Chemistry 360 Dr. Jean M. Standard Problem Set 6 Solutions. hree moles of an ideal gas expand isothermally and reversibly from 90 to 300 L at 300 K. a Calculate ΔU m, ΔS m, w per mole, and q per mole. For an ideal gas, ΔU = C v Δ, or in terms of molar quantities, ΔU m = C v,m Δ. Since the process is isothermal, Δ = 0 and so ΔU m = 0. For an isothermal reversible process, the work is w = nr ln V /V w m = R ln V /V. Substituting, w m = R ln V /V = 8.34 J mol K 300 K w m = 3003 J/mol.. herefore, the work per mole is # $ 90 L ' ln% 300 L From the first law, ΔU = q + w or in terms of molar quantities, ΔU m = q m + w m. Since ΔU m = 0 for this isothermal process, that means that q m = w m. hus, the heat per mole is q m = w m q m = 3003 J/mol. Finally, since ds = dq rev, for an isothermal process, the molar entropy change is ΔS m = q m = 3003J/mol 300 K ΔS m = 0.0J mol K. b If the expansion is carried out irreversibly by allowing the gas to expand rapidly into a vacuum, determine ΔU m, ΔS m, w per mole, and q per mole. For an ideal gas, even if the process is irreversible, ΔU m = C v,m Δ. Since Δ = 0 in this process, then ΔU m = 0. his also must be true since U is a state function and the process involves the same initial and final states as part a. he work is by definition dw = P ext dv, or in terms of molar quantities, dw m = P ext dv m. Since the expansion is into a vacuum, P ext = 0, and therefore, dw m = 0 or w m = 0. From the First Law, ΔU m = q m + w m. Since ΔU m = 0 and w m = 0, we have that q m = 0. Since S is a state function, the change in entropy for the irreversible process must be the same as for the reversible process in part a since the initial and final states are the same. hus, ΔS m = 0.0J mol K.
. A normal breath has a volume of about L. he pressure exerted by the lungs to draw air in is about 758 torr. Assuming that the outside air pressure is 760 torr, calculate the change in entropy of a breath of air when it is inhaled into the lungs. Assume that the air remains at a temperature of 5 C and that it behaves ideally. In this process, the air expands from a pressure of 760 torr to 758 torr. Assuming that the gas behaves ideally and that the process is reversible, we have that the entropy change in an isothermal reversible process is # ΔS = n R ln V %. $ ' Unfortunately, we are given the pressure change rather than the volume change. However, if the process is isothermal then we can use the ideal gas equation to get V V V = P P. Substituting this expression into the equation for the entropy change yields # ΔS = n R ln P %. $ ' Now in order to use the formula, we need to determine the moles in L of air at 5 C and an initial pressure of 760 torr = atm. Using the ideal gas law, Substituting, the entropy change is P n = PV R atm L = 0.0805 Latm/molK 98 K n = 0.0409 mol. # ΔS = n R ln P % $ P ' = 0.0409 mol 8.34 J mol K ΔS = 0.0009 J/K. # 760 torr ln% $ 758 torr '
3. he molar heat capacity of solid gold is given by the relation 3 C p,m = 5.69 7.3 0 4 + 4.58 0 6, in units of J mol K. Calculate the entropy change for heating.50 moles of gold from.0 C to 000 C at constant pressure. he entropy change for heating at constant pressure is ΔS = C p d, or since C p = n C p,m, ΔS = n C p,m d. Substituting the heat capacity for gold and integrating yields ΔS = n = 5.69 n 5.69 7.3 0 4 + 4.58 0 6 d 7.3 0 4 n ΔS = 5.69 n ln + 7.3 0 4 n ' * d d + 4.58 0 6 n d ' + 4.58 0 6 n For.5 moles of gold being heated from 95 K to 73 K, the entropy change is " ΔS = 5.69n ln % $ ' 7.3 0 4 n # + 4.58 0 6 n $ " = 5.69.5molln 73K % $ ' 7.3 0 4.5mol # 95K = 93.9 J/K.79 J/K + 8.78J/K ΔS = 00.9 J/K. " % ' # 73K 95K " + 4.58 0 6 73K.5mol $ # +. * 95K % '
