1. How heat Is Measured

2. Some specific heat capacity of substances at 25 0 C Substance Specific heat (J /g. 0 C) Water 4.184 Aluminum 0.900 Copper 0.380 Glass 0.753 Iron 0.444 Mercury 0.1389 Nitogen(N2) 1.0376 Silver 0.240 Steam 1.87 Table salt 0.875 Sand 0.787

3. Sample Problem 1  A sample of 50g of water was placed in a coffee cup calorimeter at a temperature of 20 0 C. A substance was dissolve in water lowering the temperature to 10 0 C. Calculate the energy exchange, Q, for the water. Specific heat of water = 4.18J/g. 0 C Given: m= 50g T f = 10 0 C T i = 20 0 C S = 4.184J/g. 0 C Find: Q Solution : Q = m s ( T) 50g ( 4.184J) ( -10 0 C) g. C Q = -2092J

4. Specific heat (s)  heat required to raise one gram of a substance by one degree celcius The transfer of heat from an object depends on an object’s mass, specific heat and the difference in temperature between the object and the surrounding: Mathematically: Q = m s T Where: Q  amount of heat m  mass s  specific heat T  change in temperature ( T f - T i )

5. Sample 2  How much heat is needed to raise the temperature of 5kg of water from 10 0 C to 35 0 C? Given: mass: 5 Kg x 1000g = 5000g 1 Kg T = ( Tf – Ti) = ( 35 – 10) = 25 0 C S water = 4.18 J/g. 0 C Find: Q Solution: Q = m s T = 5000g ( 4.18J/g. 0 C) ( 25 0 C) = 522500J

6. Sample #3 A 40g sample of alloy ( a mixture of metals) is heated to 90.80 0 C and then placed in water, where it cools to 23.54 0 C. The amount of heat lost by the alloy is 865J. What is the specific heat of the alloy? Given: m = 40g T = ( 23.54 0 C – 90.80 0 C) Q = 865J = -67.26 Find: s Solution : Q = m s T s = Q = 865J = 0.32J/g. 0 C m T 40g( -67.26 0 C)

7. Enthalpy

8. Enthalpy (H)  Heat content of the system Change in enthalpy ( H )  The heat transferred between the system and the surrounding under constant pressure  Also represents the difference between the enthalpy of the system before and after the process and is represented as: H = H final - H initial Note: If H f > H i H is positive ( endothermic) If H f < H i H is negative( exothermic)

9. Enthalpies of Reaction  Enthalpy change that accompanies a reaction  the sum of the enthalpies of the products of the reaction minus the sum of the enthalpies of the reactants in a reaction. The equation is given as H rxn =Σ H products - Σ H reactants Example of a thermochemical equation 2Na (s) + Cl 2(g) -  2NaCl (s) H = -822.30kJ What does the H tell you about the reaction?  The reaction release heat (exothermic)

10. Ex: of a thermochemical equation 2Na (s) + Cl2 (g) -  2NaCl (s) H = -822.30kJ What does the H tell you about the reaction?  The reaction release heat (exothermic) What happens to the enthalpy of the system in an exothermic reaction? Why?  Decreases because the sum of the enthalpies of the products is smaller than the enthalpies of the reactants

11. Example N 2(g) + O 2(g)  2 NO 2 H = + 67.8kJ What does the H tell you about the reaction?  The reaction absorbed heat Endothermic The enthalpy of the system in an endothermic reaction increases because the sum of the enthalpies of the products is greater than the enthalpies of the reactants

12. Sample Problem  When 1 mol of methane( CH 4 ) is burned at constant pressure, 890kJ of energy is released as heat. Calculate the H of the process in which 5.8g sample of methane is burned at constant pressure. Given: q = -890kJ m = 5.8g Solution: 5.8gCH 4 x molar mass CH 4 x q 5.8gCH 4 x 1molCH 4 x -890kJ = -320kJ 16.0gCH 4 1molCH 4

