Standard Enthalpy Changes of Reaction 15.1

– Define and apply the terms standard state, standard enthalpy change of formation (ΔH f ˚) and standard enthalpy change of combustion (ΔH c ˚).

 A – 298K (25˚C) around room temp  1.00 x 10 5 Pa (101.3 kPa) around room pressure Q – What are “standard” conditions?

standard enthalpy change of formation (ΔH f ˚)  ΔH f ˚= the enthalpy change that occurs when 1 mol of a substance is formed from its elements in their standard states. (see table 11 of IB data booklet)

standard enthalpy change of formation (ΔH f ˚) Enthaply of formation of any element in its standard state is ZERO !!! ○ For a given state of matter – per standard conditions ○ For a given allotrope – usually the most stable one !!! ○ Superscript  may be used to indicate standard conditions Gives a measure of the stability of a substance relative to its elements Can be used to calculate the enthalpy changes of all reactions, hypothetical or real

Sample problem 1  Q – the ΔH f ˚ of ethanol is given in Table 11 of the IB data booklet. Give the thermo chemical equation which represents the standard enthalpy of formation of eth anol.  A – Start with the chemical equation for the formation of ethanol from its component elements in their standard states.  _C(graphite) + _H 2 (g) + _O 2 (g)  _C 2 H 5 OH(l) ΔH f ˚= -277 kJmol -1 Continue by making the coefficient for ethanol 1 because ΔH f ˚ is per mole  2 C(graphite) + 3 H 2 (g) + ½ O 2 (g)  C 2 H 5 OH(l) ΔH f ˚= -277 kJmol -1

Sample problem 2  Q – Which of the following does NOT have a standard heat of formation value of zero at 25˚C and 1.00E5 Pa? Cl 2 (g) I 2 (s) Br 2 (g) Na(s)  A – Elements in their STANDARD states have a zero value. Bromine is a LIQUID in its standard state, so bromine it its gas state would have a ΔH f ˚ not equal to zero (31 kJ/mole).

Sample problem 3  Q – Which of the following DOES have a standard heat of formation value of zero at 25˚C and 1.00E5 Pa? H(g) Hg(s) C(diamond) Si(s)  A – Graphite is more stable (but not harder) than diamond so Si is the only choice in its standard state. [ΔH f ˚ C (diamond) = 1.8 kJ/mol]

Using ΔH f ˚ The following expression is used to predict the standard enthalpy change for an entire reaction. ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Why does this work? ○ Hess’s Law – see worked example in text pp

ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Sample Problem 4 Calculate the enthalpy change for the reaction C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) -105 zero 3(-394) 4(-286) kJ/mol

ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Sample Problem 4 Calculate the enthalpy change for the reaction C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) -105 zero 3(-394) 4(-286) kJ/mol  ΔH˚ reaction = (3(-394)+4(-286))-(-105)  ΔH˚ reaction = kJ/mol

ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Sample Problem 5 Calculate the enthalpy change for the reaction C 2 H 5 OH (l) + CH 3 COOH (l)  CH 3 COOC 2 H 5 (l) + H 2 O(g) kJ/mol

ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Sample Problem 5 Calculate the enthalpy change for the reaction C 2 H 5 OH (l) + CH 3 COOH (l)  CH 3 COOC 2 H 5 (l) + H 2 O(g) kJ/mol  ΔH˚ reaction = [(-2238) + (-286)] – [(-277)+(-874)]  ΔH˚ reaction = – (-1151)= kJ/mol

ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Sample Problem 6 Calculate the enthalpy change for the reaction NH 4 NO 3 (s)  N 2 O (g) + 2 H 2 O(g) kJ/mol  ΔH˚ reaction = [(+82) + 2(-285)] - (-366)]  ΔH˚ reaction = – 366 = -122 kJ/mol

Practice Problems with Heats of Formation  Read Section 15.1 pp  Do Ex 15.1 # 1-3, 8, 10, More Practice – Try Talbot – HL Practice Heat of Formation: MC 2, 14, 19, 21, OR 3a