Warm-up (11/9) WS Day 3 out on desk, please Find ΔH for: NO (g) + O (g)  NO2 (g) Use the following reactions: 2O3 (g)  3O2 (g) ΔH = -427 kJ O2 (g)  2 O (g) ΔH = 495 kJ 2NO(g) + 2O3(g)  2NO2(g) + 2O2(g) ΔH = -398 kJ -233 kJ Warm-up (11/9)

Using the data below, calculate the overall enthalpy of reaction for the combustion of butane (C4H10) – give in kJ/mol of butane -2670 kJ/mol One more practice!

Ba(OH)2 8H2O (20g) + NH4NO3 (10g) - water on block of wood Demo Time!

Questions on WS 3?

Many ways to calculate Enthalpy State function = you get the same answer regardless of the “route” you take ΔH is a state function – reaction enthalpies should be the same no matter how many steps it takes to “get there” Experimental: calorimetry (Lab 8, 9, 10) Bond Energies (6.2) Hess’ Law (6.3) – Lab 11 next week! Enthalpy of Formation (Today!) Many ways to calculate Enthalpy

The big idea Endothermic versus Exothermic Chemical reactions are all about energy in and out (enthalpy)! Bonds breaking = requires energy IN Bonds forming = energy OUT The big idea

Enthalpy of Formation – 6.4 Fill in the following equation with each element in its natural state at Room Temp. ____ + ____ + ____  CH3OH (l) Don’t forget to balance! Enthalpy of formation (ΔHf) = the enthalpy to form one mole of a substance from its basic elements in standard state at 1 atm. Endo or Exothermic? 2 C (s) + 4 H2 (g) + O2 (g)  2 CH3OH (l) Exothermic – we are making bonds (and yes, we are breaking a couple in this example too (O2 and H2), but it’s still going to almost always be exothermic because we are making more than breaking) Enthalpy of Formation – 6.4

Enthalpy of Formation – 6.4 ΔHf = -239 kJ/mol (a known value you can find in your textbook) C (s) CH3OH (l) 2 H2 (g) ½ O2 (g) -239 kJ/mol Enthalpy of Formation – 6.4

What is “standard state?” Standard State/Conditions: 1) Temperature = 25 Celsius, 298.15 K 2) Gases at 1 atm 3) Solutions at 1 M 4) Elements in their most stable form at 1 atm What is “standard state?”

But, in a full chemical reaction We have to start with reactants too! C (s) Reactants Products CH4 (g) CO2 (g) H2 (g) H2O (l) O2 (g) O2 (g) Hf given in the direction of MAKING the molecule or compound. If you want to BREAK it (CH4  C and H), then the sign would flip because you are NOT forming it, you are breaking it But, in a full chemical reaction

Find the overall ΔH of this reaction Don’t forget to balance!!! CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) ΔHf CH4 = -75 kJ/mol ΔHf H2O = -286 kJ/mol ΔHf CO2 = -393.5 kJ/mol ΔHf O2 = ??? kJ/mol ΔH of the reaction is?? ΔHf O2 is f course ZERO, because it is an element Find the overall ΔH of this reaction

Find the overall ΔH of this reaction Solution to example from last slide: Break CH4 (so flip the sign = 75 kJ) Do nothing with O2 Make 1 CO2 (so = -393.5 kJ) Make 2 H2O (so -286 kJ/mol x 2 moles = -572 kJ) Then add them up! ΔH = 75 kJ + -393.5 kJ + -572 kJ = -890.5 kJ -890.5 kJ Find the overall ΔH of this reaction

Enthalpy of Formation – 6.4 Three big things to remember: ΔH reaction = ΔH Products – ΔH Reactants OR you can flip the signs of the reactants on your own (as I did in previous example) Standard state elements Hf = 0 kJ/mol All Hf values are given in kJ/mol of substance formed (so multiply if you form or break more than one mole!) Enthalpy of Formation – 6.4

One on your own 2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s) ΔHf Fe2O3 = -826 kJ/mol ΔHf Al2O3 = -1676 kJ/mol -850. kJ One on your own

Enthalpy of Formation – 6.4 Practice problems – Worksheet 1 Check Key with me before you go on to WS 2 Enthalpy of Formation – 6.4

Use Appendix 4 (pg A19 in the back of Zumdahl) to complete worksheet 2!

Homework Day 4 REMEMBER: Lab 9 Post-lab due next class (Tuesday) Finish Unit 3 Day 4 Class work and HW Packet to practice Test 2 Corrections (by Friday) Have a great long weekend! Homework Day 4