enthalpies of formation provided to calculate the enthalpy of formation for glucose.
Example Consider the following thermochemical equation: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)DH°= –904 kJ Use the standard enthalpies of formation provided to calculate the enthalpy of formation for NO.