Calculation of standard enthalpy change for the combustion of propane.If 5 moles of propane react completely in a bomb calorimeter under standard conditions, calculate the standard enthalpy change for the combustion of propane. The balanced chemical equation for the combustion of propane is:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Given the standard enthalpy of formation at 25°C and 1 atm for C3H8(g), CO2(g), H2O(g) are -103.8 kJ/mol, -393.5 kJ/mol and -241.8 kJ/mol respectively.

Question

Calculation of standard enthalpy change for the combustion of propane.If 5 moles of propane react completely in a bomb calorimeter under standard conditions, calculate the standard enthalpy change for the combustion of propane. The balanced chemical equation for the combustion of propane is:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Given the standard

Answer 1

To calculate the standard enthalpy change for the combustion of propane, 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 of propane, H_f products  are the standard enthalpies of formation for the products, and H_f reactants  are the standard enthalpies of formation for the reactants.From the balanced chemical equation, we have:Reactants: 1 mol of C3H8 g  and 5 mol of O2 g Products: 3 mol of CO2 g  and 4 mol of H2O g The standard enthalpy of formation for O2 g  is 0 kJ/mol since it is in its elemental form.Now, we can plug the values into the equation:H_combustion = [ 3 mol  -393.5 kJ/mol  +  4 mol  -241.8 kJ/mol ] - [ 1 mol  -103.8 kJ/mol  +  5 mol  0 kJ/mol ]H_combustion = [-1180.5 kJ +  -967.2 kJ ] - [-103.8 kJ]H_combustion = -2147.7 kJ + 103.8 kJH_combustion = -2043.9 kJThe standard enthalpy change for the combustion of propane is -2043.9 kJ/mol. Since we are given 5 moles of propane, the total enthalpy change for the combustion of 5 moles of propane is:Total enthalpy change =  -2043.9 kJ/mol   5 mol = -10219.5 kJ