The homework question is :
Calculate the Delta H for the following reaction:
C6H6 + O2 -> C + H2O(l)
State whether the reaction is exothermic or endothermic.
I'm not sure where to go with this but so far I balanced out the formula to this
C6H6 + (3/2)O2 -> 6C2 + 3H2O
I found online that the standard enthalpy of formation of C6H6 is 48.95 kJ/mol but where do I go from there?
Is that the equation given for the reaction? USUALLY the products are CO2 and H2O. At any rate, here is what you do.
Yes, balance the equation. Then
dHrxn = (n*dHf procuts) - (n*dHf reactants)
Look up the dHf in your text or notes. Lots of places on the net also. By the way, for C I would not write it as C2 AND you will need to know if it is graphite or not.
dHrxn is - it is exothermic; if + endothermic.
I think it is an error. Usually we have to get CO2 instead of C as it is a combustion reaction and in theory it is an exothermic reaction
To calculate the ΔH (change in enthalpy) for the reaction, you need to use the principles of Hess's law and the standard enthalpies of formation.
Hess's law states that the ΔH for a reaction is equal to the sum of the ΔH for the formation of products minus the sum of the ΔH for the formation of reactants.
1. Start by writing out the balanced equation for the reaction:
C6H6 + (3/2)O2 -> 6C + 3H2O
2. Now, identify the standard enthalpies of formation (ΔHf) for each compound involved in the reaction. You mentioned that the standard enthalpy of formation for C6H6 (benzene) is 48.95 kJ/mol.
3. Calculate the ΔH for the formation of products:
ΔH_products = (6 * ΔHf_C) + (3 * ΔHf_H2O)
= (6 * ΔHf_C) + (3 * ΔHf_H2O(l))
4. Calculate the ΔH for the formation of reactants:
ΔH_reactants = ΔHf_C6H6 + (3/2) * ΔHf_O2
5. Finally, calculate the ΔH for the reaction by subtracting the ΔH for the formation of reactants from the ΔH for the formation of products:
ΔH = ΔH_products - ΔH_reactants
By plugging in the values and performing the calculations, you will be able to determine whether the reaction is exothermic or endothermic based on the sign (positive or negative) of ΔH.
To calculate the ΔH (change in enthalpy) for the given reaction, you will need to use Hess's Law and the standard enthalpy of formation values.
Hess's Law states that the ΔH for a reaction can be calculated by the difference in the standard enthalpy of formation (∆Hf) between the products and the reactants.
First, let's find the standard enthalpy of formation for each compound involved in the reaction. We have the value for C6H6, which is 48.95 kJ/mol. For the other compounds, refer to a reliable source, such as a thermodynamic data table or database.
Next, you need to determine the stoichiometric coefficients that relate the reactants and products. In this case, you have balanced the equation, and it represents the stoichiometry of the reaction correctly:
C6H6 + (3/2)O2 -> 6C + 3H2O
Now, you can use the standard enthalpy of formation values to calculate the ΔH for the reaction. The equation can be rewritten as follows, using the ΔHf values:
[6C + 3H2O] - [C6H6 + (3/2)O2]
Substituting the standard enthalpy of formation values into the equation:
[6(0 kJ/mol) + 3(-285.83 kJ/mol)] - [48.95 kJ/mol + (3/2)(0 kJ/mol)]
Simplifying:
-857.49 kJ/mol - 48.95 kJ/mol
ΔH = -906.44 kJ/mol
Based on the calculated ΔH value, since it is negative (-906.44 kJ/mol), the reaction is exothermic. An exothermic reaction releases heat energy to the surroundings.
Remember that the ΔH value is for the molar quantities specified in the balanced equation.