October 13, 2011

Lesson # 6 ch 6 Thermochemistry

Energy is the essence of our very existence as individuals and as a society. Energy is the capacity to do work or to produce heat.

Law of Conservation of energy - energy can be converted from one form to another but can be neither created nor destroyedPotential energy - energy due to position or composition

Kinetic energy - energy due to motion directly proportional to mass of object and to square of velocity

Chemical energyFor chemistry the system is the part of universe that is the reactants and the products. The surroundings is everything else in the universe ( reaction container, the room etc)

When energy flows out of the system to the surroundings this is considered an exothermic reaction

When energy flows into the system from the surroundings the reaction is consider an endothermic reaction.

In an exothermic reaction some of the potential energy stored in the chemical bonds is being converted into thermal energy(random kinetic energy) via heat.

The study of energy and its interconversions is called thermodynamics. The law of conservation of energy is often called the first law of thermodynamics and is stated as follows: The energy of the universe is constant.

The internal energy E of a system can be defined most precisely as the sum of the kinetic and potential energies of all the 'particles' in the system. The internal energy of a system can be changed by a flow of work, heat, or both. △E = q + w where △E represents the change in the system's internal energy, q represents heat, and w represents work.

October 13, 2011

Enthalpy = H + E =PV where E is the internal energy of the system, P is the pressure of the system, and V is the volume of the system.At constant pressure (where only PV work is allowed), the change in enthalpy △H of the system is equal to the energy flow as heat. Heat of reaction and and change in enthalpy are used interchangeably.

For a chemical reaction, the enthalpy change is given by the equation △H = Hproducts - Hreactants

Calorimetry is the science of measuring heat.

heat capacity C = heat absorbed increase in temperature specific heat capacity = heat capacity per gram (J/oC. g or J/K.g

Hess's law -the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(1) If a reaction is reversed, the sign of △H is also reversed.(2) The magnitude of △H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of △H is multiplied by the same integer.

The enthalpy changes for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants FROM the enthalpies of formation of the products.

27. Piston cylinder volume 40cm3 950J released against pressure of 650 torr(0.855 atm)Calculate new volume 950J/101.3J per atm .L = 9.38 atm . L

w = P∆V -9.38 atm . L =-650/760 atm x (Vf - 0.040L) Vf - 0.040L= 11.0 L V f = 11.0 - 0.40= 11.0L

October 13, 2011

EXOTHERMIC REACTION (see page 231)

CH4(g) + 2O2(g) ➝ CO2(g) + 2H2O (g) + Energy (heat)

pote

ntia

l ene

rgy

System1 mol CH42mol O2Reactants

3

∆(PE)

1 mol CO22 mol H2O

Products

Surrroundings

Energy released to surroundings as heat

reactants Products

October 13, 2011

ENDOTHERMIC REACTION (see page 232)

N2(g) + O2(g) + Energy (heat) ➝ 2NO(g)

pote

ntia

l ene

rgy

System

1 mol N2 1mol O2Reactants

3

∆(PE)

2 mol NO

ProductsSurrroundings

Heat absorbed FROM surroundings

Reactants Products

October 13, 2011

Problems. 17. Kinetic energy of baseball(mass of 5.25oz) at velocity of 1.00 x 102mi/hr.KE = mV2/2 5.25oz/16oz per pound xkg/2.2046 pounds x( (100 mi x 1.6093 km/mi x 1000m/km)/3600 sec))2/2=149 J

25. ideal gas 15.0 atm and 10.0L compressed at 2.0 atm constant external pressure and with constant temperatureCalculate work in units of kJ15.0 atm x 10.0 L =2.00atm x V V=75L ∆V = 65LW = P ∆V 65 x 2= 130 atm . L x 101.3J/atm .L = 13.2 kJ

27. Air and gasoline vapor in a cylinder with a piston. Original volume is 40cm Combustion releases 950 J What is new volume? constant pressure of 650 torr All energy is work to push piston Work = - p∆V 950 J = 650 torr x 1 atm x ∆V (Vf -Vo) 760 torr

