Problem 33 - Chemistry and Chemical Reactivity

Question

The formation of \(\mathrm{NO}(\mathrm{g})\) from its elements $$ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g}) $$ has a standard free energy change, \(\Delta G^{\circ},\) of \(+86.58 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{p}}\) at this temperature. Comment on the connection between the sign of \(\Delta G^{\circ}\) and the magnitude of \(\bar{K}_{\mathrm{p}^{*}}\)

Step-by-Step Solution

Answer

\( K_p \approx 6.95 \times 10^{-16} \); positive \( \Delta G^{\circ} \) means a very small \( K_p \).

Step 1: Understand the Relationship

First, recall the relationship between the standard free energy change, \( \Delta G^{\circ} \), and the equilibrium constant, \( K_p \), given by the equation: \[ \Delta G^{\circ} = -RT \ln K_p \]where \( R \) is the universal gas constant (8.314 J/mol·K), \( T \) is the temperature in Kelvin, and \( K_p \) is the equilibrium constant in terms of pressure.

Step 2: Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using: \[ T(K) = T(^{\circ}C) + 273.15 \]For \( T = 25^{\circ}C \), the temperature in Kelvin is \[ T = 25 + 273.15 = 298.15 \, K \].

Step 3: Rearrange the Equation

Rearrange the equation to solve for \( K_p \): \[ \Delta G^{\circ} = -RT \ln K_p \Rightarrow \ln K_p = -\frac{\Delta G^{\circ}}{RT} \].

Step 4: Insert Known Values

Substitute the known values into the rearranged equation:\[ \ln K_p = -\frac{86580 \, J/mol}{8.314 \, J/mol\cdot K \times 298.15 \, K} \].Note that \( \Delta G^{\circ} \) is converted from kJ to J by multiplying by 1000 since 1 kJ = 1000 J.

Step 5: Calculate \( \ln K_p \)

Perform the calculation:\[ \ln K_p = -\frac{86580}{2484.5681} \approx -34.814 \].

Step 6: Exponentiate to Find \( K_p \)

To find \( K_p \), exponentiate both sides:\[ K_p = e^{-34.814} \approx 6.95 \times 10^{-16} \].

Step 7: Interpret the Connection

Since \( \Delta G^{\circ} \) is positive, the formation of NO is nonspontaneous under standard conditions, leading to a very small \( K_p \), indicating the reaction lies far to the left.

Key Concepts

Standard Free Energy Change

The standard free energy change, often denoted as \( \Delta G^{\circ} \), is a vital concept in chemistry. It represents the amount of free energy available, or required, to drive a reaction under standard conditions, which include a pressure of 1 bar and substances in their standard states. The sign and magnitude of \( \Delta G^{\circ} \) give us critical insight into the nature of the reaction:

• If \( \Delta G^{\circ} < 0 \), the process can proceed spontaneously. It releases free energy.
• If \( \Delta G^{\circ} = 0 \), the system is at equilibrium, with no net change occurring.
• If \( \Delta G^{\circ} > 0 \), the reaction is nonspontaneous under standard conditions and requires energy input.

For the formation of NO from nitrogen and oxygen gas, \( \Delta G^{\circ} \) is positive at 86.58 kJ/mol, indicating that this reaction does not occur spontaneously at standard conditions. This understanding is crucial when predicting the directionality and feasibility of chemical reactions.

Equilibrium Constant

The equilibrium constant, \( K_p \), is a number that provides information about the ratio of products to reactants at equilibrium for a given reaction. It is expressed in terms of partial pressures when dealing with gases, which is why the subscript \( p \) is used. The connection between the equilibrium constant and free energy is established through the equation: \[ \Delta G^{\circ} = -RT \ln K_p \] Here, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. When calculating \( K_p \):

• A large \( K_p \) indicates that the products are favored at equilibrium.
• A small \( K_p \), as we find for NO formation, shows that reactants are favored, making the reaction incomplete.
• A \( K_p \) around 1 means a balanced equilibrium between reactants and products.

For the formation reaction of NO, the calculated \( K_p \) is approximately \( 6.95 \times 10^{-16} \), highlighting that the reaction heavily favors the reactants under standard conditions.

Gibbs Free Energy

Gibbs free energy is a thermodynamic potential that measures the maximal amount of work a thermodynamic system can perform at constant temperature and pressure. It incorporates both enthalpy (\( H \)) and entropy (\( S \)) changes of the system into one term:\[ G = H - TS\] Free energy helps predict whether a process will proceed spontaneously and how it would shift under different temperature and pressure conditions. The standard free energy change, \( \Delta G^{\circ} \), serves as a snapshot of these predictions under standard conditions. Key points include:

• Processes with a decrease in Gibbs free energy \((\Delta G < 0)\) are feasible.
• Gibbs free energy balances energy changes due to enthalpy and entropy.
• In the context of chemical equilibrium, it links to the equilibrium constant \( K_p \), providing insights about reaction spontaneity and composition at equilibrium.

Understanding Gibbs free energy is essential for mastering thermodynamics in chemical processes, as it shapes the journey a reaction takes from reactants to products.

Reaction Spontaneity

Reaction spontaneity refers to the tendency of a chemical reaction to occur without being driven by an external force or energy source. Whether a reaction is spontaneous can be determined by Gibbs free energy. At a glance:

• Spontaneous reactions have \( \Delta G^{\circ} < 0 \), meaning they happen independently.
• Nonspontaneous reactions, like the formation of NO (with \( \Delta G^{\circ} = +86.58 \text{ kJ/mol} \)), require energy input.
• The sign of \( \Delta G^{\circ} \) directly impacts the equilibrium constant, \( K_p \), and reflects on the favorability of the reaction's position.
  

Spontaneity doesn't account for reaction rate; it merely suggests whether a reaction can occur. While a reaction might be spontaneous based on \( \Delta G^{\circ} \), it might still proceed slowly if kinetic barriers are present, requiring catalysts to increase the speed of the reaction. Understanding the spontaneity concept is crucial when assessing energy transformations and reaction dynamics.