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

Step-by-step solution

Molar mass of ethanol ( C₂H₅OH ) = 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: C₂H₅OH(l) + 3O₂(g) 2CO₂(g) + 3H₂O(l) Change in the number of gaseous moles, n_g = 2 - 3 = -1 Now, calculating the standard enthalpy of combus