950 J = 0.855 ∆V ∆V = 1,110cm3

40cm3

new volume 1150cm3

October 13, 2011

29. Balloon filled with 39.1 mol Helium with volume of 876L at 0.0oC and 1.00 atm Temp increased to 38.0oC and expands to volume of 998L pressure constantCalculate q, w and ∆E (molar heat of He gas is 20.8J/C . molq = molar heat capacity x mol x ∆T= 20.8J/C . mol x 39.1 mol x (38.0 -0.0) =30900J= 30.9w= P∆V w= 1 x 122= -122 atm . L x101.3J/ atm . L = -12,400J = -12.4 kJ P= nRT/V 39.1 x 0.08206 x 273 / 876 =1.00 atm∆E = q+w = 30.9 + -12.4 = 18.5kJ

37. Combustion of propane C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(l) ∆H = -2221kJUsing sample exercise 6.3 assuming all propane furnishes energy at 60% efficiency what mass of propane is needed?

1.3 x 108J needed x 100/60 efficiency= 2.2 x 108J from burning propane2.2 x 108 J /2221000J/mol =100mol x 44.064g/ mol = 4400g propane

41. 4.18J/C.g x 22oC x 25g=2300J for H2OHg 550g x ∆T x 0.14 = 10700J = ∆T =10700/(0.14 x 550) = 139oC

43. Q = s x m x ∆Tm= 5.00g ∆T= 55.1-25.2 Q= 133J s=Q/m x ∆T 133/5.00 x 29.9 = 0.89 = Al

45. 4.18 x 30 x (Tf -280) = (330 -Tf) 50 x 4.18 -35112 +125.4 Tf = 68970 -209Tf

334.4 Tf = 104082 Tf = 311oC

39. the 2.5 kj/ mol represents the work done by the expansion brought on by vaporization.∆H = ∆E + ∆(PV)

42. a. Q = s x m x ∆t = 0.24 x 150.0 x 25 = 900J b. molar heat capacity of Ag = 0.24J/oC. g x 107.9g/mol = 26 J/mol c. m= Q/s x ∆t = 1.25kJ/ 0.24 x 3.2 = 1.628kg Ag

October 13, 2011

47. 5.00 g sample of Al pellets(spec heat capacity 0.89)and 10.00g sample of Fe pellets(specific heat capacity 0.45) heated to 100.0oC. Then dropped into 97.3 g water at 22.0 oCCalculate the final temperature

[5.00 x 0.89 x (100 -Tf)] +[10.00 x 0.45 x (100 - Tf)[ = [4.18 x 97.3 x(Tf -22)]445 -4.45Tf + 450 -4.5Tf = 407Tf - 8948

416Tf = 9843 Tf =23.7oC

49. 150.0 g sample of metal at 75.0oC added to 150.0 g of H2O at 15.0oC Water temperature rises to 18.3oC Calculate specific heat capacity of metal.s x 150.0 x (75.0 - 18.3) = 150.0 x 4.18J x 3.3s x 56.7 = 13.8s= .24 specific heat capacity of metal

51. calorimeter has two solutions 0.100M AgNO 3 50.0 ml and 50.0 ml 0.100M HCl original temp = 22.60oC final temp 23.40oC mass of 100.0 g specific heat capacity of 4.18Calculate the heat accompanying the reactionAg+(aq) + Cl-(aq) → AgCl(s)Q = s x m x ∆T = 4.18 x 100 x 0.80 =334J mol = 0.1M x 0.05L =0.00500 mol 334J/0.00500 =66800J/mol or 67 kJ/mol

53. Consider the dissolution of CaCl2 CaCl2(s) → Ca2+(aq) + 2Cl-(aq) ∆H = -815kJ

11.0 g sample of CaCl2 dissolved in 125 g water temp at 25.0oCCalculate final temp 11.0g /(40.08 +70.90) =0.099 mol of CaCl20.099 x -81.5kJ =-8.07kJ = 8070J/(125g+ 11.0g)=59.34/4.18=14.2∆TFinal temperature = 39.2oC

