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HL Chapter 5 Section 15.1 ppt Focus on Heats of Formation

Standard Enthalpy Changes of Reaction 15.1. HL Chapter 5 Section 15.1 ppt Focus on Heats of Formation. 15.1.1 – Define and apply the terms standard state, standard enthalpy change of formation ( Δ H f ˚) and standard enthalpy change of combustion ( Δ H c ˚).

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HL Chapter 5 Section 15.1 ppt Focus on Heats of Formation

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  1. Standard Enthalpy Changes of Reaction 15.1 HL Chapter 5 Section 15.1pptFocus on Heats of Formation

  2. 15.1.1 – Define and apply the terms standard state, standard enthalpy change of formation (ΔHf˚) and standard enthalpy change of combustion (ΔHc˚).

  3. Q – What are “standard” conditions? • A – 298K (25˚C) • around room temp • 1.00 x 105 Pa (101.3 kPa) • around room pressure

  4. 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)

  5. 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 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

  6. 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

  7. 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).

  8. 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]

  9. 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

  10. ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants Sample Problem 4 • Calculate the enthalpy change for the reaction • C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) -105 zero 3(-394) 4(-286) kJ/mol

  11. ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants Sample Problem 4 • Calculate 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

  12. ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants Sample Problem 5 • Calculate the enthalpy change for the reaction • C2H5OH (l) + CH3COOH (l)  CH3COOC2H5 (l) + H2O(g) -277 -874 -2238 -286 kJ/mol

  13. ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants Sample Problem 5 • Calculate 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

  14. ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants Sample Problem 6 • Calculate 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

  15. 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

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