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Energy Thermodynamics

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Presentation on theme: "Energy Thermodynamics"— Presentation transcript:

1 Energy Thermodynamics
Chapter 6 Energy Thermodynamics

2 6.1 Nature of Energy The ability to do work or produce heat.
Conserved - can be converted from one form to another but can neither be created nor destroyed. Work is a force acting over a distance. Potential: due to position or composition - can be converted to work. Kinetic: due to motion of the object. KE = 1/2 mv2 (m = mass, v = velocity)

3 Heat and Temperature Temperature reflects random motion of particles in a substance. Heat is the measure of energy content. Heat is energy transferred between objects because of temperature difference. State Function - property of a system that depends only on its present state. Independent of the path, or how you get from point A to B.

4 The Universe Is divided into two halves, the system and the surroundings. The system is the part you are concerned with. The surroundings are the rest. Exothermic reactions release energy to the surroundings. Endothermic reactions absorb energy from the surroundings.

5 Heat Potential energy

6 Heat Potential energy

7 Direction Every energy measurement has three parts.
A unit ( Joules or calories). A number - how many. A sign to tell direction. Negative - exothermic Positive- endothermic

8 Surroundings System Energy DE <0

9 Surroundings System Energy DE >0

10 Same rules for heat and work
Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is negative. Work done on system by surroundings is positive. Thermodynamics - The study of energy and the changes it undergoes.

11 (YDVD)

12 First Law of Thermodynamics
The energy of the universe is constant. Law of conservation of energy. q = heat w = work DE = q + w Take the systems point of view to decide signs.

13 What is work? Work is a force acting over a distance.
work = force  distance since pressure = force / area, work = pressure  volume Work can be calculated by multiplying pressure by the change in volume at constant pressure. units of L•atm

14 Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. work is negative wsystem = PV Expanding work is negative. Contracting, surroundings do work on the system w is positive. 1 L•atm = J

15 Examples Calculate the E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. EX. 6.1 What amount of work is done when 46L of gas is expanded to 64 L at 15 atm pressure? EX. 6.2 If J of heat are absorbed by the gas above. What is the change in energy? 1 L•atm = J

16 6.2 Enthalpy Where qP = H at constant pressure. Abbreviated H
H = E + PV (that’s the definition), at constant pressure. DH = DE + PDV the heat at constant pressure qp can be calculated from: DE = qp + w = qp - PDV qp = DE + P DV = DH Where qP = H at constant pressure. H = energy flow as heat (at constant pressure).

17 Examples When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is released as heat. Calculate the DH for a process in which 5.8 g sample of methane is burned at a constant pressure. EX 6.4 Consider the following reaction: 2H2(g) O2(g)  2H2O(l) DH=-572kJ How much heat is evolved when 2.56 g of hydrogen is reacted with excess oxygen? #34b



20 Calorimetry specific heat capacity
Measuring heat. We use a calorimeter. The heat capacity for a material, C, is calculated. C = heat absorbed/DT = DH/ DT specific heat capacity heat capacity per gram = J/°C•g or J/K•g molar heat capacity heat capacity per mole = J/°C•mol or J/K•mol

21 Calorimetry Constant pressure calorimeter (coffee cup calorimeter, used for solutions). heat = specific heat x m x DT heat = molar heat x moles x DT Make the units work and you’ve done the problem right. A coffee cup calorimeter measures DH. The specific heat of water is 1 cal/gºC (4.184 J/gºC) Heat of reaction= DH = s x mass x DT

22 Examples The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

23 Calorimetry Constant volume calorimeter is called a bomb calorimeter.
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. The heat capacity of the calorimeter is known and tested. Since DV = 0, PDV = 0, DE = q

24 Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb sample

25 Properties Intensive properties - not related to the amount of substance. Ex. density, specific heat, temperature. Extensive property - does depend on the amount of stuff. Ex. heat capacity, mass, heat from a reaction.

26 6.3 Hess’s Law Enthalpy is a state function.
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps. We can add equations to to come up with the desired final product, and add the DH. Two rules: If the reaction is reversed the sign of DH is changed. If the reaction is multiplied, so is DH.

27 Rules 1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g) H = 180 kJ 2NO(g)  N2(g) + O2(g) H = 180 kJ 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g)  3N2(g) + 3O2(g) H = 540 kJ

28 Examples When using Hess’s Law, work by adding the equations up to make it look like the answer. Make the other compounds cancel out. N2(g) + 2O2(g)  2NO2(g) H1 =68kJ Above reaction is carried out in two steps below: N2(g) + O2(g)  2NO(g) H2 =180kJ 2NO(g) + O2(g)  2NO2(g) H3 =-112kJ

29 2NO, O2 -112 kJ H (kJ) 180 kJ 2NO2 68 kJ N2 O2

30 Practice Given(BDVD) DHº= -1300. kJ DHº= -394 kJ DHº= -286 kJ
Calculate DHº for this reaction

31 Practice Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction

32 Practice P4(s) + 6Cl2(g)  4PCl3(g) H = -1225.6kJ
P4(s) + 5O2(g)  P4O10(s) H = kJ PCl3(g) + Cl2(g)  PCl5(g) H = kJ PCl3(g) + 1/2O2(g)  Cl3PO(g) H = kJ Calculate the H for the reaction #58 P4O10(s) + 6PCl5(g)  10Cl3PO(g)

33 6.4 Standard Enthalpies of Formation
Standard States Compound For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.

34 Standard Enthalpy of Formation
The enthalpy change that occurs in the formation of one mole of a compound for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions). Symbol DHºf There is a table in Appendix 4 (pg A21) It is a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.

35 Standard Enthalpies of Formation
Need to be able to write the equations. What is the equation for the formation of NO2 ? ½N2 (g) + O2 (g) ® NO2 (g) Have to make one mole to meet the definition. Write the equation for the combustion of methanol CH3OH. What is the equation for the formation of solid aluminum oxide?

36 Since we can manipulate the equations
We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. Ex.6.11 CH3OH(l) + 3O2(g) ® 2CO2 (g) + 3H2O(l)

37 2Al(s) + Fe2O3(s) --> Al2O3(s) + 2Fe(s)
Thermite Reaction Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction.Ex 6.10 2Al(s) + Fe2O3(s) --> Al2O3(s) + 2Fe(s) (YDVD)

38 It’s a gas! Methanol (CH3OH) is often used as a fuel in high performance engines in race cars. Using the data in table 6.2, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is pure liquid octane (C8H18)

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