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Thermochemistry First Law of Thermodynamics System, surrounding, and thermodynamic universe Heat (q), work (w) and internal energy (E) Calculation of heat.

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Presentation on theme: "Thermochemistry First Law of Thermodynamics System, surrounding, and thermodynamic universe Heat (q), work (w) and internal energy (E) Calculation of heat."— Presentation transcript:

1 Thermochemistry First Law of Thermodynamics System, surrounding, and thermodynamic universe Heat (q), work (w) and internal energy (E) Calculation of heat gained or lost by system State Functions Enthalpy of reactions Calorimetry; determination of enthalpy of reactions Hess’s law of enthalpy summation

2 Energy, Work, Heat & Enthalpy Energy – the capacity to do work or to produce heat Potential Energy: –energy associated with the relative position of a substance in a force field, such as gravitational attraction, chemical bonds, electrostatic, nuclear force, etc., or on the chemical composition Kinetic Energy: –energy associated with the translational motion of an object. Kinetic energy depends on the mass and speed of the object. E K = ½mv 2

3 Energy, Work, Heat & Enthalpy Heat: –energy associated with temperature change – heat flows from a hot object to a cold one. Heat gained or lost: q = m.s.  t –where m = mass ; s = specific heat capacity of the substance, and  t = change in temperature Enthalpy change (  H): –Heat gained or lost during a chemical reaction at constant pressure.

4 First Law of Thermodynamics Energy is not created nor destroyed during chemical or physical processes The change in the internal energy of a system (  E) depends only on the amount of heat (q) gained or lost by the system and the work (w) done on or by the system.  E = q + w For processes that involve gas expansion or compression, w = -p  V  E = q – p  V; q =  E + p  V

5 Coffee Cup Calorimeter

6 Heat Capacity of Calorimeter To determine the heat capacity of coffee cup calorimeter: –A 25.0-g sample of warm water at 40.0 o C was added to a 25.0-g sample of water in a Styrofoam coffee cup calorimeter initially at 20.0 o C. The final temperature of the mixed water and calorimeter was 29.5 o C and the specific heat capacity of water is J/g. o C. Calculate the heat capacity, C h, of the calorimeter.

7 Specific Heat Capacity To determine the specific heat capacity of a metal using coffee cup calorimeter: –A 55.0-g sample of hot metal initially at 99.5 o C was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water and calorimeter were initially at 21.0 o C. If the final temperature of mixture was 30.5 o C, calculate the total heat lost by metal and the specific heat capacity of the metal. The specific heat of water is J/(g. o C) and heat capacity of calorimeter is 10.0 J/ o C.

8 Heat of Neutralization To determine the molar enthalpy of acid-base reaction using coffee cup calorimeter. –50.0 mL of 2.0 M HCl was reacted with 50.0 mL of 2.0 M NaOH in a coffee cup calorimeter. The reaction was exothermic, which caused the temperature of the solution to increase from 22.0 o C to 35.6 o C. Assume the density of solution as 1.0 g/mL, its specific heat capacity as 4.18 J/g. o C, and the heat capacity of calorimeter as 10.J/ o C. Calculate the total amount of heat produced by the reaction. Calculate the enthalpy change (  H, in kJ/mol) for the following reaction: HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)

9 Calculation the enthalpy of reaction using Styrofoam cup calorimeter To determine the enthalpy of reaction using coffee cup calorimeter. –Suppose 100. mL of 1.0 M HCl solution is placed in a Styrofoam coffee cup calorimeter. The initial temperature of HCl solution is 22.5 o C. A g sample of magnesium ribbon is cut to short pieces and added to the acid solution in which the following exothermic reaction occurred. Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) –The heat produced by the above reaction is completely absorbed by the solution and calorimeter, which attained the highest temperature of 34.2 o C. Assume the acid solution has a density of 1.0 g/mL and its specific heat capacity as 4.0 J/g. o C, and the calorimeter has a heat capacity of 10. J/ o C. Calculate the molar enthalpy change (  H, in kJ/mol) for the above reaction.

