Understanding Thermodynamics and Thermochemistry

1. Chapter 6 Thermochemistry

2. What is Thermodynamics? • Thermodynamics is the study of the effect of work, heat and energy on a system. It is concerned with only large scale observation. • Thermochemistry is the study of the heat released or required by chemical reactions.

3. Energy, Heat and Enthalpy • The reaction type where energy is obtained are usually combustion in which a hydrocarbon fuel burns in oxygen to form carbon dioxide and water. A typical reaction is the combustion of natural gas, which is largely methane. • CH4(g)+2O2(g) CO2(g)+2H2O (l) +energy

4. Since energy is released in this reaction we include it in the product. Living organisms make use of a very sophisticated form of combustion to obtain the energy to move , grow and think. • C6H12O6(aq) + 6O2(g)--- 6CO2(g) +6H2O(l) + energy. • The glucose does not burn with a flame inside us. A series of reaction takes place at a controlled rate. The field of bioenergetics is the study of the uses of energy by organisms.

5. The conservation of energy • Energy comes in a variety of form. The kinetic energy which is the energy due to motion and the potential energy that is the energy due to position. A body of mass m at a height h above the surface of the earth has a potential energy Ep= mgh g= acceleration of free fall on earth (g=9.8m/s2) • The total energy of an object such as a planet or a ball is the sum of its kinetic and potential energies: E=Ek+Ep

6. The law of conservation of energy states that total energy of an isolated body is constant. • The internal energy of an object is a measure of its capacity to do work and supply heat. It is the total energy of all its atoms and molecules.

7. System and Surroundings • A system is the part of the world we want to study. In chemistry this is usually a reaction mixture in a flask. • A system can be open, closed or isolated. • An open system can exchange both matter and energy with the surroundings.Examples are an automobile engine, the human body, and an open flask.

8. A closed system has a fixed amount of matter, but can exchange only energy with the surroundings. Examples are electric batteries, cold packs used for athletic injuries. • An isolated system can exchange neither matter nor energy with its surroundings. Example a sealed vacuum flask.

9. Heat and work • When a system does work, such as an engine raising a weight, energy is transferred from the system to the surroundings. Energy is also transferred when a system releases heat, as in a combustion. A system with a high internal energy can do a lot of work or provide a lot of heat to the surroundings. Example atightly coiled spring, a fully charged battery and the hot steam of a turbine. A system with a low internal energy can do less work example an uncoiled spring, a nearly exhausted battery, and cold water.

10. We can change the internal energy of an open system by adding or removing matter. An automobile is an open system, and we increase its internal energy by putting gasoline in its tank. A closed system can be prepared by adding reagents and then sealing the container. Chemical reactions can occur inside the system, but all the reactants and products remain inside.

11. A second way to change internal energy of a system is by heating or cooling it. Heat is the transfer of energy that occurs as a result of temperature difference. Energy flows as heat from a hot region into a cooler region .

12. When a hot object is placed in a cool surroundings energy flows out of the system and stirs up chaotic motion of the molecules in the surroundings. Random molecular motion is called thermal motion, meaning heat stirs up the thermal motion of molecules in the system.Heat can raise the temperature of a system and it can also change the physical state , such as from liquid to vapor.

13. The third way to change the internal energy of a system is by doing work on the system or by letting the system do work on the surroundings. • Work is a transfer of energy that takes place when an object is moved against an opposing force.

14. The first Law • Internal energy is the sum of kinetic energy of all the particles in the system and potential energy arising from their interactions with one another. If molecules move rapidly their kinetic energy is high and so the internal energy. If the particles are too close example in a solid the squashing them even closer together will increase their potential energy and hence increase the internal energy.

15. Units • Kinetic energy=1/2 mv2 mass is measured in kg and speed in (meter/sec)2 kg.m2/s2 1 kg.m2/s2 = 1J 1kJ =103 J 1 calorie(cal) is the energy needed to raise the temperature of 1g of water by 1⁰C. 1cal=4.184 J

16. Internal energy changes when energy enters or leaves a system. • ∆U= Ufinal –Uinitial • A positive value of ∆U means that the internal energy of the final state of the system is higher than the initial state. The negative value of ∆U means that the final state has a lower energy than the initial state.

