Chapter 5 Thermochemistry

Transcription

1 Chapter 5 Thermochemistry Learning Outcomes: Interconvert energy units Distinguish between the system and the surroundings in thermodynamics Calculate internal energy from heat and work and state sign conventions of these quantities Explain the concept of a state function and give examples Calculate ΔH from ΔE and PΔV Relate q p to ΔH and indicate how the signs of q and ΔH relate to whether a process is exothermic or endothermic Use thermochemical equations to relate the amount of heat energy transferred in reactions in reactions at constant pressure (ΔH) to the amount of substance involved in the reaction Work Energy used to cause an object that has mass to move. w = F d Heat Energy used to cause the temperature of an object to increase. 1

2 Energy -the capacity to do work or transfer heat. 1 E k = mv 2 2 kg m 2 1 J = 1 s 2 Potential Energy, E p: Energy an object possesses by virtue of its position or chemical composition. Kinetic Energy, E k - Energy an object possesses by virtue of its motion. calorie (cal), 1 cal = J 1 kcal = 1 Calorie (Cal) Electrostatic potential energy 2

3 System and Surroundings The system includes the molecules of interest. The surroundings are everything else. In thermochemistry we study the exchange of energy between the system and surroundings. open system matter and energy can be exchanged with the surroundings closed system - exchange energy--but not matter--with the surroundings. isolated system - neither matter nor energy may be exchanged with surroundings. Transferal of Energy a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. 3

4 Transferal of Energy a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. b) As the ball falls, its potential energy is converted to kinetic energy. Transferal of Energy a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. b) As the ball falls, its potential energy is converted to kinetic energy. c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat. 4

5 Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. E = E final E initial System absorbed energy from the surroundings System released energy to the surroundings 5

6 Exchange of Heat between System and Surroundings 6

7 State Functions The internal energy of a system is independent of the path by which the system achieved that state. Internal energy, E, a state function. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its E is the same. But q and w are different in the two cases. 7

8 w = P V Work 1 L atm = J Example Calculation Calculate the work (in J) associated with the expansion of a gas from 44 ml to 63 ml at a constant pressure of 14 atm. 8

9 Enthalpy Enthalpy is a thermodynamic function equal to the internal energy plus pressure volume: H = E + PV When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + P V Since E = q + w and w = P V, we can substitute these into the enthalpy expression: H = E + P V H = (q+w) w H = q The enthalpy change, H, is defined as the heat gained or lost by the system under constant pressure. H = q p Properties of Enthalpy 1. Enthalpy is a state function. 2. Enthalpy is an extensive property. 3. Enthalpy is reversible. 4. H for a reaction depends on the state of the products and the state of the reactants. 9

10 Endothermic and Exothermic H = H final H initial or H = H products H reactants A process is endothermic when H is positive (>0). A process is exothermic when H is negative (<0). 10

11 Enthalpies of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. A thermochemical equation is an equation for which H is given: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) H = kj H 2 (g) + ½ O 2 (g) H 2 O(l) H = kj The enthalpy changes assume the coefficients are moles of the substances Calorimetry Calorimetry, the measurement of heat released or absorbed by a chemical reaction. A calorimeter is the device used to measure heat The quantity of heat transferred by the reaction causes a change in temperature of the solution. 11

12 Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1 C) is its heat capacity (C in units of J/K). q C = T We define specific heat capacity (or simply specific heat; C s or s in units of J/gK) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K. C s = s = q m T Constant Pressure Calorimetry Indirectly measure the heat change for the system Because the specific heat for water is well known (4.184 J/mol- K), we can measure q for the reaction with this equation: q = m C s T 12

13 Calorimetry Calculations The calorimeter and its contents are the surroundings, so q soln is found from the mass, heat capacity, and temperature change (q = msδt) q soln = -q rxn q soln = (specific heat of solution) (grams of solution) ΔT = -q rxn Example Calculation A metal pellet with mass g, originally at 88.4 C, is dropped into 125 g of water originally at 25.1 C. The final temperature of both the pellet and the water is 31.3 C. Calculate the heat capacity C (in J/ C) and specific heat capacity C s (in J/g C) of the pellet. The specific heat of water is J/g C. 13

14 Example Calculation When 200. g of a AgNO 3 solution mixes with 150. g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34 o C. Assume 350. g of solution and a heat capacity of J/g o C. Calculate H for the equation Ag + (aq) + I - (aq) AgI(s) Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is small. 14

15 Hess s Law Most ΔH values are labeled ΔH o, and measured under standard conditions P = 1 atm (but for gases P = 1 bar) T = usually K (25.0 o C) Concentration = 1 mol/l Hess s law states when two or more thermochemical equations are added, the enthalpy change of the resulting equation is the sum of those for the added equations. C(s) + O 2 (g) CO 2 (g) H = kj CO 2 (g) CO(g) + ½O 2 (g) H = kj C(s) + ½O 2 (g) CO(g) H = kj Hess s Law Hess s law states that If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. H is a state function 15

16 Hess s Law Example Given the thermochemical equations 2WO 2 (s) + O 2 (g) 2WO 3 (s) H = -506 kj 2W(s) + 3O 2 (g) 2WO 3 (s) H = kj calculate the enthalpy change for: 2W(s) + 2O 2 (g) 2WO 2 (s) Enthalpies of Formation An enthalpy of formation, H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Standard Enthalpy of Formation Only one enthalpy value is needed for each substance, called the standard enthalpy of formation The standard enthalpy of formation is the enthalpy change when one mole of a substance in its standard state is formed from the most stable form of the elements in their standard states 16

17 Standard State Enthalpy changes depend on the temperature and pressure at which they are measured When applying Hess s law, all values must refer to the same conditions of pressure and temperature The standard state of a substance at a specified temperature is the pure form at 1 atm pressure Tabulated values for enthalpy refer to the standard state, usually at a temperature of 25 o C Standard Enthalpy of Formation The symbol used for standard enthalpy of formation is H fo, where the o designates standard state. The product is always one mole of a single substance. The standard enthalpy of formation of the elements in their most stable form is zero. Some examples of standard enthalpies of formation: C(graphite) + O 2 (g) CO 2 (g) H 2 (g) + ½O 2 (g) H 2 O(l) H fo [CO 2 (g)] H fo [H 2 O(l)] 2Na(s) + Se(s) + 2O 2 (g) Na 2 SeO 4 (s) H fo [Na 2 SeO 4 (s)] H 2 (g) H 2 (g) H fo [H 2 (g)] = 0 17

18 Standard Enthalpies of Formation Standard enthalpies of formation, H fo, are measured under standard conditions (25 C and 1.00 atm pressure). Imagine this as occurring in 3 steps: Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. 18

19 Imagine this as occurring in 3 steps: Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. Imagine this as occurring in 3 steps: Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. 19

20 Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) The sum of these equations is: C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. H rxn0 = [(3 mol)( kj/mol) + (4 mol)( kj/mol)] - [(1 mol)( kj/mol) + (5 mol)(0 kj)] = [( kj) + ( kj)] - [( kj) + (0 kj)] = ( kj) - ( kj) = kj 20

21 Example Calculation: Enthalpy of reaction Use standard enthalpies of formation to calculate the enthalpy change for the reaction: P 4 O 10 (s) + 6H 2 O(g) 4H 3 PO 4 (s) Substance H fo (kj/mol) P 4 O 10 (s) H 2 O(g) H 3 PO 4 (s) H = n H f(products) - m H f(reactants) where n and m are the stoichiometric coefficients. Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats. 21

22 Fuels The vast majority of the energy consumed in this country comes from fossil fuels. 22