10 Calculating the enthalpy of reaction using Styrofoam cup calorimeter Reaction: Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) ; Calculations:

11 Bomb Calorimeter

12 Heat Capacity of Calorimeter To determine the heat capacity of bomb calorimeter: –When a g sample of glucose, C 6 H 12 O 6, was completely combusted in a bomb calorimeter, the temperature of the calorimeter assembly increased by 4.48 o C. If the combustion of glucose produces 14.0 kJ/g of energy, how much heat energy is absorbed by the calorimeter. Calculate the heat capacity, C h, of the calorimeter. (Assume that all of the heat produced by the combustion of glucose is absorbed by the calorimeter.)

13 Enthalpy changes of exothermic and endothermic reactions

14 Heat of Combustion To calculate the enthalpy of combustion using bomb calorimeter –When a g sample of sucrose (cane sugar) is completely combusted in a bomb calorimeter, the temperature of the calorimeter was increased by 4.50 o C. If the heat capacity of calorimeter is 3.75 kJ/ o C, how much heat was absorbed by the calorimeter? Calculate the molar enthalpy of combustion of sucrose according to the following equation: C 12 H 22 O 11 (s) + 12O 2 (g)  12CO 2 (g) + 11H 2 O (l)

15 Hess’s Law of Enthalpy of Reactions According to Hess’s law: –The net enthalpy change of a given process is independent of the number of steps taken to complete the process. –If a reaction can be broken down into several steps, then the overall enthalpy for the reaction is equal to the sum of enthalpies of individual steps.

16 Hess’s Law of Enthalpy of Summation According to Hess’s law: –For a given reaction, the overall enthalpy change is equal to the difference between the algebraic sum of enthalpies of formation of products and the algebraic sum of enthalpies of formation of reactants.  H rxn =  n p  H f [products]) –  n r  H f [reactants])

17 Enthalpy of Reactions Applying Hess’s law to calculate enthalpy of reactions. Given: –C (s) + O 2 (g)  CO 2 (g) ;  H o = -394 kJ (1) –CO (g) + ½O 2 (g)  CO 2 (g) ;  H o = -283 kJ (2) Calculate  H for the following reaction: –C (s) + ½O 2 (g)  CO (g)

18 Enthalpy Changes in Reactions #1 Examples: –C (s) + O 2 (g)  CO 2 (g) ;  H = -394 kJ –C (s) + ½O 2 (g)  CO (g) ;  H = -283 kJ –C (s) + ½O 2 (g) ____________  H f = 0.0 kJ –CO (g) + ½O 2 (g) ____________  H f = -283 kJ – CO 2 (g) ____________  H f = -394 kJ

19 Enthalpy of Formation

20 Enthalpy of Combustion Calculating enthalpy of combustion of diborane, B 2 H 6, using the molar enthalpy of formation data: Given: –(1) 4B (s) + 3O 2 (g)  2B 2 O 3 (s) ;  H f(1) = kJ –(2) 2B (s) + 3H 2 (g)  B 2 H 6 (g);  H f(2) = 36.4 kJ –(3) 2H 2 (g) + O 2 (g)  2H 2 O (g) ;  H f(3) = kJ Calculate  H for the following combustion reaction: –B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s ) + 3H 2 O (g) ;

21 Reaction Enthalpy Diagram Reaction: B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3H 2 O (g) B 2 H 6 (g) ___________  H f = 36 kJ 2B (s) + 3H 2 (g) + 3O 2 (g) __________  H f = 0.0 kJ 3H 2 O (g) ____________  H f = -726 kJ B 2 O 2 (s) ____________  H f = kJ

22 Heat of Combustion Reaction To calculate heat of reaction using molar enthalpy of formation: Given:  H f [C 3 H 8 (g) ] = -104 kJ/mol;  H f [H 2 O (g) ] = -242 kJ/mol, and  H f [CO 2 (g) ] = -394 kJ/mol; Calculate  H for the following combustion of C 3 H 8 (g) : –C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g)

23 Enthalpy Diagram Reaction: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) 3C (s) + 4H 2 (g) + 5O 2 (g) ____________  H f = 0.0 kJ C 3 H 8 (g) ____________  H f = -104 kJ 4H 2 O (g) ____________  H f = -968 kJ 3CO 2 (g) ____________  H f = kJ