55. heat capacity of bomb calorimeter? if 6.79g of methane (energy of combustion = -802kJ/mol) Temp changed 10.8oCa.[ -802 x ( 6.79g/16.033g per mol)] /10.8 = heat capacity of bomb calorimeter 31.5kJ

b. c 12.6 g sample of acetylene C2H2 increased temp by 16.9oCWhat is energy of combustion of C2H2?E combustion = heat capa. of bomb calorimeter x ∆T/(g C2H2/molar massC2H2E combustion = (31.5kJ x 16.9)/(12.6/26.018) = -1.10 x 103 kJmol

October 13, 201157. Enthalpy of combustion of solid Carbon to form CO2 = -393.7kJ/molEnthalpy of combustion of carbon monoxide to form CO2 = -283.3 kJ/molCalculate ∆H for reaction 2C(s) +O2(g) → 2CO(g)

C(s) +O2(g)→ CO2(g) ∆H = -393.7kJCO(g) +1/2O2 →CO2(g) ∆H = -283.3 kJUsing Hess's Law2C(s) +2O2(g)→ 2CO2(g) ∆H1 = 2(-393.7 kJ) 2CO2(g) →2CO(g) +O2 ∆H2 =-2(-283.3kJ)

_______________________________________________ 2C(s) +O2(g)→ 2CO(g) ∆H = ∆H1 + ∆H2 = -220.8kJ

59. NH3(g) → 1/2 N2 + 3/2 H2(g) ∆H = 46kJ

2H2(g) + O2(g) → 2H2O(g) ∆H = -484kJ

Calculate ∆H for 2N2 +6H2O →3O2 + 4NH3(g)Using Hess's Law 1/2 N2 + 3/2 H2(g) → NH3(g) ∆H = (-) 46kJ 2H2O(g) →2H2(g) + O2(g) ∆H = -(-484kJ) 4 [1/2 N2 + 3/2 H2(g) → NH3(g)] ∆H =4 (-) 46kJ 3[2H2O(g) →2H2(g) + O2(g)] ∆H = 3(484kJ) 2 N2 + 6 H2(g) → 4NH3(g) ∆H =4 (-) 46kJ 36H2O(g) →6H2(g) + 3O2(g) ∆H = 3(484kJ) _______________________________________________________________ 2N2 +6H2O →3O2 + 4NH3(g) ∆H = 1268kJ

This is not a useful reaction for producing ammonia VERY ENDOTHERMIC!

61. (1) 2O3(g) → 3O2(g) ∆H = -427kJ (2) O2(g) → 2O(g) ∆H = +495kJ(3) NO(g) + O3(g) → NO2(g) + O2(g) ∆H = -199kJCalculate ∆H for NO(g) + O(g) → NO2(g) NO(g) + O3(g) → NO2(g) + O2(g) ∆H = -199kJ 1/2[3O2(g) → 2O3(g) ] ∆H = -1/2(-427kJ)1/2[2O(g) → O2(g) ] ∆H = -1/2(+495kJ)

NO(g) + O3(g) → NO2(g) + O2(g) ∆H = -199kJ = -199kJ 1-1/2O2(g) → O3(g) ] ∆H = -1/2(-427kJ) = 213.5kJO(g) → 1/2O2(g) ] ∆H = -1/2(+495kJ) = -247.5kJ_____________________________________________________________________ NO(g) + O(g) → NO2(g) ∆H = -233kJ

ReverseReverse

October 13, 2011

#60

2ClF(g) +O2 ➞ Cl2O(g) + F2O(g) ∆ H = 167.4 kJ

2ClF3(g) + 2O2(g) ➞ Cl2O(g) + 3F2O(g) ∆H = 341.4kJ

2F2(g) + O2(g) ➞ 2F2O(g) ∆H = -43.4kJ

Calculate ∆H for the reaction ClF(g) + F2 ➞ ClF3(g)

