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The standard free energy change, denoted as \( \Delta_{\mathrm{r}} G^{\circ} \), is a measure of the energy difference between reactants and products under standard conditions. It helps predict whether a reaction will occur spontaneously. If \( \Delta G^{\circ} \) is negative, the reaction can proceed spontaneously. If it's positive, like in the formation of \( \mathrm{NO(g)} \), the reaction is not spontaneous and requires energy input.
Understanding \( \Delta G^{\circ} \) involves connecting it to the equilibrium constant \( K_{\mathrm{p}} \) using the equation:

• \( \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K_{\mathrm{p}} \)

Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. The sign of \( \Delta G^{\circ} \) plays a significant role in determining the position of equilibrium. A positive \( \Delta G^{\circ} \) usually results in a small \( K_{\mathrm{p}} \), indicating that reactants dominate at equilibrium.

Temperature plays a crucial role in chemical reactions, especially when calculating equilibrium constants. To get temperature in Kelvin, a straightforward conversion from Celsius is required.
This can be done using the formula:

• \( T = ^{\circ}C + 273.15 \)

In the given exercise, the temperature is \( 25^{\circ} \mathrm{C} \), which converts to \( 298.15 \mathrm{K} \). Using temperature in Kelvin ensures that calculations involving energy, such as free energy changes, use consistent units. It simplifies the process and helps avoid errors in calculations.

A reaction's spontaneity depends on the sign and magnitude of its free energy change. For a reaction to be spontaneous, \( \Delta G^{\circ} \) must be negative.
In this exercise, the positive \( \Delta G^{\circ} \) of \( +86.58 \mathrm{kJ/mol} \) suggests a non-spontaneous process under standard conditions.
This implies:

• Energy input is needed for the reaction to occur.
• The tendency is for the reaction to lie far to the left, favoring reactants.

When \( \Delta G^{\circ} \) is positive, achieving the desired product, like \( \mathrm{NO(g)} \), without additional energy becomes challenging. This concept is vital in predicting how and why reactions occur in nature.

Calculating the equilibrium constant \( K_{\mathrm{p}} \) is essential in understanding chemical equilibria. It quantifies the extent to which a reaction reaches equilibrium.
To find \( K_{\mathrm{p}} \), we rearrange the equation \( \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K_{\mathrm{p}} \) and solve for \( \ln K_{\mathrm{p}} \).

• Substitute known values: \( \ln K_{\mathrm{p}} = \frac{-86580}{8.314 \times 298.15} \)
• Calculate the natural logarithm: \( \ln K_{\mathrm{p}} \approx -34.97 \)
• Convert back using the exponential function: \( K_{\mathrm{p}} = e^{-34.97} \approx 7.5 \times 10^{-16} \)

The very small value of \( K_{\mathrm{p}} \) indicates that the equilibrium mixture contains predominantly reactants, confirming that forming \( \mathrm{NO(g)} \) is not favored under these conditions.