Chapter 6 Thermochemistry 199
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CHAPTER 6 THERMOCHEMISTRY Questions 13. Path-dependent functions for a trip from Chicago to Denver are those quantities that depend on the route taken. One can fly directly from Chicago to Denver, or one could fly from Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc. State functions are path-independent; they only depend on the initial and final states. Some state functions for an airplane trip from Chicago to Denver would be longitude change, latitude change, elevation change, and overall time zone change. 14. Products have a lower potential energy than reactants when the bonds in the products are stronger (on average) than in the reactants. This occurs generally in exothermic processes. Products have a higher potential energy than reactants when the reactants have the stronger bonds (on average). This is typified by endothermic reactions. 15. 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g); the combustion of gasoline is exothermic (as is typical of combustion reactions). For exothermic reactions, heat is released into the surroundings giving a negative q value. To determine the sign of w, concentrate on the moles of gaseous reactants versus the moles of gaseous products. In this reaction, we go from 25 moles of reactant gas molecules to 16 + 18 = 34 moles of product gas molecules. As reactants are converted to products, an expansion will occur because the moles of gas increase. When a gas expands, the system does work on the surroundings, and w is a negative value. 16. ∆H = ∆E + P∆V at constant P; from the definition of enthalpy, the difference between ∆H and ∆E, at constant P, is the quantity P∆V. Thus, when a system at constant P can do pressure-volume work, then ∆H ≠ ∆E. When the system cannot do PV work, then ∆H = ∆E at constant pressure. An important way to differentiate ∆H from ∆E is to concentrate on q, the heat flow; the heat flow by a system at constant pressure equals ∆H, and the heat flow by a system at constant volume equals ∆E. 17. a. The ∆H value for a reaction is specific to the coefficients in the balanced equation. Be- cause the coefficient in front of H2O is a two, 891 kJ of heat is released when 2 mol of H2O are produced. For 1 mol of H2O formed, 891/2 = 446 kJ of heat is released. b. 891/2 = 446 kJ of heat released for each mol of O2 reacted. 18. Water has a relatively large heat capacity, so it takes a lot of energy to increase the temperature of a large body of water. Because of this, the temperature fluctuations of a large body of water (oceans) are small compared to the temperatures fluctuations of air. Hence, oceans act as a heat reservoir for areas close to them which results in smaller temperature changes as compared to areas farther away from the oceans. 198 CHAPTER 6 THERMOCHEMISTRY 199 19. When the enthalpy change for a reaction is negative as is the case here, the reaction is exothermic. This means that heat is produced (evolved) as C6H12O6(aq) is converted into products. 20. A coffee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained for a reaction at constant pressure is equal to the enthalpy change (∆H) for that reaction. A bomb calorimeter is at constant volume. The heat released or gained for a reaction at constant volume is equal to the internal energy change (∆E) for that reaction. 21. Given: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H = −891 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ∆H = −803 kJ Using Hess’s law: H2O(l) + 1/2 CO2(g) → 1/2 CH4(g) + O2(g) ∆H1 = −1/2(−891 kJ) 1/2 CH4(g) + O2(g) → 1/2 CO2(g) + H2O(g) ∆H2 = 1/2(−803 kJ) H2O(l) → H2O(g) ∆H = ∆H1 + ∆H2 = 44 kJ The enthalpy of vaporization of water is 44 kJ/mol. Note: When an equation is reversed, the sign on ΔH is reversed. When the coefficients in a balanced equation are multiplied by an integer, then the value of ΔH is multiplied by the same integer. 22. A state function is a function whose change depends only on the initial and final states and not on how one got from the initial to the final state. An extensive property depends on the amount of substance. Enthalpy changes for a reaction are path-independent, but they do depend on the quantity of reactants consumed in the reaction. Therefore, enthalpy changes are a state function and an extensive property. o 23. The zero point for ΔHf values are elements in their standard state. All substances are meas- ured in relationship to this zero point. 