Chapter 6 Energy Thermodynamics 1 Energy is The
- Slides: 29
Chapter 6 Energy Thermodynamics 1
Energy is. . . The ability to do work. n Conserved. n made of heat and work. n a state function. n independent of the path, or how you get from point A to B. n Work is a force acting over a distance. n Heat is energy transferred between objects because of temperature difference. n 2
The universe is divided into two halves. n the system and the surroundings. n The system is the part you are concerned with. n The surroundings are the rest. n Exothermic reactions release energy to the surroundings. n Endo thermic reactions absorb energy from the surroundings. n 3
4 Potential energy Heat
5 Potential energy Heat
Direction n 1. 2. 3. n n 6 Every energy measurement has three parts. A unit ( Joules of calories). A number how many. and a sign to tell direction. negative - exothermic positive- endothermic
Surroundings System Energy DE <0 7
Surroundings System Energy DE >0 8
Same rules for heat and work Heat given off is negative. n Heat absorbed is positive. n Work done by system on surroundings is negative. n Work done on system by surroundings is positive. n Thermodynamics- The study of energy and the changes it undergoes. n 9
First Law of Thermodynamics The energy of the universe is constant. n Law of conservation of energy. n q = heat n w = work n DE = q + w n Take the systems point of view to decide signs. n 10
What is work? Work is a force acting over a distance. n w= F x Dd n P = F/ area n d = V/area n w= (P x area) x D (V/area)= PDV n Work can be calculated by multiplying pressure by the change in volume at constant pressure. n units of liter - atm L-atm n 11
Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. n work is negative n w = - P DV n Expanding work is negative. n Contracting, surroundings do work on the system w is positive. n 1 L atm = 101. 3 J n 12
Examples What amount of work is done when 15 L of gas is expanded to 25 L at 2. 4 atm pressure? n If 2. 36 J of heat are absorbed by the gas above. what is the change in energy? n How much heat would it take to make the same change to the gas without changing the internal energy of the gas? n 13
Enthalpy abbreviated H n H = E + PV (that’s the definition) n at constant pressure. n DH = DE + P DV n n the heat at constant pressure qp can be calculated from n 14 DE = qp + w = qp - PDV n qp = DE + P DV = DH
Calorimetry Measuring heat. n Use a calorimeter. n Two kinds n Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated n C= heat absorbed/ DT = DH/ DT n specific heat capacity = C/mass n 15
Calorimetry molar heat capacity = C/moles n heat = specific heat x m x DT n heat = molar heat x moles x DT n Make the units work and you’ve done the problem right. n A coffee cup calorimeter measures DH. n An insulated cup, full of water. n The specific heat of water is 1 cal/gºC n Heat of reaction= DH = sh x mass x DT n 16
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. n 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? n 17
Calorimetry Constant volume calorimeter is called a bomb calorimeter. n 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. n The heat capacity of the calorimeter is known and tested. n Since DV = 0, PDV = 0, DE = q n 18
Bomb Calorimeter 19 n thermometer n stirrer n full of water n ignition wire n Steel bomb n sample
Properties intensive properties not related to the amount of substance. n density, specific heat, temperature. n Extensive property - does depend on the amount of stuff. n Heat capacity, mass, heat from a reaction. n 20
Hess’s Law Enthalpy is a state function. n It is independent of the path. n We can add equations to to come up with the desired final product, and add the DH n Two rules n If the reaction is reversed the sign of DH is changed n If the reaction is multiplied, so is DH n 21
H (k. J) O 2 NO 2 -112 k. J 180 k. J NO 2 68 k. J N 2 2 O 2 22
Standard Enthalpy The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions) n Symbol DHº n When using Hess’s Law, work by adding the equations up to make it look like the answer. n The other parts will cancel out. n 23
Example n Given DHº= -1300. k. J DHº= -394 k. J DHº= -286 k. J calculate DHº for this reaction 24
Example Given DHº= +77. 9 k. J DHº= +495 k. J DHº= +435. 9 k. J Calculate DHº for this reaction 25
Standard Enthalpies of Formation Hess’s Law is much more useful if you know lots of reactions. n Made 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. n Standard states are 1 atm, 1 M and 25ºC n For an element it is 0 n There is a table in Appendix 4 (pg A 22) n 26
Standard Enthalpies of Formation Need to be able to write the equations. n What is the equation for the formation of NO 2 ? n ½N 2 (g) + O 2 (g) ® NO 2 (g) n Have to make one mole to meet the definition. n Write the equation for the formation of methanol CH 3 OH. n 27
Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. n C 2 H 5 OH +3 O 2(g) ® 2 CO 2 + 3 H 2 O n which leads us to this rule. n 28
Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. n C 2 H 5 OH +3 O 2(g) ® 2 CO 2 + 3 H 2 O n which leads us to this rule. n 29
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