Enthalpy And Hess's Law

Question

Given the enthalpies of formation: $\Delta_f H^\circ(CO_2) = -393.5$ kJ/mol, $\Delta_f H^\circ(H_2O, l) = -285.8$ kJ/mol, and $\Delta_f H^\circ(C_2H_5OH, l) = -277.7$ kJ/mol, what is the standard enthalpy of combustion of ethanol?

RankGuru · Accepted Answer

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpies of formation of products minus the sum of the enthalpies of formation of reactants, each multiplied by their stoichiometric coefficients. The balanced combustion equation for ethanol is: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$. Applying Hess's Law: $\Delta_c H^\circ = [2(-393.5) + 3(-285.8)] - [(-277.7) + 0]$. Products: $2(-393.5) = -787.0$ and $3(-285.8) = -857.4$; total products = $-787.0 + (-857.4) = -1644.4$ kJ. Reactants (excluding $O_2$ whose $\Delta_f H^\circ = 0$): $-277.7$ kJ. Therefore $\Delta_c H^\circ = -1644.4 - (-277.7) = -1366.7 \approx -1366.8$ kJ/mol. Option -1235.5 kJ/mol omits one mole of water. Option -957.1 kJ/mol uses incorrect stoichiometry for $CO_2$. Option -1644.3 kJ/mol forgets to subtract the formation enthalpy of ethanol. This is a standard NCERT Hess's Law application. Plausibility check: combustion of ethanol is highly exothermic, so a large negative value is expected.

Enthalpy And Hess's Law

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