Answers to Assigned Problems from Chapter 2

1 Answers to Assigned Problems from Chapter 2 1 mol of ice has a volume of g cm 3 = cm3 1 mol of water has a volume of g cm 3 = cm3 V(ice water) = cm3 mol 1 (PV) = 63 atm dm3 = J mol 1 H = 6025 J = U + (PV) = U ( J mol 1) U = 6025 J mol 1 = kJ mol 1 (Difference between H and U is only J mol 1.) Work done on the system = J mol 1 1 mol of water at 100 C has a volume of g cm 3 = cm3 Volume of steam = 596 g cm 3 = 30 cm3 Volume increase = 30 199 cm3 mol 1 = dm3 mol 1 (PV) = atm dm3 mol 1 = J mol 1 = kJ mol 1 H = 4063 J mol 1 = U + ( J mol 1) U = kJ mol 1 Work done by the system = w = kJ Heat the water from 10 C to 0 C: (1) q1 = CP dT = CP(T2 T1) = 753 J mol 1 Freeze the water at 0 C: (2) q2 = 6025 J mol 1 Cool the ice from 0 C to 10 C.

2 (3) q3 = 377 J mol 1 Net q = 753 6025 377 = 5649 J mol 1 = H = kJ mol 1 Heat evolved = 6937 = 11 724 J Molar mass of CH3 COCH3 = 3 + 6 + = g mol 1 Heat evolved in the combustion of 1 mol Chapter 2 2 = 11 724 58 080 700 ..J = kJ a. U = kJ mol 1 b. CH3 COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l) n(gases) = 1 H = 972 700 J mol 1 = kJ mol 1 a. Heat capacity of man = 292 600 J K 1 Temperature rise = 10 460 000 J292 600 J K 1 = C Final temperature = 37 + = C b.

3 H = 43 400 J mol 1 g mol 1 = 2411 J g 1 Mass of water required =10 460 000 J2411 J g 1 = 4340 g = kg The reaction is Zn + H2SO4 ZnSO4 + H2(g) Thus, 1 mol of gas is liberated by each mole of Zn, , by g. One hundred grams therefore liberates (100 )mol = mol of H2. The work done by the system is P V: w = P V = nH2RT = mol J K 1 mol 1 K = 3790 J = kJ The work in a sealed vessel ( V = 0) is zero. Enthalpy for this reaction at 298 K is the enthalpy of formation of two moles of water, , H (298 K) = 2 mol ( kJ mol 1) = kJ. At 800 K, using Eq. ( ), we obtain H (800 K)/J = H (298 K) + d(800 298) + 12 e(8002 2982) f (800 1 298 1), where d = 2dH2O (2dH2+ dO2), and e and f are defined similarly.

4 D/(J K 1) = 2 (2 + ) = e/(J K 2) = 2 10 3 (2 10 3 + 10 3) = 10 3 f/(J K) = 2 0 (2 104 105) = 104. Therefore, H (800 K)/J = 483640 (800 298) + 12 10 3 (8002 2982) 104(800 1 298 1) = 105 or kJ. Chapter 2 3 Molar mass of benzene = 6 + 6 = g mol 1 Heat evolved in the combustion of 1 mol = 26 5478 kJ g mol g = kJ a. U = kJ mol 1 b. C6H6(l) + 152 O2(g) 6CO2(g) + 3H2O(l) v(gases) = H = 3 274 900 J mol 1 = kJ mol 1 H = H (C2H6) + H (H2) 2[ H (CH4)] H = 2( ) = kJ mol 1 Molar mass of CH3OH = + 4 + = g mol 1 Amount of methanol = g mol 1 = mol Heat evolved = 119 5032 kJ g mol g = kJ mol 1 = cU a.

5 CU = kJ mol 1 CH3OH(l) + 32 O2(g) CO2(g) + 2H2O(l) v(gases) = cH = 726 500 J mol 1 = kJ mol 1 b. CH3OH(l) + 32 O2(g) CO2(g) + 2H2O(l) (1) H = kJ mol 1 H2(g) + 12 O2(g) H2O(l) (2) f H = kJ mol 1 C(s) + O2(g) CO2(g) (3)

6 F H = kJ mol 1 2 (2) + (3) (1) gives C(s) + 2H2(g) + 12 O2(g) CH3OH(l) (4) f H = 2( ) + = kJ mol 1 Chapter 2 4 This value is slightly different from the value listed in Appendix D. Experimental data is constantly being evaluated causing changes in listed values. This problem may have used older data. c. CH3OH(l) CH3OH(g) vH = kJ mol 1 (5) (4) + (5) gives C(s) + 2H2(g) + 12 O2(g) CH3OH(g) f H = kJ mol 1 2C(graphite) + 3H2(g) C2H6(g) (1) f H = kJ mol 1 C(graphite) + O2(g) CO2(g) (2)

7 F H = kJ mol 1 H2(g) + 12 O2(g) H2O(l) (3) f H = kJ mol 1 3 (3) + 2 (2) (1) gives C2H6(g) + 72 O2(g) 2CO2(g) + 3H2O(l) cH = kJ mol 1 First, perform a multiple regression on z = a + bx + cy using the definitions z = CP,m; x = T and y = 1/T2. The result is z = + 10 2x In other words, we find that d = J K 1 mol 1; e = 10 2 J K 2 mol 1; f = 102 J K mol 1.

8 Below, we present two plots of this function, one in the range 15 T 275, and another, in the range 10 T 25. It can be seen that the function becomes negative at T K. A negative heat capacity is obviously unphysical. This is an indication that the temperature dependence of heat capacities of solids at low temperature cannot be expressed using the model used here (see Chapter 16, Section 6). Function 2 5 C3H6(g) + 92 O2(g) 3CO2(g) + 3H2O(l) (1) cH = kJ mol 1 C(graphite) + O2(g) CO2(g) (2) f H = kJ mol 1 H2(g) + 12 O2(g) H2O(l) (3)

9 F H = kJ mol 1 3 (2) + 3 (3) (1) gives 3C(graphite) + 3H2(g) C3H6(g) f H = kJ mol 1 Perform a multiple regression on z = a + bx + cy using the definitions z = CP,m; x = T and y = 1/T2. The result is z = + 10 3x In other words, we find that d = J K 1 mol 1; e = 10 3 J K 2 mol 1; f = 104 J K mol 1. A plot of the fit is H = + = kJ mol 1 C2H5OH + 3O2 2CO2 + 3H2O (1) cH = kJ mol 1 CH3 CHO + 52 O2 2CO2 + 2H2O (2)

10 CH = kJ mol 1 Function ,mChapter 2 6 CH3 COOH + 2O2 2CO2 + 2H2O (3) cH = kJ mol 1 Reaction (a) is (1) (2); H = + = kJ mol 1 Reaction (b) is (2) (3); H = + = kJ mol 1 Let us list all the reactions involved: 1. CH2 CHCN + 154 O2(g) 3 CO2(g) + 32 H2O(g) + (1/2)N2(g); cH = kJ mol 1 2. C(graphite) + O2(g) CO2(g); fH = kJ mol 1 3.