2. Thermochemistry

3. THERMOCHEMISTRY The study of heat released or required by chemical reactions Fuel is burnt to produce energy - combustion (e.g. when fossil fuels are burnt) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) + energy

4. Energy is 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

5. Two main general forms of energy Kinetic energy (E K ) = ½ mv 2 Potential energy (E P ) = mgh Energy due to motion Energy due to position (stored energy) Energy is measured in the standard unit of Joules 1 J = 1 kg ∙ m 2 /s 2 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/s 2

6. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is an indirect measurement of the thermal or heat energy. Temperature is NOT heat Energy 90 0 C 40 0 C There is a greater amount of Heat energy in a bathub at 40 degrees Than in a coffee cup at 90 degrees! greater temperature But less heat energy

7. 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!

8. 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  t Q = heat absorbed or released m = mass of substance C = specific heat of substance  t = t final - t initial

9. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? C of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = mC  t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J

10. 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. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l)

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

12. 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”)  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H = - H products > H reactants  H = +

13. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H is positive! 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.

14. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H is negative! 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.

15. Heat, or  H, can be written as part of a chemical reaction! C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4H 2 O (g) If the reaction has a positive  H, 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  H, then the reaction releases heat, and heat is written on the left side of the arrow as a product  H = -2043 kJ or C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4H 2 O (g) + 2043 kJ

16. 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? C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4H 2 O (g) + 2043 kJ 355 grams C 3 H 8 x 1 mole C 3 H 8 = 8.07 moles C 3 H 8 44 grams 8.07 moles C 3 H 8 x 2043 kJ heat = 16,483.3 kJ heat released 1 mole C 3 H 8

17. How do we measure the heat of a reaction in an experiment…? There are three ways: 1.Using a calorimeter 2.Using Hess’ Law 3.Using a table of heats of formation Let’s look at each method….

18. 1. Using a calorimeter No heat enters or leaves! 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!

19. A 0.1964-g sample of solid quinone (C 6 H 4 O 2 ) 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 C 6 H 4 O 2 (s) + 6 O 2 (g) → 6 CO 2 (g) + 2 H 2 O (g) The heat released by the reaction is absorbed by the calorimeter: q reaction = - q calorimeter q = mc  T q = (373 grams H 2 O)(4.184 J/g 0 C)(3.2°C) = 4,994.02 J gained by calorimeter q reaction = -4,994.02 J (released by reaction)

20. This is not the  H, though! 1 C 6 H 4 O 2 (s) + 6 O 2 (g) → 6 CO 2 (g) + 2 H 2 O (g) The change in heat, or  H, 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 C 6 H 4 O 2 (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,  H = -2700 kJ (2 significant figures)

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

22. 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Using the following sets of reactions: N 2 (g) + O 2 (g)  2NO(g)  H = 180.6 kJ N 2 (g) + 3H 2 (g)  2NH 3 (g)  H = - 91.8 kJ 2H 2 (g) + O 2 (g)  2H 2 O(g)  H = -483.7 kJ Goal: NH 3 : O2 O2 : NO: H 2 O: Reverse and x 2 4NH 3  2N 2 + 6H 2  H = + 183.6 kJ Any chemical in more than one reaction - skip x2 2N 2 + 2O 2  4NO  H = 361.2 kJ x3 6H 2 + 3O 2  6H 2 O  H = -1451.1 kJ

23. 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Goal: NH 3 : O 2 : NO: H 2 O: Reverse and x2 4NH 3  2N 2 + 6H 2  H = + 183.6 kJ x2 2N 2 + 2O 2  4NO  H = 361.2 kJ x3 6H 2 + 3O 2  6H 2 O  H = -1451.1 kJ Cancel terms and take sum. 4NH 3 + 5O 2  4NO + 6H 2 O  H = - 906.3 kJ Is the reaction endothermic or exothermic?

24. 23 Determine the heat of reaction for the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Use the following reactions: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = -1550 kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ Determine the heat of reaction for the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Use the following reactions: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = -1550 kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ

25. Determine the heat of reaction for the reaction: Goal: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? Use the following reactions: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = -1550 kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 4 (g) :use 1 as is C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ H 2 (g) :# 3 as is H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 6 (g) : rev #2 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (g) + 7/2O 2 (g)  H = +1550 kJ C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = -137 kJ

26. A. Heat of Formation (  H f º) – the heat released or absorbed when one mole of a substance is formed from its elements. A. Heat of Formation (  H f º) – the heat released or absorbed when one mole of a substance is formed from its elements. EX: H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f º = -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  H f. The  H f º of an element in its standard state is zero. You Cannot make an element from elements! The  H f º 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! The heat of formation for a substance is like having its Potential energy – it is a measurement of how stable or Unstable it is! 3. Using Heats of Formation Tables

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

28. 1. Calculate the  H for: 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g)  H = ? 1. Calculate the  H for: 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g)  H = ?  H rxn =  H f º(products) –  H f º(reactants)  H rxn = [(2)(NaOH (aq) ) + (1)(H 2(g) )] – [(2)(Na (s) ) + (2)(H 2 O (l) )]

29. 1. 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g)  H = ?  H rxn =  H f º(products) –  H f º(reactants)  H rxn = [(2)(NaOH (aq) ) + (1)(H 2(g) )] – [(2)(Na (s) ) + (2)(H 2 O (l) )]  H rxn = [(2)(-469.6 kJ) + (1)(0 kJ)] – [(2)(0 kJ) + (2)(-285.84 kJ)]  H rxn = -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! 1. 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g)  H = ?  H rxn =  H f º(products) –  H f º(reactants)  H rxn = [(2)(NaOH (aq) ) + (1)(H 2(g) )] – [(2)(Na (s) ) + (2)(H 2 O (l) )]  H rxn = [(2)(-469.6 kJ) + (1)(0 kJ)] – [(2)(0 kJ) + (2)(-285.84 kJ)]  H rxn = -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!

30. 2. Calculate the  H for: C 2 H 5 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 3 H 2 O(l)  H = ? 2. Calculate the  H for: C 2 H 5 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 3 H 2 O(l)  H = ?  H = -1366.89 kJ