24 Enthalpy Diagram for the Combustion of Methane

25 Enthalpy Diagram for Combustion Reaction

26 Reactions in Aqueous Solution Enthalpy change for reactions in aqueous solution: Standard condition for gas: P = 1 atm at 25 o C Standard condition for solution: 1 M at 25 o C Under standard condition:  H f [H 3 O + (aq) ] = 0.0 kJ Given: –  H f [Mg 2 +(aq)] = -467 kJ/mol; –  H f [OH - (aq) ] = -230 kJ/mol, and  H f [Mg(OH) 2 (s) ] = -925 kJ/mol, Calculate  H for the reaction: –MgCl 2 (aq) + 2NaOH (aq)  Mg(OH) 2 (s) + 2NaCl (aq)

27 Enthalpy of Ionic Reactions Given the following enthalpy of formation (in kJ/mol):  H f [Ba 2+ (aq) ] = ;  H f [CO 3 2- (aq) ] = ;  H f [BaCO 3 (s) ] = -1219;  H f [BaSO 4 (s) ] = -1465;   H f [OH - (aq) ] = -230;  H f [H 2 SO 4 (aq) ] = and  H f [H 2 O (l)] = kJ/mol Calculate enthalpy changes for the following reactions in aqueous solution: (1) BaCl 2 (aq) + Na 2 CO 3 (aq)  BaCO 3 (s) + 2NaCl (aq) (2) HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) (3) Ba(OH) 2 (aq) + H 2 SO 4 (aq)  BaSO 4 (s) + 2H 2 O (l)

28 Global Energy Resources Biomass (mainly wood) – major sources of energy in many under-developed countries; Coal was once the major source of energy in U.S.A. and industrialized European countries; Petroleum replaces coal in the middle of 20 th Century as the major source of energy for power plants and transportation; Hydroelectric power and nuclear energy are used in certain developed countries. Geothermal energy is used as a secondary source of energy Solar energy is a secondary source of energy, but mainly for household heating.

29 Comparison of Enthalpy of Combustion ———————————————————————— Substances Energy (kJ/g fuel) ———————————————————————— Hydrogen gas (H 2 )120 Natural gas (CH 4 ) 50 Gasoline 48 Crude petroleum 43 Animal fat 38 Coal 29 Charcoal 29 Ethanol 27 Methanol 20 Paper 20 Dry biomass (wood) 16 ——————————————————— — ————

30 Petroleum and Natural Gas Origin of petroleum and natural gas: –most likely from fossilized remains of marine organisms that lived approximately 500 millions years ago Petroleum – –thick, dark liquid composed of mixture of hydrocarbons –Composition varies from one location to another, but mostly hydrocarbon compounds containing C 5 to > C 25 Natural Gas –Consists mostly methane (CH 4, >90%) and some ethane (C 2 H 6 ), propane (C 3 H 8 ) and butane (C 4 H 10 )

31 Petroleum Refining Petroleum refining –Fractional distillation of crude petroleum yields the following fractions: Gasoline (C 5 – C 10 ); Kerosene & jet fuel (C 10 – C 18 ); Diesel fuel, heating, and lubricating oil (C 15 – C 25 ), Asphalt (>C 25 ) –More gasoline is produced by pyrolytic (high temperature) cracking of larger HC compounds (> C 25 )

32 Coal Formed from fossilized plant remains that have been subjected to high temperature and pressure for many millions years Coal matures through 4 stages: –Lignite, subbituminous, bituminous, and anthracite; Composition by mass%: –Lignite: 71% C, 4% H, 23% O, 1% N, and 1% S; –Subbituminous: 77% C, 5% H, 16% O, 1% N, and 1% S; –Bituminous: 80% C, 6% H, 8% O, 1% N, and 5% S; –Anthracite: 92% C, 3% H, 3% O, 1% N, and 1% S. The relative carbon content increases and those of hydrogen and oxygen decrease as coal matures.

33 Coal as Energy Source Coal furnishes about 23% of energy needs in U.S.A. Underground coal mining is dangerous Strip mining destroys lands and the environments Coal contains sulfur and burning coal causes severe air pollution due to: –Air particulate matters, CO 2, CO, and SO 2 (from sulfur in coal) –In atmosphere, SO 2 is oxidized to SO 3, which yields acid rain (SO 3 forms H 2 SO 4 when mixed with rain water)

34 Processing Coal Coal gasification - converting coal into gaseous fuel –Treating coal with air and steam at high temperature produces mixture of CO, H 2, and CH 4. Some CO 2 and SO 2 are also formed –Mixture of CO and H 2 is also called syngas Reactions in coal gasification: –C (s) + O 2 (g)  CO 2 (g) ;  H = -394 kJ –C (s) + ½O 2 (g)  CO (g) ;  H = -111 kJ –C (s) + H 2 O (g)  CO (g) + H 2 (g) ;  H = 131 kJ –C (s) + 2H 2 (g)  CH 4 (g) ;  H = -75 kJ Note: both exothermic and endothermic reactions occur. An energy balance can be maintained by controlling the temperature, the rate of coal feed, and the flow of air and steam.

