Use the following information to find Hf of butanol [C4H9OH(l)].

C4H9OH(l) + 6 O2(g) 4 CO2(g) + 5 H2O(g)


Hcomb = -2712.9

Hf of CO2(g) = -393.5 kJ/mol

Hf of H2O(g) = -241.8 kJ/mo

To find the standard enthalpy of formation (Hf) of butanol [C4H9OH(l)], you can use the given information about the combustion reaction and known enthalpies of formation for CO2(g) and H2O(g).

The enthalpy of combustion (Hcomb) is given as -2712.9 kJ/mol, which represents the enthalpy change when one mole of butanol reacts with oxygen to produce carbon dioxide and water.

From the balanced equation:
C4H9OH(l) + 6 O2(g) → 4 CO2(g) + 5 H2O(g)

You can see that one mole of butanol will produce 4 moles of CO2 and 5 moles of H2O.

Using the enthalpy of formation values for CO2 and H2O, you can calculate the enthalpy change for the combustion reaction.

Enthalpy change = (Hf of products) - (Hf of reactants)

Enthalpy change = (4 x Hf of CO2) + (5 x Hf of H2O) - Hf of butanol

Substituting the known enthalpies of formation:
Enthalpy change = (4 x -393.5 kJ/mol) + (5 x -241.8 kJ/mol) - Hf of butanol

Simplifying the equation:
Enthalpy change = -1574 kJ/mol - 1209 kJ/mol - Hf of butanol

The enthalpy change for the combustion reaction is given as -2712.9 kJ/mol. Therefore, we can set up the equation:

-2712.9 kJ/mol = -1574 kJ/mol - 1209 kJ/mol - Hf of butanol

Rearranging the equation to solve for Hf of butanol:
Hf of butanol = -2712.9 kJ/mol + 1574 kJ/mol + 1209 kJ/mol

Calculating the value:
Hf of butanol = -929.9 kJ/mol

Hence, the standard enthalpy of formation (Hf) of butanol [C4H9OH(l)] is -929.9 kJ/mol.

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