1. Heat in Reactions

2. Thermochemistry The study of changes in heat in a chemical reaction The study of changes in heat in a chemical reaction Part of thermodynamics Part of thermodynamics

3. Reactions involve energy because: Chemical bonds are BROKEN (requires energy) think of it as a rope holding things together. It won’t break until energy is applied. Chemical bonds are BROKEN (requires energy) think of it as a rope holding things together. It won’t break until energy is applied. Chemical bonds are FORMED (releases energy) because it is an opposite process. Chemical bonds are FORMED (releases energy) because it is an opposite process.

4. Bonds Here we see bonds holding molecules together on the left & breaking and then reforming into new molecules on the right Here we see bonds holding molecules together on the left & breaking and then reforming into new molecules on the right

5. Heat Heat is the energy of moving particles transferred between objects due to a difference in temperature. Heat is the energy of moving particles transferred between objects due to a difference in temperature.

6. Heat It flows from the higher temperature object to the lower temperature object. It flows from the higher temperature object to the lower temperature object. SI unit is Joule (J) but often measured in kilojoules (kJ) SI unit is Joule (J) but often measured in kilojoules (kJ)

7. Some reactions release energy: EXOTHERMIC C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O + energy C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O + 2043 kJ Notice the heat term is on the right !

8. Some reactions absorb energy: ENDOTHERMIC C + H 2 O + energy → CO + H 2 C + H 2 O + 113 kJ → CO + H 2 Notice the heat term is on the left !

9. The heat content of any system The heat content of any system Defined by the following equation: Enthalpy = Energy of a system+ (Pressure x Volume) Defined by the following equation: Enthalpy = Energy of a system+ (Pressure x Volume) H = E + PV H = E + PV This holds mostly for gases and is often called PV work. This holds mostly for gases and is often called PV work. ENTHALPY

10.  We cannot determine absolute H, rather we can determine change in H (  H).  H = H products – H reactants  H = H products – H reactants ENTHALPY CHANGE

11.  For our purposes, the heat absorbed or released during a chemical reaction is equal to the enthalpy change for the reaction.

12. For easy comparisons, Define: standard enthalpy change (  H o ) when reactants and products are in their standard states (at 25 o C and 1 atmosphere) for now we just have to know that different conditions give us different energy changes so we need to state the conditions. when reactants and products are in their standard states (at 25 o C and 1 atmosphere) for now we just have to know that different conditions give us different energy changes so we need to state the conditions.

13. The sign of  H is informative. When  H > 0, (positive) When  H > 0, (positive) –energy is absorbed, –Endothermic –The energy term is on the left. 2NH 3 (g) + 92.4 kJ  N 2 (g) + 3H 2 (g) 2NH 3 (g) + 92.4 kJ  N 2 (g) + 3H 2 (g) Check the next slide to see another way an endothermic reaction is shown. Check the next slide to see another way an endothermic reaction is shown.

14. Notice the products are higher than the reactants

15. The sign of  H is informative. When  H < 0, (negative) When  H < 0, (negative) –energy is released, –Exothermic –The energy term is on the right. N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92.4 kJ N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92.4 kJ Check the next slide to see another way an exothermic reaction is shown. Check the next slide to see another way an exothermic reaction is shown.

16. Notice the products are lower than the reactants

18. Using Enthalpy Changes A bombardier beetle uses hydrogen peroxide to create heat and produce a steam spray. A bombardier beetle uses hydrogen peroxide to create heat and produce a steam spray. How much heat will be released if 1.0 grams of H 2 O 2 decomposes according to the following equation: How much heat will be released if 1.0 grams of H 2 O 2 decomposes according to the following equation: 2H 2 O 2  2H 2 O + O 2  H o = -190 kJ 2H 2 O 2  2H 2 O + O 2  H o = -190 kJ

19. Help in solving Change grams to moles Change grams to moles Remember your stoichiometry Remember your stoichiometry Look to see if you can use a proportion Look to see if you can use a proportion

20. Help in solving 1.0 g of H 2 O 2 / 34g/mol = 0.0294 mol 1.0 g of H 2 O 2 / 34g/mol = 0.0294 mol H 2 O 2  H 2 O + ½ O 2 H 2 O 2  H 2 O + ½ O 2 0.0294 mol x -190 kJ/mol = -5.6 kJ 0.0294 mol x -190 kJ/mol = -5.6 kJ which means 5.6 kJ energy released because the negative sign means it is an exothermic reaction. which means 5.6 kJ energy released because the negative sign means it is an exothermic reaction.

21. Practice How much heat will be transferred when 5.81 grams of graphite reacts with excess H 2 according to the following equation: How much heat will be transferred when 5.81 grams of graphite reacts with excess H 2 according to the following equation: 6 C + 3 H 2  C 6 H 6 (  H o = +49.03 kJ)? 6 C + 3 H 2  C 6 H 6 (  H o = +49.03 kJ)?

22. Hess’s Law If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps  H net =  H 1 +  H 2 +   H net =  H 1 +  H 2 + 

23. Example of Hess’s Law N 2 + 2O 2  2NO 2  H o = ? N 2 + 2O 2  2NO 2  H o = ? We do know… We do know… N 2 + O 2  2NO  H o = +181 kJ N 2 + O 2  2NO  H o = +181 kJ 2NO + O 2  2NO 2  H o = -113 kJ 2NO + O 2  2NO 2  H o = -113 kJ Rearrange equations to come up with our desired equation Rearrange equations to come up with our desired equation Add the resulting enthalpy changes Add the resulting enthalpy changes Remember to factor when needed. Remember to factor when needed.

