Thermochemistry Chapter 6 AP Chemistry Seneca Valley SHS

6.1 The Nature of Energy Kinetic and Potential Energy Energy: the capacity to do work or produce heat. Kinetic energy: energy an object has due to its motion Potential energy: energy due to position or composition

6.1 The Nature of Energy Kinetic and Potential Energy Potential energy can be converted into kinetic energy. Example: a ball of clay dropping off a building.

6.1 The Nature of Energy Temperature: the average of energy of particles in motion. temp.  heat Heat: the transfer of energy from a hotter object to a colder one. State Function: property of a system that depends only on its present condition

6.1 The Nature of Energy Systems and Surroundings System: part of the universe we are interested in. Surroundings: the rest of the universe. Exothermic: reaction that involves the release of heat to the surroundings Endothermic: reaction that absorbs energy from the surroundings

6.1 The Nature of Energy Thermodynamics Def: the study of energy and its interconversions First Law of Thermodynamics (Law of Conservation of Energy) Def: Energy cannot be created nor destroyed in ordinary chemical reactions

6.1 First Law of Thermodynamics Internal Energy Internal Energy: total energy of a system. Cannot measure absolute internal energy. Change in internal energy,  E = E final - E initial

Relating  E to Heat and Work Energy cannot be created or destroyed. Energy of (system + surroundings) is constant. when a system undergoes a physical or chemical change, the change in internal energy is given by: the heat added to or absorbed by the system plus the work done on or by the system:  E = q + w 6.1 First Law of Thermodynamics

Relating  E to Heat and Work 6.1 First Law of Thermodynamics Thermodynamic quantities are measured from the perspective of the system Example: in an endothermic process where heat is flowing into the system, q would be positive (+)

Enthalpy: Heat transferred between the system and surroundings carried out under constant pressure. One can only measure the change(  in enthalpy (H):  H = H final - H initial = q P 6.2 Enthalpy

For any reaction:  H rxn = H(products) - H (reactants) Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2CH 4 (g) + 4O 2 (g)  2CO 2 (g) + 4H 2 O(g)  H = kJ When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ Change in enthalpy depends on state: H 2 O(g)  H 2 O(l)  H = -88 kJ 6.2 Enthalpy

Heat Capacity and Specific Heat Calorimetry = science of measuring of heat flow. Calorimeter = device that measures heat flow. Heat capacity = the amount of energy required to raise the temperature of an object. Molar heat capacity = heat capacity of 1 mol of a substance. Specific heat capacity = heat capacity of 1 g of a subst. q = (specific heat)  (grams of substance)  T. 6.2 Calorimetry

Constant-Pressure Calorimetry Atmospheric pressure is constant!  H = q P q rxn = -q soln = -(specific heat of solution)  (grams) of solution)   T. 6.2 Calorimetry

Constant-Pressure Calorimetry Problem 1: When a student mixes 50.0 mL of 12.0 M HCl and 50.0 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0  C to 27.5  C. Calculate the enthalpy change per mole of water formed, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is mL, the density of the solution is 1.00 g/mL, and its specific heat is 4.18 J/g  C. 6.2 Calorimetry

The reaction is carried out under constant vol. Usually used to study combustion. q rxn = -C calorimeter T. Bomb Calorimetry (Constant-Volume Calorimetry) 6.2 Calorimetry

Hess’s lawHess’s law: if a reaction is carried out in a number of steps,  H for the overall reaction is equal to the sum of  H’s from each individual step. For example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ 6.3 Hess’s Law

Notes on Hess’s Law 1.If a reaction is reversed in order to solve a Hess’s Law problem, then the sign of the of the reaction is also reversed. 1.If a reaction is reversed in order to solve a Hess’s Law problem, then the sign of the  H of the reaction is also reversed. 2.The magnitude of  H is directly proportional to the quantities of reactants and products. If the coefficients in an equation are multiplied by an integer, the  H is also multiplied by that integer 3.Hess’s Law can be used to calculate the  H in a reaction; however, the component equations do not necessarily tell us how the overall reaction occurred. 6.3 Hess’s Law

In the above enthalpy diagram note that:  H 1 =  H 2 +  H Hess’s Law

Example (ex. 6.8 pg. 258): Diborane (B 2 H 6 ) is a highly reactive boron hydride. Calculate  H for the synthesis of diborane from its elements, according to the following equation: 2 B(s) + 3 H 2 (g) --> B 2 H 6 (g) Use the following data to help you: a)2 B (s) + 3/2 O 2(g) --> B 2 O 3 (s)  H = kJ b)B 2 H 6(g) + 3 O 2(g) --> B 2 O 3(s) + 3 H 2 O (g)  H = kJ c)H 2 (g) + 1/2 O 2(g) --> H 2 O (l)  H = -286 kJ d)H 2 O (l) --> H 2 O (g)  H = 44 kJ 6.3 Hess’s Law

6.4 Enthalpies of Formation If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H o f. Degree symbol in  H o f indicates standard conditions… Standard conditions 1 atm and 25 o C (298 K). If there is more than one state for a substance under standard conditions, the more stable one is used. Learn the distinction between standard state and standard conditions---they are not the same!

6.4 Enthalpies of Formation Standard enthalpy of formation of the most stable form of an element is zero.

6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction:

6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction Example 1: (Ex. 6.9)Use the standard enthalpies of formation listed in Table 6.2 to calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.

6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction Example 2: (Ex. 6.10) Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction, which occurs when a mixture of powdered aluminum and iron (III) oxide is ignited with a magnesium fuse: 2 Al(s) + Fe 2 O 3 (s) --> Al 2 O 3 (s) + 2 Fe(s)

End of Chapter 6 Thermochemistry To save time we will not formally cover in class (thus you won’t be tested on it!)