2ClF(g) +O2 ➞ Cl2O(g) + F2O(g) ∆ H = 167.4 kJ

Cl2O(g) + 3F2O(g) ➞ 2ClF3(g) + 2O2(g) ∆H = -341.4kJ

2F2(g) + O2(g) ➞ 2F2O(g) ∆H = -43.4kJ

2ClF(g) +2F2 ➞ 2ClF3(g)

Sum and divide by two-217.4 kJ 2

Divide coefficients by 2

ClF(g) + F2 ➞ ClF3(g) ∆H = -108.7 kJ

October 13, 2011

63. Ca(s) + 2C(graphite) ➝CaC2 ∆H = -62.8kJ ∆H Ca(s) + 1/2 O2(g) ➝ CaO ∆H = -635.5kJ CaO(s) + H2O(l) ➝ Ca(OH)2(aq) ∆H = -653.1kJ C2H2(g) + 5/2O2(g)➝2CO2(g) +H2O(l) ∆H =-1300.kJ C(graphite) + O2(g) ➝ CO2(g) ∆H= -393.5 kJCalculate ∆H for CaC2(s) +2H2O(l)➝Ca(OH)2(aq) +C2H2(g) CaC2 ➝ Ca(s) + 2C(graphite) ∆H = -(-62.8kJ ) ∆H Ca(s) + 1/2 O2(g) ➝ CaO ∆H = -635.5kJ CaO(s) + H2O(l) ➝ Ca(OH)2(aq) ∆H = -653.1kJ 2CO2(g)+H2O➝ C2H2(g) + 5/2O2(g) ∆H =-(-1300.kJ) 2[ C(graphite) + O2(g) ➝ CO2(g)] ∆H= 2(-393.5 kJ)

CaC2 ➝ Ca(s) + 2C(graphite) ∆H = +62.8kJ Ca(s) + 1/2 O2(g) ➝ CaO ∆H = -635.5kJ CaO(s) + H2O(l) ➝ Ca(OH)2(aq) ∆H = -653.1kJ 2CO2(g)+H2O➝ C2H2(g) + 5/2O2(g) ∆H =+1300.kJ 2C(graphite) + 2O2(g) ➝ 2CO2(g)] ∆H= -787k_____________________________________________________________________________

CaC2(s) +2H2O(l)➝Ca(OH)2(aq) +C2H2(g) ∆H = -712kJ

October 13, 2011

64. P4(s) + 6Cl2 (g) → 4 PCl3 (g) ∆H = - 1225.6 KJ

P4O10(S) → P4 (s) + 5O2 (g) ∆H = 2967.3 KJ

6PCl5(g) → 6PCl3(g) + 6Cl2(g) -2(-84.2 kJ X 6) = ∆H = 505.2 KJ

10PCl3 (g) + 5O2(g) → 10 Cl3PO (g) ∆H = -2857 KJ

∆H = -610.1 KJ P4O10(S) + 6PCl5(g) → 10 Cl3PO (g)

October 13, 2011

67. a. 2NH3(g) + 3O2(g) + 2CH4(g) ➝ 2HCN(g) + 6H2O(g) 2(-46) 3( 0) 2(-75) ➝ 2(+135.1) 6(-242)

∆Hof = ∆Hp -∆HR = [2(+135.1) + 6(-242)] - [2(-46) + 3(0) + 2(-75)]

= [270.2 -1452] - [-96-150] = -1182(P) - ( -242)_(R) = -940kJ

b. Ca3(PO4)2(s) + 3H2SO4(l) ➝ 3CaSO4(s) + 2H3PO4(l) -4126 + -3(814) ➝ 3(-1433) +2(-1267)