4. A quantity of ice at 0 C is allowed to melt in a large body of water also at 0 C. Calculate the standard molar entropy change for this process. 4 Since the process described involves a phase change, we can use the equation us = ΔH. he molar enthalpy of melting is also known as the enthalpy of ion, ΔH, which can be found in your textbook or in the CRC. us us = ΔH 6008 J/mol = 73.5K =.0 J mol K. 5. Calculate the change in entropy when one mole of aluminum is heated from 600 C to 700 C. he melting point of aluminum is 660 C, the enthalpy of ion is 393 J/g, the heat capacity of solid aluminum is 3.8 J mol K, and heat capacity of liquid aluminum is 34.4 J mol K. For heating a solid with a phase change from a solid to a liquid, the entropy change is ΔS = C p s d + n ΔH + C p l d. Expressing the entropy change in terms of standard molar heat capacities since that is what is given in the problem, we have ΔS = n C p,m s d + n ΔH C p,m l + n d. Assuming that the heat capacities are independent of temperature they are listed in the problem as constants, the expression becomes ΔS = n = n C p,m ΔS = n C p,m C p,m s s d + n ΔH d + n ΔH $ s ln ' + n ΔH % + n + n C p,m + n C p,m C p,m l d l $ l ln ' %. d
5. Continued 5 Substituting, ΔS = n C p,m # s ln % + n ΔH # + n C p,m l ln $ ' % $ ' # 933.5K mol 6.98 g/mol ln% + $ 873.5K ' 933.5K = mol 3.8 J mol K + mol 34.4 J mol K ΔS = 4.94 J/K. # 973.5K ln% $ 933.5K ' 393J/g 6. A mixture of 40% N O and 60% O may be used in a dentist office as an anesthetic. Assuming that the gases behave ideally, determine the entropy of mixing to produce mole of the mixture. he entropy of mixing, ΔS mix, is given as ΔS mix = R n i ln x i, where n i is the number of moles of each species and x i is the mole fraction. One mole of the mixture would correspond to 0.40 moles N O and 0.60 moles of O. he mole fraction of N O is 0.40 and the mole fraction of O is 0.60. i Substituting, ΔS mix = R i n i ln x i [ 0.40 mol ln 0.40 + 0.60 mol ln 0.60] 0.6730 mol = R n ln x + n ln x = 8.34 J/molK = 8.34 J/molK ΔS mix = 5.60 J/K.
7. A quantity of 35.0 g of water at 5.0 C is mixed with 60.0 g of water at 86.0 C. Calculate the final temperature and the entropy change of the system. he molar constant pressure heat capacity of water is 75.9 J/molK. 6 First, we need to get the final temperature of the system. We know that if two systems are placed in contact, their final temperatures will be equal and the heat gained by one system will equal the heat lost by the other system. If we call the water initially at 5.0 C system A, and the water initially at 86.0 C system B, then q A = q B. Assuming that the mixing occurs at constant pressure, we know that q p = ΔH = C p Δ = n C p,m Δ. he result given above assumes that the heat capacity is independent of temperature. Substituting, or n A C p,m H O q A = q B Δ A = n B C p,m H O Δ B. he heat capacity of water cancels out on both sides of the equation. Also, the moles on either side can be written in terms of masses, n = w M, where w is the mass and M is the molecular weight. Substituting, n A C p,m H O Δ A = n B C p,m H O Δ B n A Δ A = n B Δ B w A M H O Δ A = w B M H O Δ B. he molecular weights cancel out, leaving the equation Solving for the final temperature, or w A Δ A = w B Δ B,. or w A final init,a = w B final init,b w A + w B final = w A init,a + w B init,b, final = w A init,a + w B init,b w A + w B = + 60.0 g 86.0 C 35.0 g 5.0 C final = 75.05 C. 35.0 g + 60.0 g
7. Continued 7 he entropy change can be calculated as the sum of the entropy changes of systems A and B, ΔS = ΔS A + ΔS B. At constant pressure, the entropy change for heating or cooling a system is ΔS = C final p d = init n C final p,m d. init Substituting this into the equation above for systems A and B, ΔS = ΔS A + ΔS B final n A C p,m = d + init,a final init,b n B C p,m $ = n A C p,m ln ' final % + n $ BC p,m ln ' final init,a % init,b 35.0 g = 8.05g/mol 75.9J mol K $ 348.0 K ' ln % 98.5K 60.0 g $ 348.0 K ' + ln 8.05g/mol % 359.5K =.70 J/K - 0.70 J/K ΔS =.00 J/K. d 75.9J mol K 8. Which system has the higher absolute entropy? a g solid Au at 064 K or g liquid Au at 064 K g liquid Au According to the hird Law, for a given substance, the liquid will always have higher absolute entropy than the solid at the same temperature. b mole CO at 5 C and atm or mole CO at 5 C and atm mole CO At the same temperature and pressure the triatomic molecular has more energy states available to it; therefore, gaseous CO has higher entropy than gaseous CO. c mole Ar at 5 C and atm or mole Ar at 5 C and 0.0 atm. mole Ar at 5 C and 0.0 atm Both samples are at the same temperature, so the gas with the lower pressure will have the larger volume and hence the larger entropy.
