Thermochemistry

1. Thermochemistry

2. Fuel is burnt to produce energy - combustion (e.g. when fossil fuels are burnt) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + energy THERMOCHEMISTRY The study of heat released or required by chemical reactions

3. Energyis the capacity to do work – there are many types of energy! • Thermal energy is the energy associated with the movement of molecules in a substance – it is the same as kinetic energy! • Chemical energy is the energy stored within the bonds of chemical substances • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Electrical energy is the energy associated with the flow of electrons • Potential energy is the energy available by virtue of an object’s position 6.1

4. Two main general forms of energy • Energy is measured in the standard unit of Joules • 1 J = 1 kg ∙ m2/s2 • mass must be in kg • velocity must be in m/s • height must be in meters • g = acceleration due to gravity must be in m/s2 Potential energy (EP) = mgh Kinetic energy (EK) = ½ mv2 Energy due to position (stored energy) Energy due to motion

5. Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is an indirect measurement of the thermal or heat energy. Temperature is NOT heat Energy 400C greater temperature But less heat energy There is a greater amount of Heat energy in a bathub at 40 degrees Than in a coffee cup at 90 degrees! 900C

6. UNITS OF ENERGY S.I. unit of energy is the joule (J) Heat and work ( energy in transit) also measured in joules The calorie (cal) is another metric unit for energy – 1 cal = 4.184 J A Food calorie, with a capital C – is equal to a 1000 chemistry calories: 1 Food Calorie (1 Calorie) = 1000 calories A Candy bar with 480 Calories actually contains 480,000 chemistry calories!

7. The specific heat (C) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. To measure the Heat (q) absorbed or released by a substance: q = m C Dt Q = heat absorbed or released m = mass of substance C = specific heat of substance Dt = tfinal - tinitial

8. How much heat is given off when an 869 g iron bar cools from 940C to 50C? C of Fe = 0.444 J/g •0C Dt = tfinal – tinitial = 50C – 940C = -890C q = mCDt = 869 g x 0.444 J/g •0C x –890C = -34,000 J

9. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

10. Fireworks exploding is an exothermic reaction Burning fossil fuels is an exothermic reaction Photosynthesis is an endothermic reaction (requires energy input from sun) Ice melting is an endothermic process!

11. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. (comes from Greek for “heat inside”) DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH = - DH = +

12. H2O (s) H2O (l) DH = 6.01 kJ Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH is positive! 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

13. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH is negative! 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

14. C3H8(g) + 5 O2(g) 3 CO2(g) + 4H2O (g) Heat, or DH, can be written as part of a chemical reaction! If the reaction has a positive DH, then the reaction needs heat, and heat is written on the left side of the arrow as a reactant If the reaction has a negative DH, then the reaction releases heat, and heat is written on the left side of the arrow as a product DH = -2043 kJ or C3H8(g) + 5 O2(g) 3 CO2(g) + 4H2O (g) + 2043 kJ

15. C3H8(g) + 5 O2(g) 3 CO2(g) + 4H2O (g) + 2043 kJ We can treat heat, then, like a reactant or product…. And perform stoichiometry problems… How many kJ of heat are released when 355 grams of propane are burned with excess oxygen? 355 grams C3H8 x 1 mole C3H8 = 8.07 moles C3H8 44 grams 8.07 moles C3H8 x 2043 kJ heat = 16,483.3 kJ heat released 1 mole C3H8

16. How do we measure the heat of a reaction in an experiment…? • There are three ways: • Using a calorimeter • Using Hess’ Law • Using a table of heats of formation • Let’s look at each method….

17. 1. Using a calorimeter • A calorimeter is an insulated device used to capture all of the heat either absorbed or released by a reaction! • The reaction is usually surrounded by water….why? • Water is stable, and has a high specific heat • It changes temperature slowly! • q reaction = - q surroundings • By measuring the heat that the water absorbs or releases, we can calculate the heat of the reaction! No heat enters or leaves!