∆Hof = ∆Hp -∆HR [ --4299 +(-2534)] - [- (-4126-2442] = -6833-(-6568) = -265kJ

c. NH3(g) + HCl(g) ➝ NH4Cl(s) -46 -92 -314

∆Hof = ∆Hp -∆HR [-314] - [-46-92] =-176kJ

October 13, 2011

69. Ostwald process for producing nitric acid using following reactions 4NH3(g) + 5O2(g) ➝ 4NO(g) + 6H2O(g)

∆Hof = ∆Hp -∆HR [4(90) +6(-242)] - [4(-46)+5(0)] = [-1092] - [-184] =--908kJ

2NO(g) +O2➝2NO2(g)

∆Hof = ∆Hp -∆HR [2(+34)] - [2(+90) + 0] =-112kJ

3NO2(g) +H2O(l) ➝2HNO3(aq) + NO(g)

∆Hof = ∆Hp -∆HR [2(-207) +(+90)] - [3(+34) +(-286)] =[-324] - [-184] = -140kJ

4NH3(g) + 5O2(g) ➝ 4NO(g) + 6H2O(g) ∆Hof = -908kJ

2[2NO(g) +O2➝2NO2(g)] ∆Hof = -112kJ

[3NO2(g) +H2O(l) ➝2HNO3(aq) + NO(g) ] ∆Hof = -140kJ

H2O(g) →H2O(l)

3[4NH3(g) + 5O2(g) ➝ 4NO(g) + 6H2O(g) ] ∆Hof = -908kJ

6[2NO(g) +O2➝2NO2(g)] ∆Hof = -112kJ

4[3NO2(g) +H2O(l) ➝2HNO3(aq) + NO(g) ] ∆Hof = -140kJ

4H2O(g) →4H2O(l)

12NH3(g) + 15O2(g) ➝ 12NO(g) + 18H2O(g) ] ∆Hof = -908kJ

12NO(g) +6O2➝12NO2(g)] ∆Hof = -112kJ

12NO2(g) +4H2O(l) ➝8HNO3(aq) + 4NO(g) ] ∆Hof = -140kJ

4H2O(g) → 4H 2O(l) ∆Hof = -44kJ

______________________________________________________________________12NH3(g) +21O2(g) → 8HNO 3(aq) + 14H2O(g) + 4NO(g) All reactions are exothermic so overall reaction is EXOTHERMIC!

∆Hof = ∆Hp -∆HR

October 13, 2011

71. Reusable space shuttle rockets us aluminum and ammonium perchlorate as fuel.

calculate ∆Ho for 3Al(s) + 3NH4ClO4(s)→ Al2O3(s) + AlCl3 +3NO(g) +6H2O(g)

∆Ho = ∆Hp -∆HR Al2O3(s) + AlCl3 +3NO(g) +6H2O(g)

∆Hp= [-1676 +(-704) +3(90) +6(-242)] -∆HR [3(0) + 3(-295) [-3562] -[-885] = -2677kJ

73. 2ClF3(g) + 2NH3(g) → N2(g) + 6HF(g) + Cl2(g) ∆Ho = -1196kJ

Calculate ∆Hof for ClF3(g

∆Ho = ∆Hp -∆HR -1196 = [ 0 + 6(-271) + 0] -[2 (-46) +∆Ho

f ClF3 ] 2 -1196= [-1626 ] - [-92 +∆Ho

f ClF3] -1196 +1626 -92= -∆Hof ClF3

2mol 2mol

1196 + 92 -1626 =∆Hof ClF3 = -169kJ for ∆Hof for ClF3(g)

2mol

75. Ethanol enthalpy of combustion

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) [(2 x -395.5) +3(-286)] -[-278] x 1g/46.og per mole= 29.8kJ/g

October 13, 2011

77. C3H8(g) + 5O2(g)→3CO2(g) +4H2O(l) [3(-393.5) +4(-286)] - [-104] = - 2221kJ/44.1= -50.4 kJ C3H8/g

C8H18(l) + 12-1/2 O2(g)→ 8CO2+ 9H2O(l) [8(-393.5) +9(-286)] -[-269 table 2=5453/114] = -47.8kJC8H18/g

Disadvantage for propane is it needs to be pressurized (safety issue)to liquefy (storage quantity!!)

