Unit 6 Thermodynamics Reaction Enthalpies Part 2 1 6.8 Enthalpies of Formation 6.9 Hess’ Law 6.5 Energy of Phase Changes
2 Learning objectives At the end of the lesson, students should be able to 6.8 Calculate the enthalpy change for a chemical or physical process based on the standard enthalpies of formation. 6.9 Represent a chemical or physical process as a sequence of steps. Explain the relationship between the enthalpy of a chemical or physical process and the sum of the enthalpies of the individual steps. 6.5 Explain changes in the heat q absorbed or released by a system undergoing a phase transition based on the amount of the substance in moles and the molar enthalpy of the phase transition.
Method 3: 6.8 Standard Enthalpy Formation, Definition: The enthalpy change when 1 mole of a pure compound is formed from its constituent elements in their standard states under standard conditions. (298K and 1atm) values are important as they • give a measure of the stability of a substance relative to its elements. • can be used to calculate the enthalpy changes of all reactions, either hypothetical or real. (ΔHf q values of substances will be provided if needed for calculation) 3
Method 3: 6.8 Standard Enthalpy Formation, Definition: The enthalpy change when 1 mole of a pure compound is formed from its constituent elements in their standard states under standard conditions. (298K and 1atm) 4
Quick check 1. N2(g) + 3H2(g) → 2NH3(g) 2. Na+ (aq) + OH- (aq) → NaOH(aq) 3. 2H2(g) + O2(g) → 2H2O(l) 5 Are the following correct equations for enthalpy change of formation?
of an element • # $ of an element in its standard form is zero. Quick Check… 6 1. 2.
Using to determine = ∑ n (products) - ∑ n (reactants) 7 Stoichiometric coefficient
Using to determine 8 Example 1 Use the information below, calculate % $ for the following reaction. SnO2(s) + 2CO(g) → Sn(s) + 2CO2(g) = ∑ n (products) - ∑ n (reactants) # $[SnO2(s)] = -581kJmol-1 # $[CO2(g)] = -394kJmol-1 # $[CO(g)] = -111kJmol-1 % $ = (ΔHf q Sn(s) + 2 ΔHf q CO2(g)) – (ΔHf q SnO2(s) + 2ΔHf q CO(g)) = [0 + 2(-394)] – [(-581) + 2(-111)] = +15kJmol-1 Be careful of the negative signs!
Quick Check 1 9 Use the information below, calculate % $ for the following reaction. 2H2S (g) + 3O2(g) → 2SO2(g) + 2H2O(g) = ∑ n (products) - ∑ n (reactants) # $[SO2 (g)] = -296.8 kJmol-1 # $[H2O(g)] = -241.8 kJmol-1 # $[H2S(g)] = -20.2 kJmol-1
Quick Check 2 10 Calculate # $ for glucose. C6H12O6(g) + 6O2(g) → 6CO2(g) + 6H2O(g) ' $ = -2552 kJmol-1 = ∑ n (products) - ∑ n (reactants) # $[CO2 (g)] = -393.5 kJmol-1 # $[H2O(g)] = -241.8 kJmol-1
Method 4: 6.9 Hess’ Law • Sometimes difficult to measure ∆H of a reaction directly. • Many reaction enthalpies cannot be determined experimentally because the reaction cannot be performed in the laboratory. • Chemists have developed a method which uses an indirect route. à chemical reaction is broken down into a series of steps. à ∆H of a particular reaction is calculated from the known enthalpy of these smaller steps. 11
6.9 Hess’ Law 12 • states that the enthalpy change of a chemical reaction is independent of the route, provided the starting conditions and final conditions, and reactants and products, are the same. à means regardless of one steps or a series of steps, ∆H is the same.
13 A + B ® C + D ΔHr = ? M + N ΔH1 ΔH2 ΔHr A + B C + D Reaction: Series of steps: A + B ® M + N ΔH1 = x kJmol-1 M + N ® C + D ΔH2 = y kJmol-1 Overall ΔHr = ΔH1 + ΔH2 • Sum of ΔH of each smaller step.
