50 mol of h2o vapor from o o 130 c to 100 c s1 n

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Unformatted text preview: rried out at 298.15K (ΔH(2)). In the third step, the products are heated isobarically from 298.15K to 348.15K (ΔH(3)). The standard reaction enthalpy at 348.15 K is obtained as the sum of the enthalpy changes associated o with first, second and third steps. ΔRH 348.15 K (kJ/mol) = ΔH(1) + ΔH(2) + ΔH(3) o ΔH(1) = Cpm 298.15 K (CH4(g)) x (298.15K – 348.15K) + 2 x Cpmo 298.15 K (Cl2(g)) x (298.15K – 348.15K) ΔH(1) = (35.3 + 2 x 33.9 J/mol.K) x (298.15K – 348.15K) = 103.1 J/mol.K x - 50.00 K = - 5155 J/mol Δ H(1) = - 5.155 kJ/mol ΔH(2) = ΔRHo298.15 K (kJ/mol) = - 53.6 kJ/mol ΔH(3) = Cpmo 298.15 K (CCl4(l)) x (298.15K – 348.15K) + 2 x Cpmo 298.15 K (H2(g)) x (298.15K – 348.15K) ΔH(3) = (131.3 + 2 x 28.8 J/mol.K) x (348.15K – 298.15K) = 188.9 J/mol.K x 50.00 K Δ H(3) = 9.445 kJ/mol ΔRHo348.15 K (kJ/mol) = - 5.155 kJ/mol – 53.6 kJ/mol + 9.445 kJ/mol Δ RHo348.15 K (kJ/mol) = - 49.3 kJ/mol 2 Question 8: (10 points) o o Given the value for the standard vaporization enthalpy of H2O at 100. C and 1 atm (ΔvapH = 40.7 kJ/mol), calculate the change in entropy for a system composed of 1.50 mol of water as a result of the reversible o (very slow) decrease in temperature from the initial state characterized by 1.50 mol H2O(g) at 130. C and o 1.00 atm to a final state characterized by 1.50 mol H2O(l) at 30. C and 1.00 atm. The relevant constant o o pressure molar heat capacities are CPm (H2O(l)) = 75.3 J/K.mol and CPm (H2O(g)) = 33.6 J/K.mol. The...
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