24. a. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H = ? Utilizing Hess’s law: Reactants → Standard State Elements ∆H = ∆Ha + ∆Hb = 75 + 0 = 75 kJ Standard State Elements → Products ∆H = ∆Hc + ∆Hd = –394 – 572 = –966 kJ Reactants → Products ∆H = 75 – 966 = –891 kJ b. The standard enthalpy of formation for an element in its standard state is given a value of zero. To assign standard enthalpy of formation values for all other substances, there needs to be a reference point from which all enthalpy changes are determined. This reference point is the elements in their standard state which is defined as the zero point. So when using standard enthalpy values, a reaction is broken up into two steps. The first 200 CHAPTER 6 THERMOCHEMISTRY step is to calculate the enthalpy change necessary to convert the reactants to the elements in their standard state. The second step is to determine the enthalpy change that occurs when the elements in their standard state go to form the products. When these two steps are added together, the reference point (the elements in their standard state) cancels out and we are left with the enthalpy change for the reaction. c. This overall reaction is just the reverse of all the steps in the part a answer. So ∆H° = +966 – 75 = 891 kJ. Products are first converted to the elements in their standard state which requires 966 kJ of heat. Next, the elements in the standard states go to form the original reactants [CH4(g) + 2 O2(g)] which has an enthalpy change of −75 kJ. All of the signs are reversed because the entire process is reversed. 25. No matter how insulated your thermos bottle, some heat will always escape into the surroundings. If the temperature of the thermos bottle (the surroundings) is high, less heat initially will escape from the coffee (the system); this results in your coffee staying hotter for a longer period of time. 26. From the photosynthesis reaction, CO2(g) is used by plants to convert water into glucose and oxygen. If the plant population is significantly reduced, not as much CO2 will be consumed in the photosynthesis reaction. As the CO2 levels of the atmosphere increase, the greenhouse effect due to excess CO2 in the atmosphere will become worse. 27. Fossil fuels contain carbon; the incomplete combustion of fossil fuels produces CO(g) instead of CO2(g). This occurs when the amount of oxygen reacting is not sufficient to convert all the carbon to CO2. Carbon monoxide is a poisonous gas to humans. 28. Advantages: H2 burns cleanly (less pollution) and gives a lot of energy per gram of fuel. Water as a source of hydrogen is abundant and cheap. Disadvantages: Expensive and gas storage and safety issues Exercises Potential and Kinetic Energy 1 1 kg m 2 29. KE = mv2; convert mass and velocity to SI units. 1 J = 2 s 2 1lb 1 kg Mass = 5.25 oz × × = 0.149 kg 16 oz 2.205 lb 1.0 × 102 mi 1 h 1 min 1760 yd 1 m 45 m Velocity = × × × × = h 60 min 60 s mi 1.094 yd s 2 1 1 45 m KE = mv2 = × 0.149 kg × = 150 J 2 2 s CHAPTER 6 THERMOCHEMISTRY 201 2 2 1 1 1.0 m 1 1 2.0 m 30. KE = mv2 = × 2.0 kg × = 1.0 J; KE = mv2 = × 1.0 kg × 2 2 s 2 2 s = 2.0 J The 1.0-kg object with a velocity of 2.0 m/s has the greater kinetic energy. 31. a. Potential energy is energy due to position. Initially, ball A has a higher potential energy than ball B because the position of ball A is higher than the position of ball B. In the final position, ball B has the higher position so ball B has the higher potential energy. b. As ball A rolled down the hill, some of the potential energy lost by A has been converted to random motion of the components of the hill (frictional heating). The remainder of the lost potential energy was added to B to initially increase its kinetic energy and then to increase its potential energy. 9.81 m 196 kg m2 32. Ball A: PE = mgz = 2.00 kg × × 10.0 m = = 196 J s2 s2 At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal 196 J. 9.81 m At point II: PE = mgz = 4.00 kg × × 3.00 m = 118 J s2 KE = Etotal − PE = 196 J − 118 J = 78 J Heat and Work 33.