Questions: Calculate the equilibrium constant from the standard free energy change. Using standard thermodynamic data, calculate the equilibrium constant at 298.15 K for the following reaction. N2(g) + 2 O2(g) -> 2 NO2(g) K =

Solution Steps

The standard free energy change for a reaction can be calculated using the standard free energies of formation (\(\Delta G_f^\circ\)) of the reactants and products. The reaction is:

\[ \mathrm{N}_{2}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \]

The standard free energy change for the reaction is given by:

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

Using standard thermodynamic data:

• \(\Delta G_f^\circ (\mathrm{NO}_2(\mathrm{~g})) = 51.31 \, \text{kJ/mol}\)
• \(\Delta G_f^\circ (\mathrm{N}_2(\mathrm{~g})) = 0 \, \text{kJ/mol}\) (since it is in its standard state)
• \(\Delta G_f^\circ (\mathrm{O}_2(\mathrm{~g})) = 0 \, \text{kJ/mol}\) (since it is in its standard state)

Thus,

\[ \Delta G^\circ = [2 \times 51.31 \, \text{kJ/mol}] - [1 \times 0 \, \text{kJ/mol} + 2 \times 0 \, \text{kJ/mol}] \]

\[ \Delta G^\circ = 102.62 \, \text{kJ/mol} \]

Since the equilibrium constant calculation requires \(\Delta G^\circ\) in Joules, we convert:

\[ \Delta G^\circ = 102.62 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 102620 \, \text{J/mol} \]

The relationship between the standard free energy change and the equilibrium constant is given by:

\[ \Delta G^\circ = -RT \ln K \]

Where:

• \(R = 8.314 \, \text{J/(mol·K)}\) (universal gas constant)
• \(T = 298.15 \, \text{K}\) (temperature)

Rearranging to solve for \(K\):

\[ K = e^{-\Delta G^\circ / (RT)} \]

Substituting the values:

\[ K = e^{-102620 / (8.314 \times 298.15)} \]

\[ K = e^{-41.4011} \]

\[ K \approx 1.073 \times 10^{-18} \]

Final Answer