17. When we do work on a system for instance, by winding a spring inside it its internal energy rises. If we do 100kJ of work; then the system increases its store of energy by 100kJ and we write as ∆U=+100kJ • Changes in internal energy= energy supplied to system as work • ∆U=w

18. If we transfer 40kJ of energy to a large beaker of water by standing it on ahot surface; then the internal energy of the water increases by 40kJ and we write ∆U=+40kJ. • Change in internal energy= energy supplied to system as heat • ∆U=q • Change in internal energy=energy supplied to system as heat + energy transferred to system as work

19. If we supply 40kJ of heat (q=+40kJ)and do 25kJ of work on the system (w=+25kJ), then the total change in internal energy is • ∆U= q+w =40 +25 kJ =+65 kJ • If, in another process ,40kJ of heat leaks out of the system while we are doing 25kJ of work on it we would write q= -40kJthe minus sign indicates that heat has left the system

20. ∆U =q+w = (-40) + 25 kJ =-15 kJ

21. Class Practice • A battery drives an electric motor in an aquarium pump, and we consider the battery plus the motor as the system. During a certain period, the system does 555kJ of work on the pump and releases 124kJ of heat into the surroundings. What is the change in internal energy of the system?

22. Transferring Energy as Work • How to measure changes in the internal energy brought about by doing work? • Work done = distance moved х Opposing force • Opposing force = area х external pressure • When we combine these two expressions, we get • work done= distance moved х area х external pressure • The distance moved multiplied by the area is the change in volume that occurs during the expansion, • work done = change in volume х external pressure

23. The above relation can be expressed in symbols. The change in volume (final – initial) is ∆V, and the external pressure is Pexternal. • w= -Pexternal∆V • The negative sign indicates that w is negative (there is a decrease in internal energy that is the energy leaves the system) and the positive means the system expands. • 1 L-atm = 101.325 J

24. Class Practice • The hot gaseous products of combustion in an internal combustion engine expand by 2.2 L against an external pressure of 1.0 atm. How much work is done by the expansion?

25. Transferring Energy as Heat • Suppose we arrange for a reaction to take place inside a sealed container: • At constant volume ∆U=q • If 75 kJ of heat is supplied into the container full of gas. If the volume could not change then all the heat supplied to it would be used to raise the internal energy of the system, and ∆U=+75kJ • When the gas is free to change its volume some of the energy supplied is used to raise the temperature and the rest leaks back to the surroundings as work. This is art constant pressure.

26. If 10kJ is used to expand the system ( w= -10kJ) in this case though q= +75kJ, the change in internal energy ∆U = q+w =75-10kJ =+65kJ • At constant pressure ∆U = q+ w =q - P∆V

27. Enthalpy • A quantity called enthalpy is a convenient procedure that takes into account the expansion work automatically • ∆H=∆U + P∆V • ∆H=q-P∆V+P∆V • ∆H=q • A change in internal energy can be identified with the heat supplied at constant volume. A change in enthalpy is equal to the heat supplied at constant pressure

28. Class practice • In a certain reaction at constant pressure ,50kJ of heat left the system and 30kJ of energy left the system as expansion work to make room for the products. What are the values of ∆H and ∆U for the reaction? • -50kJ • -80kJ

29. Exothermic and Endothermic reaction • An exothermic process is one that releases heat; when ∆H< 0 signifies exothermic process.(At constant pressure). • An endothermic reaction absorbs heat; • ∆H>0 signifies endothermic process.

30. Heat Capacity • Heat capacity, C is the heat required to raise the temperature of an object through 1K; • Heat capacity=heat supplied/ temperature rise • C= q / ∆T • Unit J/K if temperature is expressed in Kelvin.

