Question

Calculate the standard enthalpy change for the combustion of 1 mole of propane gas (C3H8) at standard conditions using the following balanced chemical equation: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)Given the standard enthalpy of formation values at 298K for C3H8(g), CO2(g), and H2O(l) as -103.8, -393.5, and -285.8 kJ/mol respectively, calculate the standard enthalpy change for the combustion reaction.

Answer 1

To calculate the standard enthalpy change for the combustion of 1 mole of propane gas, we can use the following equation:H_combustion =  H_f products  -  H_f reactants where H_combustion is the standard enthalpy change for the combustion reaction, H_f products  is the sum of the standard enthalpies of formation of the products, and H_f reactants  is the sum of the standard enthalpies of formation of the reactants.For the given balanced chemical equation:C3H8 g  + 5O2 g  -> 3CO2 g  + 4H2O l The standard enthalpy of formation values at 298K are:C3H8 g : -103.8 kJ/molCO2 g : -393.5 kJ/molH2O l : -285.8 kJ/molFirst, we calculate the sum of the standard enthalpies of formation of the products:3 *  -393.5 kJ/mol  + 4 *  -285.8 kJ/mol  = -1180.5 kJ/mol - 1143.2 kJ/mol = -2323.7 kJ/molNext, we calculate the sum of the standard enthalpies of formation of the reactants:1 *  -103.8 kJ/mol  + 5 * 0 kJ/mol = -103.8 kJ/molNote that the standard enthalpy of formation of O2 g  is 0 kJ/mol because it is an element in its standard state.Now, we can calculate the standard enthalpy change for the combustion reaction:H_combustion = -2323.7 kJ/mol -  -103.8 kJ/mol  = -2219.9 kJ/molTherefore, the standard enthalpy change for the combustion of 1 mole of propane gas is -2219.9 kJ/mol.