Unit4Day3-LaBrake Monday, November 11, 2013 1:15 PM Vanden Bout/LaBrake/Crawford CH301 THERMODYNAMICS Quantifying Heat Flow – Chemical Change UNIT 4 Day 3 CH301 Vanden Bout/LaBrake Fall 2013 Important Information HW10 DUE T 9AM CH302 Vanden Bout/LaBrake Spring 2012 Unit4Day3-LaBrake Page 1
What are we going to learn today? Use calorimetry to calculate ΔH rxn Use different methods to calculate ΔH rxn Define Heats of Formation, Hess’s Law, and Bond Energies CH301 Vanden Bout/LaBrake Fall 2013 QUIZ: iClicker Question 1 A bomb calorimeter measures heat at constant volume, which is equivalent to a) ΔU b) ΔH c) Work CH301 Vanden Bout/LaBrake Fall 2013 System and State A system is the part of the universe on which we want to focus our attention. The surroundings are everything else The universe is the system and the surroundings Universe = system + surroundings We also describe chemical changes with beginning and end states A change in a chemical reaction is described as Δ State = State end – State beginning CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 2
We also describe chemical changes with beginning and end states A change in a chemical reaction is described as Δ State = State end – State beginning CH301 Vanden Bout/LaBrake Fall 2013 Enthalpy for a Chemical Change ENERGY REACTION PATH 2 CH 3 OH + 3 O 2 2 CO 2 + 4 H 2 O + heat CH301 Vanden Bout/LaBrake Fall 2013 Enthalpy for a Chemical Change Thermochemical equation A chemical reaction written with the corresponding enthalpy change CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Δ H = -890 kJ mol rxn -1 2CH 4 (g) + 4O 2 (g) 2CO 2 (g) + 4H 2 O(l) Δ H = -1780 kJ mol rxn -1 CO 2 (g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g) Δ H = +890 kJ mol rxn -1 CH301 Vanden Bout/LaBrake Fall 2013 POLL: iClicker Question 2 How much heat is released when 10 g of CH 4 is combusted? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Δ H = -890 kJ mol rxn -1 A. -890 kJ B. 890 kJ C. -556 kJ D. 556 kJ Unit4Day3-LaBrake Page 3
CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Δ H = -890 kJ mol rxn -1 A. -890 kJ B. 890 kJ C. -556 kJ D. 556 kJ CH301 Vanden Bout/LaBrake Fall 2013 Standard Enthalpies, Δ H rxn Reaction enthalpies are based on all reactants and products in their standard state (1 bar pressure). Tabulated data can be assumed to be at 25°C. Standard conditions are represented by CH301 Vanden Bout/LaBrake Fall 2013 Enthalpy Calculations It is possible to calculate the enthalpy of a chemical change using tabulated data. We can do this because Enthalpy is a State Function These type of calculations allow us to Estimate reaction enthalpy Run an experiment that is typically too expensive Predict the spontaneity of a reaction CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 4
Enthalpy Calculations There are three strategies for calculating Enthalpy 1. Hess’s Law 2. Hess’s Law using Heats of Formation 3. Bond Energies CH301 Vanden Bout/LaBrake Fall 2013 Activity Please open your course pack to page 107. Thermo Unit – Reaction Enthalpies CH301 Vanden Bout/LaBrake Fall 2013 POLL: iClicker Question 4 Calculate the reaction enthalpy for the combustion of methanol. A. -2233 kJ mol -1 B. -1516 kJ mol -1 C. -3793 kJ mol -1 D. -1277 kJ mol -1 CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 5
CH301 Vanden Bout/LaBrake Fall 2013 Hess’s Law Hess’s Law The combination of a series of chemical reactions to estimate the change in enthalpy for an overall net reaction What is the standard change in enthalpy for the following reaction? C (s) + 0.5 O 2(g) CO (g) GIVEN: Δ H º = + 283 kJ mol rxn -1 CO 2(g ) CO (g ) + ½O 2(g ) Δ H º = - 393 kJ mol rxn -1 C (s) + O 2(g) CO 2(g) CH301 Vanden Bout/LaBrake Fall 2013 Hess’s Law GOAL: C (s) + 0.5 O 2(g) CO (g) Δ H º = + 283 kJ mol rxn -1 CO 2(g ) CO (g ) + 0.5 O 2(g ) Δ H º = - 393 kJ mol rxn -1 C (s) + O 2(g) CO 2(g) CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 6
