HL CHAPTER 5 SECTION 15.1

PPTFOCUS ON HEATS OF FORMATION

Standard Enthalpy Changes of Reaction15.1

15.1.1 – Define and apply the terms standard state, standard enthalpy change of formation (ΔHf˚) and standard enthalpy change of combustion (ΔHc˚).

A – 298K (25˚C) around room temp

1.00 x 105 Pa (101.3 kPa) around room pressure

Q – What are “standard” conditions?

standard enthalpy change of formation (ΔHf˚)

ΔHf˚= 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 (ΔHf˚)

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

conditionsGives a measure of the stability of a substance

relative to its elementsCan be used to calculate the enthalpy changes of all

reactions, hypothetical or real

Sample problem 1 Q – the ΔHf˚ 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 ethanol.

A – Start with the chemical equation for the formation of ethanol from its component elements in their standard states.

_C(graphite) + _H2(g) + _O2(g) _C2H5OH(l) ΔHf˚= -277 kJmol-1

Continue by making the coefficient for ethanol 1 because ΔHf˚ is per mole

2 C(graphite) + 3 H2(g) + ½ O2(g) C2H5OH(l) ΔHf˚= -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?Cl2(g)

I2(s)

Br2(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 ΔHf˚ 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. [ΔHf˚ C (diamond) = 1.8 kJ/mol]

Using ΔHf˚The following expression is used to predict the standard enthalpy change for an entire reaction.

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Why does this work?

○ Hess’s Law – see worked example in text pp 147-148

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Sample Problem 4Calculate the enthalpy change for the reaction

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

-105 zero 3(-394) 4(-286) kJ/mol

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Sample Problem 4Calculate the enthalpy change for the reaction

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

-105 zero 3(-394) 4(-286) kJ/mol

ΔH˚reaction = (3(-394)+4(-286))-(-105) ΔH˚reaction = -2221 kJ/mol

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Sample Problem 5Calculate the enthalpy change for the reaction

C2H5OH (l) + CH3COOH (l) CH3COOC2H5 (l) +

H2O(g)

-277 -874 -2238 -286 kJ/mol

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Sample Problem 5Calculate the enthalpy change for the reaction

C2H5OH (l) + CH3COOH (l) CH3COOC2H5 (l) +

H2O(g)

-277 -874 -2238 -286 kJ/mol

ΔH˚reaction = [(-2238) + (-286)] – [(-277)+(-874)] ΔH˚reaction = -2524 – (-1151)= -1373 kJ/mol

ΔH˚reaction= ΔHf˚products - ΔHf˚reactants

Sample Problem 6Calculate the enthalpy change for the reaction

NH4NO3(s) N2O (g) + 2 H2O(g)

- 366 + 82 - 285 kJ/mol

ΔH˚reaction = [(+82) + 2(-285)] - (-366)] ΔH˚reaction = - 488 – 366 = -122 kJ/mol

Practice Problems with Heats of Formation

Read Section 15.1 pp 147-149 Do Ex 15.1 # 1-3, 8, 10, 11

-------------------------------------------------------More Practice – Try Talbot – HL Practice

Heat of Formation: MC 2, 14, 19, 21, OR 3a