What is the balanced equation for the combustion of methanol, and how do you calculate ΔH°, ΔS°, and ΔG° at 25°C? Is it spontaneous?

The balanced equation for the combustion of methanol (CH₃OH) is:

2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O (g)

To calculate the standard enthalpy change (ΔH°), standard entropy change (ΔS°), and standard Gibbs free energy change (ΔG°) at 25°C, we can use standard thermodynamic values for the reactants and products:

1. **ΔH° Calculation:**

ΔH° for the reaction can be calculated using standard enthalpy of formation values (ΔH°f) from a table:

• ΔH°f (CH₃OH) = -238.6 kJ/mol
• ΔH°f (CO₂) = -393.5 kJ/mol
• ΔH°f (H₂O) = -241.8 kJ/mol

Substituting these values into the formula:

ΔH° = [2(ΔH°f CO₂) + 4(ΔH°f H₂O)] – [2(ΔH°f CH₃OH) + 3(ΔH°f O₂)]

ΔH° = [2(-393.5) + 4(-241.8)] – [2(-238.6) + 3(0)]

Calculating this gives:

ΔH° = [-787 – 967.2] – [-477.2] = -1270.2 + 477.2 = -793 kJ

2. **ΔS° Calculation:**

ΔS° can be calculated using standard entropy values (S°):

• S° (CH₃OH) = 126.0 J/(mol·K)
• S° (O₂) = 205.0 J/(mol·K)
• S° (CO₂) = 213.7 J/(mol·K)
• S° (H₂O) = 188.8 J/(mol·K)

Using the formula:

ΔS° = [2(S° CO₂) + 4(S° H₂O)] – [2(S° CH₃OH) + 3(S° O₂)]

ΔS° = [2(213.7) + 4(188.8)] – [2(126.0) + 3(205.0)]

Calculating gives:

ΔS° = [427.4 + 755.2] – [252 + 615] = 1182.6 – 867 = 315.6 J/K

3. **ΔG° Calculation:**

Finally, we can calculate ΔG° using the relationship ΔG° = ΔH° – TΔS°.

At 25°C (298 K):

ΔG° = -793 kJ – (298 K)(315.6 J/K * 1 kJ/1000 J)

ΔG° = -793 kJ – (298 * 0.3156) kJ = -793 – 94.011 (approximately) = -887.0 kJ

4. **Spontaneity:**

A reaction is spontaneous if ΔG° < 0. Since ΔG° is -887.0 kJ, the combustion of methanol is indeed spontaneous at 25°C.