Enthalpy of Formation DHrxn has been tabulated for many different reactions. Often tabulated according to the type of chemical reaction or process DHvap.

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Presentation transcript:

Enthalpy of Formation DHrxn has been tabulated for many different reactions. Often tabulated according to the type of chemical reaction or process DHvap (heat of vaporization) liquid to gas DHfusion (heat of fusion) solid to liquid DHcombustion (heat of combustion)

C (graphite) + 2 H2 (g)  CH4 (g) Enthalpy of Formation Enthalpy of formation (DHf) also called heat of formation enthalpy change associated with the formation of a compound from its constituent elements in their elemental forms DHf for methane (CH4) would be the enthalpy change associated with the following reaction: C (graphite) + 2 H2 (g)  CH4 (g)

Enthalpy of Formation The magnitude of any DH depends on factors such as temperature pressure physical state (s, l, g) of reactants and products To compare DHf for different reactions, we define a set of standard conditions called the standard state.

Enthalpy of Formation The standard state of a substance is its pure form at 1 atm and 25oC (298 K). The DHf measured under standard state conditions is known as the standard enthalpy of formation (DHof) the enthalpy change for the reaction that forms 1 mole of a compound from its elements when all substances are in their standard states

Enthalpy of Formation The DHof for ethanol (C2H5OH) is depicted in the following thermochemical equation. 2 C (graphite) + 3 H2 (g) + 1/2 O2(g)  C2H5OH (l) DHof = - 277.7 kJ/mol Notice: standard state of C = graphite H = H2 (g) O = O2 (g) can use fractional coefficients

Enthalpy of Formation The DHof for C2H5OH (or many other compounds prepared from their elements in the standard state) can be looked up in a table. Table 5.3 Appendix C of text

Enthalpy of Formation By definition, the standard enthalpy of formation of the most stable form of any element is zero. There is no formation reaction when the element is already in its standard state. DHof for C (graphite) = 0 DHof for H2 (g) = 0 DHof for O2 (g) = 0

Enthalpy of Formation Example: Write a balanced chemical equation showing the formation of 1 mole of C2H5Cl (g) from its constituent elements in their standard states.

Enthalpy of Formation Example: Write a complete thermochemical equation showing the formation of 1 mole of solid sodium bicarbonate from its constituent elements in their standard states.

Enthalpy of Formation Remember: A complete thermochemical equation must have two components: balanced chemical equation DHrxn (DHof)

Enthalpy of Formation You will have problems similar to the previous two examples on your exam. You must be able to write a balanced chemical equation for the formation of 1 mole of a product from its constituent elements in their standard states. You must be able to look up the DHof for the product in the table that I give you.

Enthalpy of Formation Who cares about DHof??? Hess’s Law can be used to calculate the enthalpy change for any reaction for which we know DHof for all reactants and products. DHorxn = S n DHof (products) - S n DHof (reactants) where n indicates the number of moles of each reactant or product

Enthalpy of Formation Example: Calculate the DHorxn for the combustion of propane gas (C3H8). C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Enthalpy of Formation

Enthalpy of Formation Example: Calculate the heat of formation of glucose using the following thermochemical equation and the standard heats of formation given below. C6H12O6 (s)  2 C2H5OH (l) + 2 CO2 (g) DHo = - 69 kJ

Enthalpy of Formation

Enthalpy of Formation Your exam will include problems similar to the ones in the last two examples. Given DHof for all reactants and products, you must be able to find DHorxn. Given the DHorxn and DHof for all but one reactant or product, you will be able to find the DHof for that reactant or product.