18. A 0.1964-g sample of solid quinone (C6H4O2) is burned in a bomb calorimeter that contains 373 grams of water. The temperature of the water inside the calorimeter increases by 3.2°C. Calculate the energy of combustion of quinone per mole. • First – write a balanced chemical equation! • 1 C6H4O2 (s) + 6 O2 (g) → 6 CO2 (g) + 2 H2O (g) • The heat released by the reaction is absorbed by the calorimeter: • q reaction = - q calorimeter • q = mcDT • q = (373 grams H2O)(4.184 J/g0C)(3.2°C) = 4,994.02 J gained by calorimeter • q reaction = -4,994.02 J (released by reaction)

19. This is not the DH, though! • 1 C6H4O2 (s) + 6 O2 (g) → 6 CO2 (g) + 2 H2O (g) • The change in heat, or DH, is the energy released for the reaction the way it was written! • We only used .1964 grams of the chemical! • The reaction calls for one mole of the chemical! • 1 mole C6H4O2 (s) = 108 grams • So I set up a ratio: • -4,994.02 J/.1964 g = X/108 g • X = -2,746,202.4 J = -2746.2024 kJ • So, DH = -2700 kJ (2 significant figures)

20. 2. Using Hess’ Law What if a reaction is too costly or dangerous to conduct, but we still want to calculate its DH? 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) We can use algebra to manipulate other reactions to look like the desired reaction – making sure we change the energies as well! This is called Hess’ Law: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ

21. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3 2N2 + 6H2 H =+183.6 kJ NH3: Reverse and x 2 Any chemical in more than one reaction - skip O2 : 2N2 + 2O24NO H = 361.2 kJ NO: x2 H2O: x3 6H2 + 3O26H2O H = -1451.1 kJ

22. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H =+183.6 kJ NH3: Reverse and x2 O2 : NO: x2 2N2 + 2O24NO H = 361.2 kJ H2O: x3 6H2 + 3O26H2O H = -1451.1 kJ Cancel terms and take sum. + 5O2 H = -906.3 kJ + 6H2O 4NH3  4NO Is the reaction endothermic or exothermic?

23. Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ

24. Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g) C2H6(g) H = -137 kJ

25. 3. Using Heats of Formation Tables A. Heat of Formation (Hfº) – the heat released or absorbed when one mole of a substance is formed from its elements. EX: H2(g) + ½ O2(g)  H2O(l) Hfº = -289 kJ The reactant elements are in their “standard state” – their most stable form at 25 ºC and 1 atm. This is usually indicated with a 0 by the Hf. The Hfº of an element in its standard state is zero. You Cannot make an element from elements! The heat of formation for a substance is like having its Potential energy – it is a measurement of how stable or Unstable it is!

26. How do we use the table to figure out the H For a reaction? The H of a rxn is equal to the sum of the Hfº’s ofthe products minus the sum of the Hfº’s of the reactants. (Each product’s or reactant’s Hfº must be multiplied by its coefficient.) Hrxn = S Hfº(products) – S Hfº(reactants) YOU MUST PRINT OFF THE TABLE FROM MY WEBSITE – IT WAS NOT INCLUDED IN YOUR PACKET!

27. 1. Calculate the H for: 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g) H = ? Hrxn = S Hfº(products) – S Hfº(reactants) Hrxn = [(2)(NaOH(aq)) + (1)(H2(g))] – [(2)(Na(s)) + (2)(H2O(l))]

28. 1. 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g) H = ? Hrxn = S Hfº(products) – S Hfº(reactants) Hrxn = [(2)(NaOH(aq)) + (1)(H2(g))] – [(2)(Na(s)) + (2)(H2O(l))] Hrxn = [(2)(-469.6 kJ) + (1)(0 kJ)] – [(2)(0 kJ) + (2)(-285.84 kJ)] Hrxn = -367.52 kJ Notice that the table from my website has some values listed twice – and they are slightly different – you might get slightly differing answers based on the values that you use – that is fine!

29. 2. Calculate the H for: C2H5OH (l) + 3 O2(g)  2 CO2(g) + 3 H2O(l) H = ? H = -1366.89 kJ