Test02SOL-docB

# Hence pext dv ncvm dt this equation is integrated as

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Unformatted text preview: entical to the initial state for the process described in Question 8. Similarly, the final state is the same for the processes described in Questions 8 and 9. Entropy is a state function; hence, its change is independent of the path as long as the same initial and final states are considered. So, ΔS for questions 8 and 9 must have the same value. Δ S = - 161 J/K 3 Question 10: (10 points) 0.3177 g phenol (C6H5OH(s)) is placed in an adiabatic constant volume bomb calorimeter of heat capacity, o - 1 CCAL = 8.30 kJ. C (this heat capacity accounts for both the calorimeter parts and the water in which the bomb is placed). Combustion of phenol with an excess oxygen at 298.15 K leads to formation of CO2(g) and H2O(l). Using the data in the thermodynamic table below, calculate the temperature change in the calorimeter. o Substance M (g/mol) ΔfH 298.15 K (kJ/mol) C6H5OH (s) - 165.00 94.12 CO2 (g) - 393.51 44.04 H2O (l) - 285.83 18.015 O2 (g) - - - 32.00 The combustion reaction for phenol is written as: C6H5OH (s) + 7 O2 (g) à༎ 6 CO2 (g) + 3 H2O (l) The calorimeter equation is q = qCAL + qRXN = 0 since the calorimeter is adiabatic and the heat of reaction is transferred to the calorimeter only (no leak). o The heat of reaction is given by: qRXN = n(RXN) x ΔRU 298.15K where n(RXN) is the number of moles of o reactions that are carried out in the calorimeter and ΔRU 298.15K is the standard reaction energy change for the reaction equation as written above. ΔRUo298.15K is related to ΔRHo298.15K using: ΔRUo298.15K = ΔRHo298.15K – RT Δng = ΔRHo298.15K – (- 1) x RT since there is a net loss of one mole of gas for the combustion reaction as written above. Using the standard formation enthalpies of reactants and products, we obtain: ΔRHo298.15K = 6xΔfHo298.15K (CO2 (g)) + 3xΔfHo298.15K (H2O (l)) - ΔfHo298.15K (C6H5OH (s)) ΔRHo298.15K = 6x(- 393.51 kJ/mol) + 3x(- 285.83 kJ/mol) – (- 165.00 kJ/mol) Δ RHo298.15K = - 3053.55 kJ/mol ΔRUo298.15K = - 3053.55 kJ/mol + 0.0083145 kJ/mol.K x 298.15K Δ RUo298.15K = - 3051.07 kJ/mol n(RXN) = (0.3177 g phenol / 94.12 g phenol /mol phenol) x (1 mol RXN/mol phenol) n(RXN) = 0.0033754 mol RXN o The heat of reaction is calculate...
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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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