The standard enthalpy of combustion of liquid acetone \((CH_3)_2CO\) is \(-1790 \, \text{kJ/mol}\) at \(25^\circ C\). What is the \(\Delta H^\circ_f, 298\) of acetone in \(\text{kJ/mol}\)?

Given:
- \(\Delta H^\circ_f, 298 \, (\text{H}_2\text{O}(l)) = -286 \, \text{kJ/mol}\)
- \(\Delta H^\circ_f, 298 \, (\text{CO}_2(g)) = -393.5 \, \text{kJ/mol}\)

Options:
a) 895
b) -145
c) 145
d) 0
e) 447.5

The ΔHf,298 (standard enthalpy of formation at 298K) of acetone is -145 kJ/mole, calculated by using the known standard enthalpies of water and carbon dioxide.

The answer is -145 kJ/mole. The standard enthalpy of combustion is the energy change when one mole of a substance burns completely in oxygen under standard conditions. In this case, we know the enthalpy of combustion of acetone, and the enthalpies of formation of water and carbon dioxide at 298 K.

The description works through the principle that the enthalpy change for a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In this case, the combustion reaction yields water and carbon dioxide as products, and acetone is the reactant. Therefore, the enthalpy change for the acetone combustion is equal to (2 ΔHf,298 H2O(l) + 3 ΔHf,298 CO2(g)) - (ΔHf,298 acetone). Plugging in known values gives (-2(286) + 3(-393.5)) - ΔHf,298 acetone = -1790. From this, ΔHf,298 acetone = -145 kJ/mole.

Hence, the ΔHf,298 of acetone is -145 kJ/mole.

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