13. Self-Check10.5 p. 302 The reaction that occurs in the heat packs to treat sports injuries is 4 Fe(s) + 3O 2 (g) ---> 2Fe 2 O 3 (s) H= -1652kJ How much heat is released when 1.00 g Fe(s) is reacted with excess O 2 ? Solution: molar mass Fe = 55.85g/mol 1.00g Fe x 1mol Fe = 1.79x 10 -2 mol Fe 55.85gFe 1.79x 10 -2 mol Fe x -1652kJ = -7.39kJ 4 molFe

15. Objective: Explain the energy involve during the phase change Change of phase  Occurs when a substance change from one state to another Note: Change of phase always occur with a change in the amount of heat of a substance Heat travels spontaneously from a warmer body to a colder body

16. What happens during phase change?  During phase change, heat either comes out of the material or goes into the material but its temperature remains the same Terms for different phase changes Solid to liquid  melting (fusion) Solid to gas  sublimation Liquid to solid  freezing (solidification) Liquid to gas  vaporization Gas to liquid  condensation Gas to solid  deposition

17. vaporization condensation sublimation deposition melting freezing (fusion) (solidification) SOLID LIQUID GAS

18.  The amount of heat absorbed by one mole of a substance when it changes from solid to liquid  The quantity of heat absorbed during fusion is equal to the heat released during freezing

19.  The amount of heat released by one mole of a substance to change from liquid to solid.

20. Molar heat of vaporization  Amount of heat required by one mole of a substance to change from liquid to gas

21. Molar heat of condensation  Amount of heat released by one mole of a substance to change from gas to liquid

22. Explain the kinetic energy of the water molecule in the heating curve  When all the ice has turned to liquid and heat continues to flow into the water, the heat goes into increasing the kinetic energy of the molecules resulting in an increase in temperature

25.  a process that occurs by itself. Which means no external force is needed to continue the process Ex: Sugar dissolving in hot coffee rusting of iron

26.  is a process that occurs when an external force is continuously applied Ex: riding in a playground swing electrolysis of water

27. Note:  Most exothermic reactions are spontaneous  Most endothermic reactions are non spontaneous However there are some endothermic reactions that are nonspontaneous Ex: the melting of ice at room temperature or above 0 0 C

28. Consider the melting of ice cube  Melting of an ice cube is spontaneous if the temperature is above the melting point  Nonspontaneous is below the melting point

29.  Degree of freedom that particles have  Degree of disorder in a system  The greater the degree of randomness or disorder, in a system, the greater is its entropy Which has a greater entropy liquid or gas?  Gas has greater entropy than pure liquids  Entropy is measured by J/K

30. How to predict the S using the rules Ex: C 6 H 12 O 6 (s) ---> 2C 2 H 5 OH(l) + 2CO 2 (g) What is the sign of the S for the above reaction?  positive because the entropy of the final product is higher Sometimes predicting the sign of S is not possible Ex: CO(g) + H 2 O(g) ---> CO 2 (g) + H 2 (g) Predicting S in this case is impossible

31.  Symbol ( S)  Is the difference between the final entropy and the initial entropy of a system. S = S final - S initial  S is positive if there is an entropy increase Ex: solid ---> liquid or gas  S is negative if there is an entropy decrease Ex: when a precipitate for Ag + (aq) + Cl - (aq)  AgCl(s) The entropy of the solid is lower than the original aqueous ions

32. Rules to predict the sign of S  Generally entropy increases when  A reaction breaks up a larger molecule into smaller molecular fragments  A reaction occurs in which there is an increase in the mole of gas in the product  A process where a solid changes to a liquid or gas or a liquid to a gas

33. Exercises  2NH 4 NO 3 (s) --->2N 2 (g) + 4H 2 O(g) + O 2 (g)  2SO 2 (g) + O 2 (g) --> 2SO 3 (g)  C 12 H 22 O 11 (aq)  C 12 H 22 O 11 (s)  CO 2 (g) + H 2 0(g) ---> CO 2 (g) + H 2 (g)

