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Test02SOL-docA

# 17 jk s12 57437717 jk 1974 jk since the change

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Unformatted text preview: 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 = 2xΔfHo298.15K (CO2 (g)) + 3xΔfHo298.15K (H2O (l)) - ΔfHo298.15K (C2H5OH (l)) ΔRHo298.15K = 2x(- 393.51 kJ/mol) + 3x(- 285.83 kJ/mol) – (- 267.69 kJ/mol) = - 1376.82 KJ/mol ΔRUo298.15K = - 1376.82 kJ/mol + 0.0083145 kJ/mol.K x 298.15K = - 1374.34 kJ/mol n(RXN) = (0.3177 g ethanol / 46.07 g ethanol /mol ethanol) x (1 mol RXN/mol ethanol) n(RXN) = 0.006986 mol RXN o The heat of reaction is calculated from ΔRU 298.15K using: qRXN = (0.006986 mol RXN) x (- 1374.34 kJ/mol) = - 9.4775 kJ qCAL is calculated using qCAL = CCAL ΔT where ΔT is the temperature rise in the calorimeter as a result of the exothermic combustion of 0.3177 g ethanol. Hence, CCAL ΔT = - qRXN and ΔT = - qRXN / CCAL ΔT = (+13.56 kJ) / (8.30 kJ/ oC- 1 ) = 1.1419 K With three significant...
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