16. From the standard enthalpies of formation, calculate the ƒ´Horxn for the reaction: (ƒ´Hof = -151.9 kJ/mol for C6H12.)

C6H12(l) + 9 O2(g) „³ 6 CO2 (g) + 6 H2O(l)

Haven't you seen this before?

dHrxn = (n*dHf products) - (n*dHf reactants)
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To calculate the standard enthalpy change (ΔHorxn) for a reaction using standard enthalpies of formation (ΔHof), you can use the following equation:

ΔHorxn = ΣnΔHof(products) - ΣnΔHof(reactants)

Where:
- ΔHorxn is the standard enthalpy change for the reaction
- ΣnΔHof(products) is the sum of the standard enthalpies of formation of the products, each multiplied by their respective stoichiometric coefficient (n)
- ΣnΔHof(reactants) is the sum of the standard enthalpies of formation of the reactants, each multiplied by their respective stoichiometric coefficient (n)

In this case, we are given the standard enthalpy of formation for C6H12 as -151.9 kJ/mol. However, we don't have the standard enthalpies of formation for CO2(g) and H2O(l). To obtain these values, you can refer to a reference table or look them up in a chemistry database.

Let's assume that the standard enthalpy of formation for CO2(g) is -393.5 kJ/mol and for H2O(l) is -285.8 kJ/mol.

Using the equation above, we can calculate the ΔHorxn:

ΔHorxn = [6ΔHof(CO2) + 6ΔHof(H2O)] - [ΔHof(C6H12) + 9ΔHof(O2)]

ΔHorxn = [6*(-393.5 kJ/mol) + 6*(-285.8 kJ/mol)] - [(-151.9 kJ/mol) + 9*(0 kJ/mol)]

ΔHorxn = [-2361 kJ/mol - 1715 kJ/mol] - [-151.9 kJ/mol]

ΔHorxn = -4076 kJ/mol + 151.9 kJ/mol

ΔHorxn = -3924.1 kJ/mol

Therefore, the ΔHorxn for the given reaction is approximately -3924.1 kJ/mol.