31. Specific heat capacity • The heat capacity divided by the mass of the sample in grams. • Heat capacity = mass x specific heat capacity • C= m x Cs • If we know the temperature rise it undergoes during an experiment, then the heat transferred to it is obtained as • Heat supplied = mass(g) x specific heat capacity(J/⁰C.g)x change in temperature(⁰ C) • q = m x Cs x ∆T

32. Class Practice • A swimming pool contains 2x104L of water. How much energy in Joules is required to raise the temperature of the water from 20⁰ C to 25⁰C?The specific heat capacity of water is 4.184 J/⁰ C.g • 4.2x 10⁵ kJ

33. Thermochemistry of Physical Change • Vaporization: This is an endothermic process because energy has to be supplied to overcome the forces that hold molecules together in a liquid. The enthalpy change per mole of molecules when a liquid vaporizes is called the enthalpy of vaporization, ∆Hvap. For water at 100⁰C • ∆Hvap= Hvapor – Hliquid = =40.7kJ/mol

34. Class Practice • An electric heater is used to raise the temperature of a sample of water to its boiling point in a constant pressure calorimeter. We bring the water to a boiling point and the continue heating until 35 g of water has vaporized. The calculated time taken the vaporization alone requires 79 kJ of heat. What is the enthalpy of vaporization of water at 100⁰ C?

35. Melting and sublimation • The enthalpy changes that accompanies melting is called the enthalpy of fusion, • ∆H fusion= Hliq –Hsolid • The enthalpy of fusion of water at 0⁰C is 6.0kJ/mol. This means we have to supply 6.0kJ of energy as heat to melt one mole of water. • The enthalpy of freezing is the change in enthalpy per mole when a liquid turns into a solid. For water at 0⁰C the enthalpy of freezing is -6.0kJ/mol • ∆H(reverse process)= − ∆H(forward process)

36. Sublimation is the direct conversion of a solid into its vapor. The enthalpy of sublimation, ∆Hsub=Hvapor – Hsolid • ∆Hsub = ∆Hfusion + ∆Hvap

37. The temperature of a sample is constant at its boiling and melting points, even though heat is being applied.

38. Reaction Enthalpies • A thermochemical equation is a combination of a chemical equation with the enthalpy change for the reaction as written. The reaction enthalpy is the change in enthalpy per mole for the stoichiometry numbers of moles of reactants in the chemical equation.

39. Class practice • When 2.31 g of phosphorus reacts with chlorine to form phosphorus trichloride,PCl3, in a calorimeter of heat capacity 2.16kJ/⁰C, the temperature of the calorimeter rises by 11.06⁰C. Write the thermochemical equation for the reaction.

40. Hess’s Law • A reaction that is exothermic in one direction is endothermic in the other direction: • C₆H12O₆(aq) +6O2(g) 6CO2(g) + 6H2O(l) • ∆H⁰=−2808kJ • This is exothermic reaction. • The reverse of this reaction , the formation of glucose and oxygen from carbon dioxide and water, is endothermic: • 6CO2(g) + 6H2O(l) C₆H12O₆(aq) +6O2(g) • ∆H⁰=+2808 kJ

41. According to Hess’s Law, thermochemical equations for the individual steps of a reaction sequence may be combined to obtain the thermochemical equation for the overall reaction. • explanation

42. Class practice • Gasoline, which contains octane as one component, may burn to carbon monoxide if the air supply is restricted. Derive the standard reaction enthalpy for the incomplete combustion of octane: • 2C8H18(l) +17O2(g) --- 16CO(g) +18H2O(l)

43. Enthalpies of combustion • The heat absorbed or given off by a reaction can be treated like a reactant or product in a stoichiometry relation.

44. Class practice • How much propane should a backpacker carry: do we really need to carry a kilogram of gas? Calculate the mass of propane that you would need to burn to obtain 350kJ of heat, which is just enough energy to heat 1 L of water from 27⁰C to boiling at sealevel (if we ignore all heat losses). The thermochemical equation is • C3H₈(g) + 5O2(g) - 3CO2(g) + 4H2O(l) ∆H⁰ =−2220kJ

45. The specific enthalpy tells how much heat can be obtained per gram of fuel. The enthalpy density indicates how much heat can be obtained per liter of fuel.

46. Homework • Page 258 • 6.8, 6.18,6.20,6.26,6.32