Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data:

2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) ΔH∘ = -6534kJ
ΔH∘f (CO2) = -393.5kJ/mol
ΔH∘f (H2O) = - 285.8kJ/mol

To calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, we need to use the given equation and the enthalpies of formation (ΔH∘f) for CO2 and H2O.

The balanced equation for the reaction is:
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l)

From the equation, we can see that for every 12 moles of CO2 formed, 2 moles of C6H6 are consumed. Therefore, we can write the equation in terms of moles of CO2 as follows:
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l)

Now we can calculate the ΔH∘f for benzene, C6H6:

1. Calculate the moles of CO2 formed:
12 moles of CO2 are formed from 2 moles of C6H6.

2. Calculate the enthalpy change for the formation of 12 moles of CO2:
ΔH∘ = - 6534 kJ (given)

3. Calculate the enthalpy change for the formation of 1 mole of CO2:
ΔH∘ (CO2) = ΔH∘ / 12 = - 6534 kJ / 12 = - 544.5 kJ/mol

4. Calculate the enthalpy change for the formation of 1 mole of C6H6:
From the balanced equation, we can see that for every 2 moles of C6H6, 12 moles of CO2 are formed. Therefore, the enthalpy change for the formation of 1 mole of C6H6 is:
ΔH∘f (C6H6) = ΔH∘ (CO2) x (2/12) = - 544.5 kJ/mol x (2/12) = - 90.75 kJ/mol

Therefore, the ΔH∘f (enthalpy of formation) for benzene, C6H6, is -90.75 kJ/mol.

To calculate ΔH∘f (standard enthalpy of formation) for benzene, C6H6, we'll use the given chemical equation and the enthalpy values for CO2 and H2O.

The balanced chemical equation is:
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l)

We know the enthalpy change (ΔH∘) for the reaction is -6534 kJ/mol. This value represents the enthalpy change for the formation of 12 moles of CO2 and 6 moles of H2O.

We also know the enthalpy of formation (ΔH∘f) for CO2 is -393.5 kJ/mol and for H2O is -285.8 kJ/mol. These values represent the enthalpy change for the formation of 1 mole of CO2 and 1 mole of H2O, respectively.

From the chemical equation, we can see that to form 12 moles of CO2 and 6 moles of H2O, we need 2 moles of C6H6.

ΔH∘f (C6H6) = [ΔH∘ × (moles of products)] / (moles of C6H6)

Plugging in the values, we get:
ΔH∘f (C6H6) = [-6534 kJ × (12 CO2 + 6 H2O)] / (2 C6H6)

Simplifying further:
ΔH∘f (C6H6) = [-6534 kJ × (12 × -393.5 kJ/mol + 6 × -285.8 kJ/mol)] / (2 mol)

Calculating the expression in the brackets:
ΔH∘f (C6H6) = [-6534 kJ × (-4722 kJ + -1714.8 kJ)] / 2 mol
ΔH∘f (C6H6) = [-6534 kJ × (-6436.8 kJ)] / 2 mol

Evaluating the expression:
ΔH∘f (C6H6) = 42096295.2 kJ/mol

Therefore, the standard enthalpy of formation ΔH∘f for benzene, C6H6, is approximately 42096 kJ/mol.

dHorxn = (n*dHo formation) - (n*dHo proucts).