35 Coal Gasification Process Coal + Steam + Air Heat CH 4 + CO, CO 2, H 2, H 2 O (+ sulfur containing impurities) Separate Treatment CO + H 2 O  CO 2 + H 2 CO + 3H 2  CH 4 + H 2 O After removal of CO 2 and H 2 O, the remaining mixture contains CH 4 and syngas (CO + H 2 )

36 Coal to Coal Slurry Coal is pulverized and mixed with water to form a thick slurry Coal slurry burns like residual oil, with less CO and SO 2 produced

37 Hydrogen Fuel Combustion of H 2 : –H (g) + ½O 2 (g)  H 2 O (l) ;  H = -286 kJ –H 2 (g) + ½O 2 (g)  H 2 O (g) ;  H = -242 kJ Combustion of H 2 produces 2.5 times more energy per gram than natural gas Combustion of H 2 only produces water. However, production, storage and transportation of the gas pose major problems

38 Problems in Hydrogen Production H 2 does not exist in the free form (like N 2 or O 2 ) All existing methods to produce H 2 gas involve endothermic reactions or high temperature: –Steam reformation: CH 4 (g) + H 2 O (g)  CO (g) + 3H 2 (g) ;  H = 206 kJ –Electrolysis: H 2 O (l)  H 2 (g) + ½O 2 (g) ;  H = 286 kJ –Thermochemical decomposition of H 2 O: 2HI (g)  H 2 (g) + I 2 (s) ; (425 o C) 2H 2 O (l) + SO 2 (g) + I 2 (s)  H 2 SO 4 (aq) + 2HI (aq) ; (90 o C) H 2 SO 4 (aq)  SO 2 (g) + H 2 O (l) + ½O 2 (g) ; (825 o C) Net Reaction: H 2 O (l)  H 2 (g) + ½O 2 (g) ;

39 Problems in Storage & Transportation of H 2 H 2 decomposes to H-atoms on metal surface H-atoms are small enough to be absorbed into the metal lattices and makes the metal to become brittle H 2 requires a much larger container for storage and transportation Storage or transportation of H 2 under high pressure (say as liquefied gas) poses high explosion hazard Energy produced per unit volume by H 2 is only one-third that produced by natural gas under similar conditions An alternative method suggested for H 2 storage/transportation is to convert it into solid metal hydrides, MH 2.

40 Alternative Fuels Hydrogen gas is most efficient and clean fuel It produces the most energy per gram, Transportation and storage are difficult It requires a much large fuel tank than gasoline; May be compressed into liquid form, but poses explosion hazard An alternative way is to convert it into a solid metal hydride (MH 2 ) – one that is able to release H 2 when heated. Most likely the hydrogen will be used to power fuel cells that will be installed in automobiles

41 Alternative Fuels Oil shale –Shale rocks contain complex HC called kerogen –Huge deposits in western states, especially Colorado; –Kerogen is not fluid like petroleum - cannot be pumped; –Rocks containing fuel must be heated to >250 o C to decompose the kerogen into smaller HC molecules; –Process produces large quantities of waste rocks – a negative environmental impact.

42 Alternative Fuels Ethanol: –Produced by fermentation of sugar (from sugar cane or corn) –Pure ethanol produces about 27 kJ/g of energy –Use as fuel supplement –Added to gasoline as gasohol (which contains ca. 10% ethanol) –Pure ethanol not suitable for motor fuel in USA, especially in winter, because it does not vaporize easily in cold climate. –Pure ethanol is widely used as motor fuel in Brazil where the climate is warmer.

43 Alternative Fuels Methanol: –Methanol produces about 20 kJ/g of energy when completely combusted –It has been used in race cars as a mixture of 85% methanol and 15% gasoline. –California is evaluating methanol as motor fuel –Arizona and Colorado are considering similar step.

44 New Energy?


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