Question

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

Answer 1

Answer:

-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.