24. Practice Given the following enthalpy changes Given the following enthalpy changes Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2  H o = -27 kJ AND Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2  H o = -27 kJ AND C + CO 2  2 CO  H o = +172 kJ C + CO 2  2 CO  H o = +172 kJ Calculate the  H for the following reaction Calculate the  H for the following reaction 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2

25. Calorimetry Calorimetry experiments determine the enthalpy (heat) changes of reactions Calorimetry experiments determine the enthalpy (heat) changes of reactions Determined by measuring temperature changes to the outside environment, using the Law of Conservation of Energy. Determined by measuring temperature changes to the outside environment, using the Law of Conservation of Energy.

26. More Calorimetry 4.184 Joules = 1 calorie 4.184 Joules = 1 calorie 4.184 Kilojoules = 1 Calorie (Cal) 4.184 Kilojoules = 1 Calorie (Cal) The Cal is also known as the Food Calorie and is what you see on Nutritional Labels of the foods you buy. The Cal is also known as the Food Calorie and is what you see on Nutritional Labels of the foods you buy.

27. A Bomb Calorimeter

28. A Simple Calorimeter

29. Heat Capacity Heat capacity is the amount of heat it takes to raise the temperature of an object by one degree Celsius. Heat capacity is the amount of heat it takes to raise the temperature of an object by one degree Celsius. The heat capacity per one gram is called the specific heat. The heat capacity per one gram is called the specific heat. Specific heat of liquid water is 4.184 J / (g C o ) Specific heat of liquid water is 4.184 J / (g C o ) Iron 0.45 J / (g C o ) Iron 0.45 J / (g C o )

30. Determining the Amount of Heat Released Use the formula Use the formula q = m x C x  T q is the quantity of heat (J) q is the quantity of heat (J) m is the mass of the water surrounding the reaction (1mL = 1 g) m is the mass of the water surrounding the reaction (1mL = 1 g) C is the specific heat capacity (4.184 J / (g C o ) for water) C is the specific heat capacity (4.184 J / (g C o ) for water)  T is the change in temperature (final – initial)  T is the change in temperature (final – initial)

31. Practice How much heat is released when 5 grams of NH 4 NO 3 is dissolved in 50 mL of water? How much heat is released when 5 grams of NH 4 NO 3 is dissolved in 50 mL of water? What is the specific heat of Aluminum if the temperature of a 28.4 gram sample increased by 8.1 o C when 207 J of heat is added? What is the specific heat of Aluminum if the temperature of a 28.4 gram sample increased by 8.1 o C when 207 J of heat is added?

32. Applying the equation slide 1 The equation q = m x C x  T can be combined with the Law of Conservation of Energy to give us help with solving some problems. The equation q = m x C x  T can be combined with the Law of Conservation of Energy to give us help with solving some problems. If a piece of hot metal is placed in a cup of cold water, we all know that the temperature of the metal will decrease and the temperature of the water will increase. If a piece of hot metal is placed in a cup of cold water, we all know that the temperature of the metal will decrease and the temperature of the water will increase.

33. Applying the equation slide 2 So for the heat gain of the water we can use the equation: q = m x C x  T where q equals the heat gained by the water. So for the heat gain of the water we can use the equation: q = m x C x  T where q equals the heat gained by the water. And for the heat lost by the metal we can use the equation: q = m x C x  T where q equals the heat lost by the metal. And for the heat lost by the metal we can use the equation: q = m x C x  T where q equals the heat lost by the metal. According to the Law: The heat lost by the metal has to equal the heat gained by the water. SO this leads to ……. According to the Law: The heat lost by the metal has to equal the heat gained by the water. SO this leads to …….

34. Applying the equation slide 3 Heat Loss = Heat Gain or q lost = q gain Heat Loss = Heat Gain or q lost = q gain Since q = m x  T x C we can change this equation to: Since q = m x  T x C we can change this equation to: m metal x  T metal x C metal = m water x  T water x C water This equation allows us to find a lot of different values: Specific Heat of a metal, temperature change of a system & more. This equation allows us to find a lot of different values: Specific Heat of a metal, temperature change of a system & more.

35. Working our new equation Remember to watch which variables are being added. In other words, remember to keep all the data for the water together on the same side of the equation and do the same for the other material as well. Remember to watch which variables are being added. In other words, remember to keep all the data for the water together on the same side of the equation and do the same for the other material as well. If you are having problems with the algebra, don’t forget to watch the solved problems done in class. If you are having problems with the algebra, don’t forget to watch the solved problems done in class.

36. Calories in Food Caloric content in food is determined by food make up and digestibility. Caloric content in food is determined by food make up and digestibility. Food is made up of Food is made up of –Carbohydrates 17 kJ/g or 4 Cal/g –Protein 17 kJ/g or 4 Cal/g –Fat 38 kJ/g or 9 Cal/g

37. Calories in Food Calories in food are temporarily stored in the liver and used when needed Calories in food are temporarily stored in the liver and used when needed You need to walk ~ 43 minutes to burn off a 300 Calorie ice cream cone, depending on your size. You need to walk ~ 43 minutes to burn off a 300 Calorie ice cream cone, depending on your size. If Calories are not burned they are stored long term as fats in the body. If Calories are not burned they are stored long term as fats in the body.

38. Energy Sources Fossil Fuels: Fossil Fuels: –Petroleum –Natural Gas –LPG & LNG –Coal –Kerosene Renewable Energy Renewable Energy –Solar –Wind –Geothermal –Nuclear –Fuel Cell –Wave Some are more viable than others.