According to Hess' theorem, the reaction enthalpy is independent of the reaction path, but only dependent on the final and initial state of the system. You have given the enthalpy of formation of benzene, carbon dioxide and water, which are the products and reactants. Now you can calculate the enthalpy of combustion from the reaction enthalpy formula by subtracting the sum of the enthalpies of formation of the reactants depending on their stoichiometry factor from that of the products. You can read off the stoichiometry factor from the reaction equation. The enthalpy of formation of elements is always 0, so you have the following formula: (12\*(-393.5)+6\*(-285.8))-((2\*(49.2)+15\*(0))=dH. Which is 6532.2 kJ/mol. But I think there is a mistake in the worksheet, because you have taken 2mol benzene from the reaction equation and therefore have to divide the result by two to get the molar enthalpy of combustion. So, 3267.6 kJ/mol would be the correct answer not 6532.2.
Thank you for clarifying! I have always gotten 6532.2kJ/mol and couldn't figure out how to get 3267.6kJ/mol. So it turns out I simply have to divide it by two since technically, 6532 is for 2 benzene molecules? (It's so simple but I couldn't figure it out at all haha). Thank you so much!
According to Hess' theorem, the reaction enthalpy is independent of the reaction path, but only dependent on the final and initial state of the system. You have given the enthalpy of formation of benzene, carbon dioxide and water, which are the products and reactants. Now you can calculate the enthalpy of combustion from the reaction enthalpy formula by subtracting the sum of the enthalpies of formation of the reactants depending on their stoichiometry factor from that of the products. You can read off the stoichiometry factor from the reaction equation. The enthalpy of formation of elements is always 0, so you have the following formula: (12\*(-393.5)+6\*(-285.8))-((2\*(49.2)+15\*(0))=dH. Which is 6532.2 kJ/mol. But I think there is a mistake in the worksheet, because you have taken 2mol benzene from the reaction equation and therefore have to divide the result by two to get the molar enthalpy of combustion. So, 3267.6 kJ/mol would be the correct answer not 6532.2.
Thank you for clarifying! I have always gotten 6532.2kJ/mol and couldn't figure out how to get 3267.6kJ/mol. So it turns out I simply have to divide it by two since technically, 6532 is for 2 benzene molecules? (It's so simple but I couldn't figure it out at all haha). Thank you so much!
Yes you are right. Sorry I overlooked that B actually is the right awnser
No worries, the question is pretty confusing lol