79. Methane to heat a home heat of combustion is -891kJ/mol 4.19 x106 kJ to heat a homeHow many liters needed ? 4.19 x 106/891kJ/mol =4703mol x 22.42L/mol =1.05 x 105L

85. Neutralization of HCl and NaOH -56kJ/mol0.200L 0.400M HCl reacts with 0.150L 0 .500MNaOH - 0.075mol (NaOH limiting reagent) x -56kJ/mol= -4.2kJ

95. a . C12H22O11 + 12O2→ 12CO2 + 11H2O b. ∆E = 24.00kJ x 342.30g sucrose per mol = 5630kJ/mol 1.46gc.PV =nRT At constant P and T P∆V=nRT (∆n=12O 2 -12CO2 gas products - gas reactants) so nRT=0 and P∆V= 0∆H = ∆E + P∆V = ∆E +RTn∆H = ∆E = 5630kJ/mol

October 13, 2011

97. Solar home electricity ( sun supplies a bout 1.0 kw/m2 x0.13 efficiency =0.13kw/m2 =0.13/s x 8 =1.04Kwh/day per m2 40kwh/day 1.04kwh/day/m =38m2 needed

99. Chocolate cake 400 Calories = 1000 thermochemical calories x 400 need to get rid of 400,000 calories x4.184kJ/calorie= 1.67 x 106 kJ PE =mgh how many8 inch steps must he 180 pounds climb to get rid of the cake?180/2.2046= 81.6kg x 9.81x 8x 2.54/100cm per meter =163KJ per step so 1.63 102kJ/step 1.67x106kJ/1.63x 102kJ/step= 10245 steps

101. 1.00 mol sample of water at -30oC. Heated to gaseous water at 140oCalculate q

specific heat capacity of ice is 2.03J/oC.g specific heat capacityof water is 4.18J/oC.g specific heat capacityof steam is 2.02J/oC.g

H2O(s) → H2O(l) ∆Hfusion = 6.02kJ/mol (at 0oC)

H2O(sl) → H2O(g) ∆Hvaporization = 40.7kJ/mol (at 100oC)

18.02g/mol x 2.03J/oC.g x 30o = 1.100kJ

∆Hfusion = 6.02kJ/mol (at 0oC)18.02g/mol x 4.18J/oC.g x 100o =7.530kJ

∆Hvaporization = 40.7kJ/mol (at 100oC)18.02g/mol x 2.02J/oC.g x (140o-100=40) = 1.5kJtotal q = 56.9kJ

October 13, 2011

103. N2(g) +2O2(g) ➔ 2NO2(g) enthalpy change = 67.7kJ/2 mol following balanced equation with whole number coefficients

nN2 = PV/RT 3.5 atm x 0.250L/0.08206 x 373k =0.0286 mol N2no2 = PV/RT 3.5 atm x .450L /0.08206 x 373 =0.0515mol O2oxygen is limiting reagent giving 0.0515mol of NO2 x 67.7kj/mol/2mol=1.74kJ

105. cubic piece of Uranium metal (specific heat capacity of 0.117 at 200.0oC dropped into 1.00L deuterium oxide ("heavy water") specific heat capacity =4.211 at 25.5oC temperature of "heavy water" rose to 28.5 oCdensity of Uranium is 19.05g/cm3 and density of deuterium oxide is 1.11g/ml.find edge length of uranium cube?

heat given by Uranium cube equals heat gained by "heavy water"

Q= m s ∆T s is specific heat capacity0.117 x m x (200-28.5) = 4.211 x (28.5-25.5) x 1000g m=density x Volume0.117 x (dxV) x (200-28.5) = 4.211 x (28.5-25.5) x 1000g0.117 x V x (200-28.5) = 4.211 x (28.5-25.5) x 1000g/d[U]20.1V =12633/19.05V= 663/20.1V=33cm3 volume of cube = cube root of volume = 3.2 cm