SOLUTIONS to Problem Set I - EEE 2010

H mol e solution to problem 1236 svna li 12 cl2

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Unformatted text preview: 3)2(20.51 H2O) (Table C.4) ΔH1 = 6 (−285.83 kJ) ΔH2 = −302.9 kJ; ΔH3 = − (−2110.8 kJ); ΔH4 = −47.84 kJ ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4; ΔH = 45.08 kJ This value is for 1 mol of the hydrated copper nitrate. On the basis of 1 second Q=1830kJ/sec Q = n1.ΔH / mol; e. Solution to problem 12.36 (SVNA) Li + 1/2 Cl2 + (n+2) H2O ---> LiCl(n+2 H2O) (1) 2(H2 + 1/2 O2 ---> H2O) (2) LiCl.2H2O ---> Li + 1/2 Cl2 + 2H2 + O2 (3) -------------------------------------------------------------------------------------LiCl.2H2O + n H2O ---> LiCl(n+2 H2O) ΔH2 = 2 (−285.83 kJ); ΔH3 = 1012.65 kJ (Table C.4) Since the process is isothermal, ΔH = ΔH1 + ΔH2 + ΔH3 Since it is also adiabatic, ΔH = 0 Therefore, ΔH1 = −ΔH2 − ΔH3; ΔH1 = −440.99 kJ Interpolation in the table on pg. 457 shows that the LiCl is dissolved in 8.878 moles H2O xLiCl = 1/9.878; xLiCl = 0.1012 f. Solution to problem 12.37(SVNA) Ca + Cl2 + n H2O ---> CaCl2 (n H2O) CaCl2(s) ---> Ca + Cl2 -------------------------------------------CaCl2(s) + n H2O ---> CaCl2 (n H2O) From Table C.4: ΔHfCaCl2:= −795.8⋅kJ Plot (1)-(2) vs. ni (no. of moles) ΔHf (1) ΔHfCaCl2 (2) ΔHtilde g. Solution to problem 12.38(SVNA) CaCl2 ---> Ca + Cl2 (1) 2(Ca + Cl2 + 12.5 H2O ---> CaCl2 (12.5 H2O) (2) CaCl2 (25 H2O) ---> Ca + Cl2 + 25 H2O (3) -----------------------------------------------------------------------------------CaCl2 (25 H2O) + CaCl2 ---> 2 CaCl2 (12.5 H2O) ΔH1 = 795.8 kJ (Table C.4) ΔH2 = 2 (−865.295 kJ); ΔH3 = 871.07⋅kJ Q = ΔH; Q = −63.72 kJ Ans. ΔH = ΔH1 + ΔH2 +ΔH3; h. Solution to problem 12.39(SVNA) The process may be considered in two steps: Mix at 25 °C, then heat/cool solution to the final temperature. The two steps together are adiabatic and the overall enthalpy change is 0. Calculate moles H2O needed to form solution: n = 34.911 moles of H2O per mole CaCl2 in final solution Moles of water added per mole of CaCl2.6H2O; n-6 = 28.911 Basis: 1 mole of CaCl2.6H2O dissolved CaCl2.6H2O(s) ---> Ca + Cl2 + 6 H2 + 3 O2 (1) Ca + Cl2 + 34.991 H2O --->CaCl2 (34.911 H2O) (2) 6(H2 + 1/2 O2 ---> H2O) (3) --------------------------------------------------------------------------------------CaCl2.6H2O + 28.911 H2O ---> CaCl2 (34.911 H2O) ΔH1 = 2607.9⋅kJ; ΔH3 = 6 (−285.83⋅kJ) (Table C.4) ΔH2 = −871.8⋅kJ (Pb. 12.37) ΔH298 = ΔH1 + ΔH2 + ΔH3 for reaction at 25 °C ΔH298 = 21.12 kJ; msoln = (110.986 + 34.911⋅18.015) gm; msoln = 739.908 gm ΔH298 + CP⋅ΔT = 0; CP = 3.28 kJ/ (kg °C); ΔT = −ΔH298 / (msoln⋅CP) ΔT = −8.702°C; T = 25°C + ΔT; T = 16.298 °C Ans i. Solution to problem 12.43(SVNA) m1 = 150⋅lb (H2SO4); m2 = 350⋅lb (25% soln.) H1=8 BTU/lbm; H2 = -23 BTU/lbm (Fig. 12.17) Final Solution: (100% m1 + 25% m2) / (m1+m2) = 47.5% m3= m1+m2 H3= -90 BTU/lbm (Fig. 12.17) Q = m3H3 – (m1H1 + m2H2) Q = -38150 BTU Ans. j. Solution to problem 12.44(SVNA) Enthalpies from Fig. 12.17 x1 = 0.5; x2 = 1 − x1; H = −69 BTU/lbm (50 % soln) H...
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