Standard Enthalpy of Formation

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Presentation transcript:

Standard Enthalpy of Formation College Chemistry

Enthalpies of Formation If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof . Standard conditions (standard state): Most stable form of the substance at 1 atm and 25 oC (298 K). Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

Enthalpies of Formation If there is more than one state for a substance under standard conditions, the more stable one is used. Standard enthalpy of formation of the most stable form of an element is zero.

Enthalpies of Formation Substance DHof (kJ/mol) C(s, graphite) O(g) 247.5 O2(g) N2(g)

Enthalpies of Formation C3H8(g) + 5O2(g)  3CO2(g) + 4H2O() Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction Note: n & m are stoichiometric coefficients. Calculate heat of reaction for the combustion of propane gas giving carbon dioxide and water. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O()

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O() ∆Hof (kJ/mol):

Hess’s Law Hess’s law: if a reaction is carried out in a number of steps or if a product can’t easily be made form its standard state, H for the overall reaction is the sum of H for each individual step. For example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ 2H2O(g)  2H2O(l) H= - 88 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ

Another Example of Hess’s Law Given: C(s) + ½ O2(g)  CO(g) DH = -110.5 kJ CO2(g)  CO(g) + ½ O2(g) DH = 283.0 kJ Calculate DH for: C(s) + O2(g)  CO2(g)