Hess’s Law Suppose the controlled reaction of the oxygen in air with Methane could produce Methanol, which is a clean burning liquid fuel source. Find the standard reaction enthalpy for the formation of 1 mole of CH 3 OH (l) from methane and oxygen, given the following information: Δ H º = +206.10 kJ CH 4(g) + H 2 O (g) CO (g) + 3 H 2(g) Δ H º = -128.33 kJ 2 H 2(g) + CO (g) CH 3 OH (l) Δ H º = -483.64 kJ 2H 2(g) + O 2(g) 2H 2 O (g) CH301 Vanden Bout/LaBrake Fall 2013 Hess’s Law GOAL: 2 CH 4(g) + O 2(g) 2 CH 3 OH (l) Δ H º = +206.10 kJ CH 4(g) + H 2 O (g) CO (g) + 3 H 2(g) Δ H º = -128.33 kJ 2 H 2(g) + CO (g) CH 3 OH (l) Δ H º = -483.64 kJ 2 H 2(g) + O 2(g) 2H 2 O (g) CH301 Vanden Bout/LaBrake Fall 2013 POLL: iClicker Question 5 Use the enthalpies of formation from the table on page 108 to calculate the enthalpy for the same reaction. A. -2233 kJ mol -1 B. -1516 kJ mol -1 C. -3793 kJ mol -1 D. -1277 kJ mol -1 CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 7
Standard Enthalpy of Formation, Δ H f Standard Enthalpy of Formation Δ H for the formation of 1 mole of a compound from its elements in their most stable form at standard conditions 2 C (gr) + 3 H 2(g) + 0.5 O 2(g) 1 C 2 H 5 OH (l) Δ H f º = -277.67 kJ mol -1 CH301 Vanden Bout/LaBrake Fall 2013 POLL: iClicker Question 6 What is the standard enthalpy of formation for O 2(g) ? A. -277.67 kJ mol -1 B. 277.67 kJ mol -1 C. -6.02 kJ mol -1 D. 6.02 kJ mol -1 0 kJ mol -1 E. CH301 Vanden Bout/LaBrake Fall 2013 Standard Enthalpy of Formation, Δ H f We can use the standard enthalpy of ° ) to calculate the standard formation ( Δ H f ° ) enthalpy of reaction ( Δ H f This is possible because Δ H is a state function (independent of pathway), where Δ H = Δ H final – Δ H initial ° = Σ n Δ H f ° ° Δ H r products - Σ n Δ H f reactants Unit4Day3-LaBrake Page 8
Δ Δ Δ This is possible because Δ H is a state function (independent of pathway), where Δ H = Δ H final – Δ H initial ° = Σ n Δ H f ° ° Δ H r products - Σ n Δ H f reactants CH301 Vanden Bout/LaBrake Fall 2013 Example Calculate the standard enthalpy of combustion of methanol from the provided data. ° = -239 kJ mol -1 Δ H f, CH3OH ° = -394 kJ mol -1 Δ H f, CO2 ° = -286 kJ mol -1 Δ H f, H2O CH301 Vanden Bout/LaBrake Fall 2013 POLL: iClicker Question 7 Use the bond energy table on page 108 to calculate the enthalpy for this same reaction. A. Approximately -1277 kJ mol -1 x 0 B. Approximately -1277 kJ mol -1 x 1 C. Approximately -1277 kJ mol -1 x 2 D. Approximately -1277 kJ mol -1 x 3 CH301 Vanden Bout/LaBrake Fall 2013 Bond Enthalpies Bond Enthalpy The heat required to break a mole of bonds at constant pressure. ° = Σ BE reactants - Σ BE products Δ H r Unit4Day3-LaBrake Page 9
Bond Enthalpies Bond Enthalpy The heat required to break a mole of bonds at constant pressure. ° = Σ BE reactants - Σ BE products Δ H r CH301 Vanden Bout/LaBrake Fall 2013 Example ° for Estimate the Δ H r CCl 3 CHCl 2(g) + 2HF (g) CCl 3 CHF 2(g) + 2HCl (g) BE C-Cl = 338 kJ mol -1 BE H-F = 567 kJ mol -1 BE C-F = 484 kJ mol -1 BE H-Cl = 43 kJ mol -1 CH301 Vanden Bout/LaBrake Fall 2013 What have we learned today? The transfer of heat energy into or out of a system at constant pressure is a state function called Enthalpy. The change in Enthalpy can be determined experimentally using a coffee cup calorimeter at constant pressure. Then change in Enthalpy can be calculated based on a variety of tabulated data: Heats of formation/Other Heats of Reaction/Bond Energies CH302 Vanden Bout/LaBrake Fall 2012 Unit4Day3-LaBrake Page 10
Learning Outcomes Write a formation chemical equation for a compound Calculate change in enthalpy for a reaction based on calorimetry data Calculate change in enthalpy for a reaction based on tabulated data (Hess’s law, formation data, bond energy data). CH301 Vanden Bout/LaBrake Fall 2013 Unit4Day3-LaBrake Page 11
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