Problem 27 - Chemistry and Chemical Reactivity

Question

The standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ}\), for the formation of \(\mathrm{NO}(\mathrm{g})\) from its elements is \(+86.58 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature for the equilibrium $$1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})$$ Comment on the sign of \(\Delta G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

Step-by-Step Solution

Answer

Kp is approximately \(7.5 \times 10^{-16}\); the reaction is non-spontaneous at 25°C.

Step 1: Recall the Relationship Between ΔG° and Kp

The relationship between the standard free energy change \(\Delta_{\mathrm{r}} G^{\circ}\) and the equilibrium constant \(K_{\mathrm{p}}\) is given by the equation:\[ \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K_{\mathrm{p}} \]where \(R\) is the gas constant \(8.314\, \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\), and \(T\) is the temperature in Kelvin.

Step 2: Convert Given Temperature to Kelvin

The temperature needs to be converted from Celsius to Kelvin:\[ T = 25 ^{\circ} \mathrm{C} + 273.15 = 298.15\, \mathrm{K} \]

Step 3: Plug Known Values into the Equation

Substitute the given \(\Delta_{\mathrm{r}} G^{\circ}\) value and the derived temperature into the relation:\[ 86.58\, \mathrm{kJ} / \mathrm{mol} = 86580\, \mathrm{J} / \mathrm{mol} \]This yields:\[ 86580 = -(8.314)(298.15) \ln K_{\mathrm{p}} \]

Step 4: Solve for ln(Kp)

Rearrange and solve for \(\ln K_{\mathrm{p}}\):\[ \ln K_{\mathrm{p}} = \frac{-86580}{8.314 \times 298.15} \]Calculate:\[ \ln K_{\mathrm{p}} \approx -34.97 \]

Step 5: Compute Kp using the Exponential Function

Convert \(\ln K_{\mathrm{p}}\) back to \(K_{\mathrm{p}}\) using the exponential function:\[ K_{\mathrm{p}} = e^{-34.97} \]Calculating this gives:\[ K_{\mathrm{p}} \approx 7.5 \times 10^{-16} \]

Step 6: Interpret the Significance of ΔG° and Kp

The positive \(\Delta_{\mathrm{r}} G^{\circ}\) indicates the reaction is non-spontaneous under standard conditions. A very small value for \(K_{\mathrm{p}}\) (much less than 1) suggests the equilibrium strongly favors the reactants over the product - \(\mathrm{NO}(\mathrm{g})\) is not significantly formed under these conditions.

Key Concepts

Standard Free Energy Change

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 Conversion

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.

Reaction Spontaneity

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.

Chemical Equilibrium Calculations

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.