Question

Calculate the standard enthalpy change for the reaction in which 6 moles of calcium oxide reacts with 3 moles of carbon dioxide to produce 2 moles of calcium carbonate. The equation for the reaction is:CaO(s) + CO2(g) → CaCO3(s)Given the standard enthalpy of formation values: ΔHf°[CaO(s)] = -635.09 kJ/mol, ΔHf°[CO2(g)] = -393.51 kJ/mol, and ΔHf°[CaCO3(s)] = -1207.44 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H reaction  =  Hf products  -  Hf reactants First, we need to determine the moles of each substance involved in the reaction. The balanced equation is:3 CaO s  + 3 CO2 g   3 CaCO3 s Now, we can calculate the enthalpy change for the reaction:H reaction  = [3 * Hf CaCO3 ] - [3 * Hf CaO  + 3 * Hf CO2 ]H reaction  = [3 *  -1207.44 kJ/mol ] - [3 *  -635.09 kJ/mol  + 3 *  -393.51 kJ/mol ]H reaction  = [-3622.32 kJ] - [ -1905.27 kJ  +  -1180.53 kJ ]H reaction  = -3622.32 kJ + 3085.80 kJH reaction  = -536.52 kJThe standard enthalpy change for the reaction is -536.52 kJ.