Calculate the standard enthalpy of formation of liquid methanol, CH3OH (l), using the following information: C (graphite) + O2 LaTeX: / longrightarrow ⟶ CO2 (g) LaTeX: / Delta Δ H° = - 393.5 kJ/mol H2 (g) + O2 LaTeX: / longrightarrow ⟶ H2O (l) LaTeX: / Delta Δ H° = - 285.8 kJ/mol CH3OH (l) + O2 (g) LaTeX: / longrightarrow ⟶ CO2 (g) + 2H2O (l) LaTeX: / Delta Δ H° = - 726.4 kJ/mol

-238.54 kJ/mol.

Explanation:

We need to calculate the standard enthalpy of formation of liquid methanol, CH₃OH (l) that has the equation:

C (graphite) + 2H₂ (g) + ½ O₂ (g) → CH3OH (l) ΔHf° = ?? kJ/mol.

using the information of the three equations:

(1) C (graphite) + O₂ (g) → CO₂ (g), ΔHf₁° = - 393.5 kJ/mol.

(2) H2 (g) + ½ O₂ (g) → H₂O (l), ΔHf₂° = - 285.8 kJ/mol.

(3) CH₃OH (l) + 3/2 O₂ (g) → CO₂ (g) + 2H₂O (l), ΔH₃° = - 726.56 kJ/mol.

We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation:

equation (1) be as it is:

(1) C (graphite) + O₂ (g) → CO₂ (g), ΔHf₁° = - 393.5 kJ/mol.

equation (2) should be multiplied by (2) and also the value of ΔHf₂°:

(2) 2H2 (g) + O₂ (g) → 2H₂O (l), ΔHf₂° = 2x (-285.8 kJ/mol).

equation (3) should be reversed and also the value of ΔH₃° should be multiplied by (-1):

(3) CO₂ (g) + 2H₂O (l) → CH₃OH (l) + 3/2 O₂ (g), ΔH₃° = 726.56 kJ/mol.

By summing the modified equations, we can get the needed equation and so:

The standard enthalpy of formation of liquid methanol, CH₃OH (l) = ΔHf₁° + 2 (ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol) + 2 (-285.8 kJ/mol) - ( - 726.56 kJ/mol) = - 238.54 kJ/mol.