34. Hess Law  States that if the reaction occurs in two or more steps, the enthalpy for the reaction is the sum of the enthalpies of the individual steps  Why is it useful? because it allows us to calculate heats of reaction that are difficult or inconvenient to measure in a calorimeter

37. Gibbs Free Energy  thermodynamic entity that takes into account both entropy and enthalpy  Useful for determining whether a process is spontaneous or not  Defined as G = H – T S Where: H  enthalpy T  temperature S  entropy

38. For a process at constant temperature the change in free energy is G = H - T S G -> represents the maximum energy that is free to do useful work to the surrounding. - G  means that the system does useful work to the surrounding  means spontaneous process Ex: movement of water in a water fall is spontaneous Why? Because it does a useful work in turning the wheels of a turbine

39. + G  indicates that work has to be done on the system for the process to take place  means that energy will have to be absorbed from the surrounding  means non spontaneous process Ex: maintenance of life It requires sustained input of work or energy that we get from food If G = 0  The reaction is at equilibrium

40. Example CoCl2(g)  CO(g) = Cl2(g) 1.Calculate the delta G at 127 0 C. 2.Calculate the temperature when the above reaction is at equilibrium Solution: G = H – T S Solve for H using standard enthalpies of formation (See Thermodynamics Table) H = [ sum H fproducts ] – [ sum H f reactants ] = [ 1mol x (-110.5kJ) + 1mol ( 0kJ) ] – [ 1mol x 220kJ ] mol mol mol = -110 kJ – (-220kJ) H= + 110kJ

41.  Determine delta S for the reaction using standard molar entropies and Hess Law of Summation ( See Thermodynamics table) S = [ sum S products ] – [ sum S reactants ] = [1 mol x 197.5 J + 1 mol x 223J] – [1 mol x 283.7 J] mol.K mol.K mol.K = 420.5J/K - 283.7J/K = 136.8J/K ( convert this to kJ) 136.8J/K x 1kJ = 0.1368kJ/K 1000J S = 0.1368kJ/K Solve G G = 110.5kJ – 400K ( 0.1368kJ/K G =+55.78kJ

42. 2) When delta G = 0, solve T Solution G = H - T S 0 = 110.5kJ – T ( 0.1368kJ/K) (0.1368kJ/K)T = 110.5kJ T = 110.5kJ 0.1368kJ/K T = 807.7K

43. Calculating Energy Changes:Solid to Liquid ( Chapter14: Solids and Liquids) Calculate the energy required to melt 8.5g of ice at 0 0 C. The molar heat of fusion of ice is 6.02kJ/mole Solution: 8.5gH2O x 1mol H2O x 6.02kJ = 2.8kJ 18gH2O 1molH2O

44. Calculating Energy changes( Liquid to Gas)  Calculate the energy in kJ to heat 25g of liquid water from 25 o c to 100 o c and change it to steam at 100 o C. The specific heat capacity of liquid water is 4.184J/g o C and the molar heat of vaporization of water is 40.6kJ/mol.  Solution: Q = ms T (25g)(75 o C)(4.184J) = 7.8x10 3 J x 1kJ = 7.8kJ(to boil) g o C 1000J 25gH2O x 1molH2O =1.4mol H2O x 40.6kJ = 57kJ( vaporization) 18gH2O 1mol H2O The total energy is the sum of the two steps 7.8kJ + 57kJ = 65kJ

45. Heating and Cooling Curve

46.  Fig 20.3 Heating curve for water 125 F. 100 D. E. E-FWater vapor 75 Liquid-gas C-D 50 B-C- melting 25.B.C 0 -25 A.

47. Cooling Curve  Describes the change in temperature and the amount of heat during the cooling process  The slope of the curve is decreasing when heat is removed from the material

48.  Fig 20.4 Cooling Curve of water 1 2 3 4 5 100 Liquid 75 Gas 50 Gas-liquid 25 Liquid-solid solid 0 H o Vap -25 H o fus Stage1:Gaseous water cools. Stage 2: Gaseous water condenses Stage3: Liquid water cools. Stage4: Liquid water freezes Stage5: solid water cools