Using the information below, calculate $\Delta$ H°_f for CH₃OH(l) . 2CH₃OH(l) + 3O₂(g) $\rightarrow$ 2CO₂(g) + 4H₂O(l) , $\Delta$ H° = -1453 kJ
$\Delta$ H°_f for CO₂(g) = -393.5 kJ/mol
$\Delta$ H°_f for H₂O(l) = -286 kJ/mol

Question

Using the information below, calculate

\Delta

H°_f for CH₃OH(l) . 2CH₃OH(l) + 3O₂(g)

\rightarrow

2CO₂(g) + 4H₂O(l) ,

\Delta

H° = -1453 kJ

\Delta

H°_f for CO₂(g) = -393.5 kJ/mol

\Delta

H°_f for H₂O(l) = -286 kJ/mol

Quizplus · Accepted Answer

To find the standard enthalpy of formation for CH₃OH(l), use the given reaction and enthalpies of formation for CO₂(g) and H₂O(l). The reaction enthalpy is -1453 kJ. Using the formula: ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants), solve for ΔH°f(CH₃OH). The calculation gives -239 kJ/mol.

Using the Information Below, Calculate Δ\DeltaΔ H°f for CH3OH(l) →\rightarrow→ 2CO2(g) + 4H2O(l)

Using the Information Below, Calculate $\Delta$ H°_f for CH₃OH(l) $\rightarrow$ 2CO₂(g) + 4H₂O(l)