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1 Chapter 6 EnergyThermodynamics 2 Energy is... n …the ability to do work or produce heat. n …conserved. n …defined as kinetic or potential. n …a state.

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Presentation on theme: "1 Chapter 6 EnergyThermodynamics 2 Energy is... n …the ability to do work or produce heat. n …conserved. n …defined as kinetic or potential. n …a state."— Presentation transcript:

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2 1 Chapter 6 EnergyThermodynamics

3 2 Energy is... n …the ability to do work or produce heat. n …conserved. n …defined as kinetic or potential. n …a state function (independent of the path, or how you get from point A to B...depends only on its present state ). n Work is a force acting over a distance. n Heat is energy transferred between objects because of temperature difference.

4 3 The universe… n …is divided into two halves. n …the system and the surroundings. n The system is the part you are concerned with (as a chemist). n The surroundings are the rest. n Exothermic reactions release energy to the surroundings. n Endothermic reactions absorb energy from the surroundings.

5 4 Potential energy Heat

6 5 Potential energy Heat

7 6 Direction n Every energy measurement has three parts: 1. A unit ( Joules, kJ or Calories). 2. A number. 3. A sign to tell direction of energy flow. n negative - exothermic n Positive - endothermic

8 7 System Surroundings Energy E <0 Energy is leaving the system!

9 8 System Surroundings Energy E >0 Energy is entering the system!

10 9 Same rules for heat and work n 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.

11 10 First Law of Thermodynamics n The energy of the universe is constant. n Law of conservation of energy. n q = heat and w = work E = q + w (Internal Energy = E) E = q + w (Internal Energy = E) E is the sum of the KE and PE of all particles in a system…which can be changed by a flow of work and/or heat! E is the sum of the KE and PE of all particles in a system…which can be changed by a flow of work and/or heat! n Take the systems point of view to decide signs.

12 11 What is work? n Work is a force acting over a distance. w= F x d w= F x d n P = F/ area n Typically, we will let the physicist worry about the work!

13 12 Work needs a sign… n If the volume of a gas increases, the system has done work on the surroundings (it pushed against the surroundings). n work is negative (or lost to the surr.) w = - P V w = - P V n Expanding work is negative. n Contracting, surroundings do work on the system…w is positive.

14 13 Enthalpy (for nerds) n abbreviated H n H = E + PV (thats the definition) n Calculated at a constant pressure. H = E + P V H = E + P V n the heat at constant pressure q p can be calculated from E = q p + w = q p - P V E = q p + w = q p - P V q p = E + P V = H q p = E + P V = H

15 14 Enthalpy for Chemists Since q p = E + P V = H…then we can say q p = H Since q p = E + P V = H…then we can say q p = H n In a nutshell: at constant pressure, the change in enthalpy of the system is equal to the energy flow as heat. Practical applications: the change in enthalpy is given by the equation H = H products - H reactants Practical applications: the change in enthalpy is given by the equation H = H products - H reactants

16 15 Calorimetry n Science of measuring heat. n Use a calorimeter: Two types… n #1: Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated C= heat absorbed/ T = H/ T C= heat absorbed/ T = H/ T n specific heat capacity = C/mass (units J/ºCg or J/Kg)

17 16 Calorimetry n molar heat capacity = C/moles heat = specific heat x m x T heat = specific heat x m x T heat = molar heat x moles x T heat = molar heat x moles x T n Make the units work and youve done the problem right. A coffee cup calorimeter measures H. A coffee cup calorimeter measures H. n An insulated cup, full of water. n The specific heat of water is 1 cal/gºC Heat of reaction= H = Cp x mass x T Heat of reaction= H = Cp x mass x T

18 17 Examples n 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?

19 18 Calorimetry n #2 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. Since V = 0, P V = 0, E = q Since V = 0, P V = 0, E = q

20 19 Bomb Calorimeter n thermometer n stirrer n full of water n ignition wire n Steel bomb n sample

21 20 Properties n Intensive property - not related to the amount of substance. n density, specific heat, temperature. n Extensive property - does depend on the amount of substance. n Heat capacity, mass, heat from a reaction.

22 21 Hesss Law n Enthalpy is a state function (independent of the path). We can add equations to come up with the desired final product, and then add the H values to arrive at the final enthalpy change! We can add equations to come up with the desired final product, and then add the H values to arrive at the final enthalpy change! n Two rules #1 - If the reaction is reversed the sign of H is changed #2 - If the reaction is multiplied, so is H

23 22 State Function n One two-step pathway… N 2 + O 2 2NOΔH = 180 kJ 2NO + O 2 2NO 2 ΔH = -112 kJ n A different single-step pathway… N 2 + 2O 2 2NO 2 ΔH = 68 kJ Regardless of the pathway…68 kJ are absorbed in this endothermic reaction!

24 23 N2N2 +2O 2 O2O2 +2NO 68 kJ 2NO kJ -112 kJ H (kJ) N2N2 +2O 2 Two-step One-step

25 24 Standard Enthalpy n The enthalpy change for a reaction under standard conditions (25ºC, 1 atm, 1 M solutions) Symbol Hº Symbol Hº n When using Hesss Law, work by adding the equations up to obtain the target equation. n The other species will cancel out…and you will sum the enthalpy values!

26 25 Procedure for solving… n Goal: Establish a pathway to the Target Equation by rearranging the given equations n Hints: –Work backwards from the target equation –Look for unique formulas (which only appear once) –Avoid formulas which appear multiple times until you get started

27 26 Hº= -394 kJ Hº= -286 kJ Hesss Law Example n Given calculate Hº for this reaction (target equation) calculate Hº for this reaction (target equation) Hº= kJ

28 27 Standard Enthalpies of Formation n Hesss Law is useful when you are given multiple sets of reactions…. However, the standard enthalpy of formation ( H f º) for a compound is defined as However, the standard enthalpy of formation ( H f º) for a compound is defined as –the change in enthalpy that accompanies the formation of one mole of a compound from its elements in their standard states.

29 28 Standard Enthalpies of Formation n Standard states are 1 atm, 1M and 25ºC n See p. 246 of our text. n For an element it is 0 (zero)! –Why? It comes to us already formed! n There is a table of values listed in Appendix 4 (pg A19-22).

30 29 Standard Enthalpies of Formation n First, you need to be able to write the balanced equation from the elements. n What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g) NO 2 (g) 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.

31 30 Since we can manipulate the equations… n We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O n which leads us to this rule:

32 31 Since we can manipulate the equations… n We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O n which leads us to this rule.

33 32 Solve the problem: C 2 H 5 OH(l) +3O 2 (g) 2CO 2 (g) + 3H 2 O(l) ΔHf: (kJ/mol) ΔH = [( kJ)+(3-286kJ)]-[-278kJ + 30kJ] ΔH = -1367kJ of energy (released) per mole of ethanol burned.

34 33 Chapter 6 problems: n


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