Question

Calculate ΔG for the following chemical reaction at 298 K.



2C6H6 (l) + 15O2 (g) ⟶ 12CO2 (g) + 6H2O (g)

Answer 1

To calculate the standard Gibbs free energy change (ΔG°) for the reaction

\[
2C_6H_6 (l) + 15O_2 (g) \rightarrow 12CO_2 (g) + 6H_2O (g)
\]

at 298 K, we can use the following formula:

\[
\Delta G° = \Delta H° - T\Delta S°
\]

Where:
- \(\Delta H°\) is the standard enthalpy change,
- \(T\) is the temperature in Kelvin,
- \(\Delta S°\) is the standard entropy change.

To find \(\Delta G°\), we need the values of \(\Delta H°\) and \(\Delta S°\) for the reactants and products at standard conditions (298 K and 1 atm).

### Step 1: Gather Thermodynamic Data

**Standard Enthalpies of Formation (\(ΔH_f^\circ\))**:
- \(ΔH_f^\circ (C_6H_6, l) = 49.04 \, \text{kJ/mol}\)
- \(ΔH_f^\circ (O_2, g) = 0 \, \text{kJ/mol}\) (elements in their standard state)
- \(ΔH_f^\circ (CO_2, g) = -393.5 \, \text{kJ/mol}\)
- \(ΔH_f^\circ (H_2O, g) = -241.8 \, \text{kJ/mol}\)

### Calculation of \(ΔH^\circ\):
Using Hess's law:

\[
\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})
\]

For our reaction:

\[
\Delta H^\circ = \left[12 \times (-393.5) + 6 \times (-241.8)\right] - \left[2 \times 49.04 + 15 \times 0\right]
\]

Calculating:

\[
\Delta H^\circ = \left[12 \times (-393.5) + 6 \times (-241.8)\right] - [2 \times 49.04]
\]

\[
= \left[-4722 + (-1450.8)\right] - [98.08]
\]

\[
= -6172.8 - 98.08 = -6270.88 \, \text{kJ}
\]

### Step 2: Gather Standard Entropy Values (\(S^\circ\))

**Standard Entropy Values (\(S^\circ\))**:
- \(S^\circ (C_6H_6, l) = 173.0 \, \text{J/(mol K)}\)
- \(S^\circ (O_2, g) = 205.0 \, \text{J/(mol K)}\)
- \(S^\circ (CO_2, g) = 213.8 \, \text{J/(mol K)}\)
- \(S^\circ (H_2O, g) = 188.8 \, \text{J/(mol K)}\)

### Calculation of \(ΔS^\circ\):

\[
\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})
\]

For our reaction:

\[
\Delta S^\circ = \left[12 \times 213.8 + 6 \times 188.8\right] - \left[2 \times 173.0 + 15 \times 205.0\right]
\]

Calculating:

\[
\Delta S^\circ = \left[2565.6 + 1132.8\right] - [346 + 3075]
\]

\[
= 3698.4 - 3421 = 277.4 \, \text{J/K}
\]

Thus, in kJ, \( \Delta S^\circ \) is:

\[
\Delta S^\circ = 0.2774 \, \text{kJ/K}
\]

### Step 3: Calculate ΔG°

Now we can use the values to calculate \(ΔG^\circ\):

Using \(T = 298 \, K\):

\[
\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ
\]

\[
\Delta G^\circ = -6270.88 \, \text{kJ} - (298 \, K \times 0.2774 \, \text{kJ/K})
\]

Calculating \(T\Delta S^\circ\):

\[
298 \, K \times 0.2774 \, \text{kJ/K} = 82.66 \, \text{kJ}
\]

Thus:

\[
\Delta G^\circ = -6270.88 \, \text{kJ} - 82.66 \, \text{kJ} = -6353.54 \, \text{kJ}
\]

### Final Result

\[
\Delta G^\circ = -6353.54 \, \text{kJ}
\]

This is the standard Gibbs free energy change for the combustion of benzene at 298 K.