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Thermochemistry. n Thermochemistry is the study of _________________ during chemical reactions.

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Presentation on theme: "Thermochemistry. n Thermochemistry is the study of _________________ during chemical reactions."— Presentation transcript:

1 Thermochemistry

2 n Thermochemistry is the study of _________________ during chemical reactions.

3 Energy n Capacity to do ______________ or supply ______________ n 2 Types of Energy –______________ - stored energy –______________ - energy of motion n Energy is weightless, odorless, and tasteless

4 Energy n Kinds of ______________ and their arrangements in a substance determines the amount of ______________ stored in that substance.

5 Heat Energy n Energy ______________ from one object to another causing a temperature change. n It cannot be detected - only the ______________ by heat can be measured.

6 Heat Energy n Thermometers measure the average ______________, not heat. n Heat flows from a ______________ object to a ______________ object.

7 Heat Energy n All chemical reactions and changes in physical states have either an ______________ of heat or a ______________ of heat.

8 Heat Energy n ______________ - What we focus on. n ______________ - everything else. n Heat flow is given in relation to the ______________

9 System and Surroundings n The ______________ includes the molecules we want to study (here, the hydrogen and oxygen molecules). n The ______________ are everything else (here, the cylinder and piston).

10 Heat Energy n Law of Conservation of Energy - energy cannot be ______________ nor ______________, it just changes from one form to another.

11 2 Processes of Heat Energy n ______________ - a process in which the system ______________ heat. –Value is ______________. –Symbol - +  H n ______________ - a process in which the system ______________ heat. –Value is ______________. –Symbol - -  H

12 Units of Heat n Calorie - food calorie n ______________ - quantity of heat required to raise the temperature of 1 gram of water 1°C n Joule - more commonly used (J) –1 C = 1000 cal –4.184 J = 1 cal

13 Heat Capacity n The amount of energy needed to ______________ the temperature of an object 1°C

14 Specific Heat n Amount of ______________ needed to raise 1 gram of a substance 1°C – q = m C  T –q = heat change –m = mass –C = specific heat –  T = change in temperature

15 Calorimetry n Measurement of heat changes n ______________ - heat content of a system at constant pressure n Heat released = Heat absorbed n ______________ is the device used to measure the change

16 Calorimeter

17 q lost = q gained n If q = m C  T, then: n (m C  T) lost = (m C  T) gained

18 How much heat is needed to warm 250 g of water (about 1 cup) from 22°C (about room temperature) to near its boiling point, 98°C? The specific heat of water is 4.18 J/g ºC.

19 n Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.082 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0°C.

20 In the laboratory, a student heats a 100 g sample of metal to 101.5 º C. The student then adds the metal to a calorimeter that has 100 mL of water at 22.5  C. Calculate the specific heat of the metal if the final temperature of the metal-water mixture is 38.7  C.

21 Thermochemical Equations n An equation that includes the ______________.

22 Reactant n Heat as a reactant will give a ______________ q value. n Heat as a reactant will be an ______________ process.

23 Product n Heat as a product will give a ______________ q value. n Heat as a product will be an ______________ process.

24 n 2H 2 O (l)  2H 2(g) + O 2(g)  H rxn = 572 kJ n 572 kJ + 2H 2 O (l)  2H 2(g) + O 2(g)

25 n Al 2 O 3(s)  2Al (s) + 3/2 O 2(g)  H rxn = -1676 kJ n Al 2 O 3(s)  2Al (s) + 3/2 O 2(g) + 1676 kJ

26 Heat of Reaction n Conditions are standard conditions. –1 atm or 101.3 kPa and 25°C n  H rxn

27 Heat of Combustion n Complete ______________ of 1 mole of a substance. n  H comb

28 Heat of Fusion n Heat ______________ by one mole of a solid to change it to a liquid. n  H fus

29 Heat of Solidification n Heat ______________ by one mole of a liquid to change it to a solid. n  H solid

30 n  H fus = -  H solid n Same process just in opposite direction.

31 Heat of Vaporization n Amount of heat needed to change 1 mole of a ______________ to a ______________. n  H vap

32 Heat of Condensation n Amount of heat needed to change 1 mole of ______________ to a ______________. n  H cond

33 n  H vap = -  H cond n Same process just in opposite direction

34 Heat of Solution n Change in heat when dissolving a ______________ in a solvent. n  H soln

35 Hess’s Law of Summation Hess’s law states that “If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

36 n If we know the individual reactions that make up the final equation, we can determine the  H rxn for the final equation.

37 Individual Steps C 3 H 8 (g)  3C (graphite) + 4H 2 (g) 3C (graphite) + 3O 2 (g)  3CO 2 (g) 4H 2 (g) + 2O 2 (g)  4H 2 O (l)

38 C 3 H 8 (g)  3C (graphite) + 4H 2 (g) 3C (graphite) + 3O 2 (g)  3CO 2 (g) 4H 2 (g) + 2O 2 (g)  4H 2 O (l) _______________________________ C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4 H 2 O (l)

39 Standard Heats of Formation An enthalpy of formation,  H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

40 Standard Heats of Formation Standard enthalpies of formation,  H f, are measured under standard conditions (25°C and 1.00 atm pressure).

41 n When we do not know the individual steps of a reaction, we can use the standard heats of formation to calculate  H.  H =  n  H f(products) -  m  H f(reactants) where n and m are the stoichiometric coefficients.

42 C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4 H 2 O (l)  H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ


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