Example 14 Step 1: Adding the equations for the 2 small steps will give us the equation for the overall reaction. • You can reverse the equations à ΔH will change sign • You can multiply by factors à ΔH will also need to be multiplied by the same factor Step 2: Add the ΔH values together.
Example 15
Quick Check! 16
Summary 4 methods to Determine DHr 1) Conduct Calorimetry Experiments 2) Use Average Bond Enthalpies 3) Use Standard Enthalpies of Formation 4) Hess’ Law 17
Recall: Heating Curve (G10 Unit 1) 18 melting vaporisation freezing condensation
19 During phase changes, temperature remains a constant. • Energy is taken in to break attractions between particles. • Energy is released to form attractions between particles • No ∆ in T, No ∆ in KE Solid à Liquid à Gas • E absorbed by system, Esys increases Ø Heating curve is Endothermic • ∆ in T, ∆ in KE Gas à Liquid à Solid • E released by the system, Esys decreases Ø Cooling curve is Exothermic • ∆ in T, ∆ in KE
Calculate q 20 q = mcsolid∆T q = nCsolid∆T q = mcliquid∆T q = nCliquid∆T q = mcgas∆T q = nCgas∆T
Molar Heat of Fusion, ∆Ho fus 21 q = n∆Ho fus • Melting is an endothermic process • ∆Ho fus is always positive • Freezing is an exothermic process • q = - n∆Ho fus • ∆Ho fus, is the heat absorbed as 1 mole of a solid liquefies. • Energy is required to break attractions as a substance moves from solid to liquid phase.
Example 1 How much heat is required to melt 2.50 moles of Al at 290K? Al melts at 933K, ∆Ho fus for Al is 10.7 kJ/mol, and the molar heat capacity of Al is 24.3 J/mol·K. 22 Hint: 1) Al needs to be heated from 290K to its melting point and 2) then continue heating till Al melts completely.
Example 1 How much heat is required to melt 2.50 moles of Al at 290K? Al melts at 933K, ∆Ho fus for Al is 10.7 kJ/mol, and the molar heat capacity of Al is 24.3 J/mol·K. Solution: nAl = 2.50 mol Ti = 290K Tf = 933K CAl = 24.3 J/mol·K ∆Ho fus for Al = 10.7 kJ/mol 23 Step 1: Heating to melting point q = nCAl∆T = (2.50 mol) (24.3 J/mol·K) (933-290)K = (2.50 mol) (24.3 J/mol·K) (643)K = 39062 J = 39.1 kJ (3s.f.) Step 2: Melting q = n∆Ho fus = (2.50mol)(10.7 kJ/mol) = 26.8 kJ (3s.f.) Step 3: add q values q = (39.1 + 26.8) kJ = 65.9 kJ
24 q = n∆Ho vaporisation • ∆Ho vap, is the heat absorbed as 1 mole of a liquid becomes gaseous. • Energy is required to break attractions as a substance moves from liquid to gas phase. Molar Heat of Vaporisation, ∆Ho vap • Vaporisation is an endothermic process • ∆Ho vap is always positive • Condensation is an exothermic process • q = - n∆Ho vap
Quick Check! How much heat is required to vaporize 2.00 moles of H2O at 298.0K? ∆Ho vap for H2O is 41 kJ/mol, and the molar heat capacity of H2O is 75.4 J/mol·K. 25 Hint: 1) H2O needs to be heated from 298K to its noiling point and 2) then continue heating till H2O vaporizes completely.
Quick Check How much heat is required to vaporize 2.00 moles of H2O at 298.0K? ∆Ho vap for H2O is 41 kJ/mol, and the molar heat capacity of H2O is 75.4 J/mol·K. Solution: 26
∆Ho vap vs ∆Ho fus 27 ∆Ho fus ∆Ho vap ∆Ho vap > ∆Ho fus • More energy is required to break attractions in the liquid to gas transition than solid to liquid transition. • More energy is required to form attractions in the gas to liquid transition than liquid to solid transition.
Summary • Enthalpy of phase changes • Heating and cooling curves • Calculating q when a substance undergoes phase change • ∆Ho vap and ∆Ho fus 28