If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The | H_f°| of ethanol at 298.15 K is ______ 10^2 kJ mol^ -1 . (Nearest integer) Given: Standard enthalpy for combustion of graphite =-393.5 kJ mol^ -1 Standard enthalpy of formation of water (l)=-285.8 kJ mol^ -1 Molar mass in g mol^ -1 of C, H, O are 12, 1 and 16 respectively
A. A
Molar mass of ethanol (C_2H_5OH) = 2(12) + 6(1) + 16 = 46 g mol^ -1 Number of moles of ethanol burnt, n = 3.365 46 = 0.07315 mol Since the combustion occurs in a bomb calorimeter (constant volume), the heat produced corresponds to the change in internal energy ( U_c^ ). U_c^ = - 99.472 0.07315 = -1359.8 kJ mol^ -1 The balanced chemical equation for the combustion of ethanol is: C_2H_5OH(l) + 3O_2(g) 2CO_2(g) + 3H_2O(l) Change in the number of gaseous moles, n_g = 2 - 3 = -1 Now, calculating the standard enthalpy of combustion ( H_c^ ): H_c^ = U_c^ + n_g RT H_c^ = -1359.8 + (-1) (8.314 10^ -3 kJ K ^ -1 mol ^ -1 ) 298.15 K H_c^ = -1359.8 - 2.48 = -1362.28 kJ mol^ -1 The standard enthalpy of combustion can also be written in terms of enthalpies of formation: H_c^ = H_f^ ( products ) - H_f^ ( reactants ) H_c^ = [2 H_f^ (CO_2, g) + 3 H_f^ (H_2O, l)] - [ H_f^ (C_2H_5OH, l) + 3 H_f^ (O_2, g)] Substituting the given values (note that H_f^ of O_2(g) is 0): -1362.28 = [2(-393.5) + 3(-285.8)] - H_f^ (C_2H_5OH, l) -1362.28 = [-787.0 - 857.4] - H_f^ (C_2H_5OH, l) -1362.28 = -1644.4 - H_f^ (C_2H_5OH, l) H_f^ (C_2H_5OH, l) = -1644.4 + 1362.28 = -282.12 kJ mol^ -1 The magnitude is | H_f^ | = 282.12 kJ mol^ -1 = 2.8212 10^2 kJ mol^ -1 Rounding to the nearest integer, we get 3. Answer: 3
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