9. he thermite reaction involves solid aluminum powder reacting with ironiii oxide hematite to produce aluminum oxide and iron. he reaction is so exothermic that the iron produced is molten liquid phase. Write the balanced chemical reaction. Determine for the process using the tables in the appendix of your textbook and the standard molar entropy of liquid Fe, 34.76 J/molK. Assume standard pressure and 5 C. 8 he balanced reaction is Al s + Fe O 3 s Al O 3 s + Fe l. For the chemical reaction shown above, the molar entropy change is Al O 3 + S m Fe S m Fe O 3 S m Al. At 5 C, the standard molar entropies are available in the appendix of the text. Substituting, Al O 3 + S m Fe S m Fe O 3 S m Al = 50.9 + 34.76 87.40 8.33 in J mol K = 3.6 J mol K. 0. Using the tables in the appendix of your textbook, determine the standard molar entropies of formation of the following compounds at 5 C. a H O l he formation reaction is H g + O g H O l. For the chemical reaction shown above, the molar entropy of formation is H O,l S m H S m O. Substituting the standard molar entropies at 5 C from the appendix, H O,l S m H S m O = 69.9 30.68 05.4 in J mol K = 63.34 J mol K.
0. Continued 9 b H O g he formation reaction is he molar entropy of formation is H g + O g H O g. H O,g S m H S m O. Substituting the standard molar entropies at 5 C from the appendix, H O,g S m H S m O = 88.83 30.68 05.4 in J mol K = 44.4 J mol K. c CuSO 4 s he formation reaction is he molar entropy of formation is Cu s + S s + O g CuSO 4 s. CuSO 4 S m Cu S m S S m O. Substituting the standard molar entropies at 5 C from the appendix, CuSO 4 S m Cu S m S S m O = 09 33.5 3.08 05.4 in J mol K = 366 J mol K.
0. Continued 0 d H 3 PO 4 s he formation reaction is he molar entropy of formation is 3 H g + P s + O g H 3 PO 4 s. H 3 PO 4 3 S m H S m P S m O. Substituting the standard molar entropies at 5 C from the appendix, H 3 PO 4 3 S m S m H P S m O = 0.50 3 30.68 4.09 05.4 in J mol K = 536.89 J mol K. e C s, diamond he formation reaction is he molar entropy of formation is C s, graphite C s, diamond. C, diamond S m C, graphite. Substituting the standard molar entropies at 5 C from the appendix, C, diamond S m C, graphite =.38 5.74 in J mol K = 3.36 J mol K.
. Using the tables in the appendix of the textbook, determine the standard molar entropy change at 5 C for the combustion reaction CH 4 g + O g CO g + H O g For the chemical reaction shown above, the molar entropy change is. CO + S m H O,g S m CH 4 S m O.. At 5 C, the standard molar entropies are available in the appendix of your text. Substituting, CO + S m H O,g S m CH 4 S m O = 3.74 + 88.83 86.6 05.4 in J mol K = 5.4 J mol K.