Test02SOL-docB

15 k kjmol h1 h2 h3 o h1 cpm 29815 k ch4g

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Unformatted text preview: when calculating standard reaction enthalpies. d) The temperature change observed in an adiabatic bomb calorimeter is always positive whenever an exothermic reaction takes place. e) The heat of reaction in an adiabatic bomb calorimeter is equal to the reaction enthalpy change. f) The standard formation enthalpy of iodine is only equal to zero when iodine is in the pure solid molecular state (I2(s)) under standard pressure. Question 7: (10 points) Using the thermodynamic data reported at 298.15K (shown below), determine the standard reaction enthalpy at 398.15K for the reaction: CH4(g) + 2 Cl2(g) à༎ CCl4(l) + 2 H2(g) o o Substance Cpm 298.15 K (J/K.mol) ΔfH 298.15 K (kJ/mol) CH4(g) - 74.8 35.3 Cl2(g) - - - 33.9 CCl4(l) - 128.4 131.3 H2(g) - - - 28.8 The standard reaction enthalpy is given at 298.15 K by: ΔRHo298.15 K (kJ/mol) = [ΔFHo298.15 K (CCl4(l)) + 2 ΔFHo298.15 K (H2(g))] – [ΔFHo298.15 K (CH4(g)) + 2 ΔFHo298.15 K (Cl2(g))] ΔRHo298.15 K (kJ/mol) = [- 128.4 kJ/mol] – [- 74.8 kJ/mol] ΔRHo298.15 K (kJ/mol) = - 53.6 kJ/mol Calculation of the standard reaction enthalpy at 398.15K can be carried out imagining a cycle where: In the first step, reactants are cooled isobarically from 398.15K to 298.15K (ΔH(1)). In the second step, the reaction is carried out at 298.15K (ΔH(2)). In the third step, the products are heated isobarically from 298.15K to 398.15K (ΔH(3)). The standard reaction enthalpy at 398.15 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 – 398.15K) + 2 x Cpmo 298.15 K (Cl2(g)) x (298.15K – 398.15K) ΔH(1) = (35.3 + 2 x 33.9 J/mol.K) x (298.15K – 398.15K) = 103.1 J/mol.K x - 100.00 K = - 10310 J/mol Δ H(1) = - 10.31 kJ/mol ΔH(2) = ΔRHo298.15 K (kJ/mol) = - 53.6 kJ/mol ΔH(3) = Cpmo 298.15 K (CCl4(l)) x (298.15K – 398.15K) + 2 x Cpmo 298.15 K (H2(g)) x (298.15K – 398.15K) ΔH(3) = (131.3 + 2 x 28.8 J/mol.K) x (398.15K – 298.15K) = 188.9 J/mol.K x 100.00 K Δ H(3) = 18.89 kJ/mol ΔRHo348.15 K (kJ/mol) = - 10.31 kJ/mol – 53.6 kJ/mol + 18.89 kJ/mol Δ RHo348.15 K (kJ/mol) = - 45.0 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 (...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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