Problem 81 - Chemistry and Chemical Reactivity

Question

Calculate \(\Delta_{f} G^{\circ}\) for \(\mathrm{HI}(\mathrm{g})\) at \(350^{\circ} \mathrm{C},\) given the following equilibrium partial pressures: \(P\left(\mathrm{H}_{2}\right)=\) 0.132 bar, \(P\left(\mathrm{I}_{2}\right)=0.295\) bar, and \(P(\mathrm{HI})=1.61\) bar. At \(350^{\circ} \mathrm{C}\) and \(1 \mathrm{bar}, \mathrm{I}_{2}\) is a gas. $$1 / 2 \mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$$

Step-by-Step Solution

Answer

\( \Delta_{f}G^{\circ} \approx -21.8 \ kJ/mol \).

Step 1: Write the expression for equilibrium constant (Kp)

For the given reaction \( \frac{1}{2} \mathrm{H}_2(\mathrm{g}) + \frac{1}{2} \mathrm{I}_2(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g}) \), the equilibrium constant \( K_p \) for the reaction at the given conditions can be expressed in terms of the partial pressures: \[ K_p = \frac{{(P_{HI})^2}}{{P_{H_2} \cdot P_{I_2}}} \]

Step 2: Substitute the given partial pressures into Kp equation

Insert the given equilibrium partial pressures into the \( K_p \) expression: \[ K_p = \frac{{(1.61)^2}}{{0.132 \times 0.295}} \]

Step 3: Calculate the value of Kp

Calculate \( K_p \) using the values:\[ K_p = \frac{(1.61)^2}{0.132 \times 0.295} \approx \frac{2.5921}{0.03894} \approx 66.57 \]

Step 4: Relate Kp to Gibbs free energy change

The standard Gibbs free energy change is related to the equilibrium constant by the equation:\[ \Delta_{f}G^{\circ} = -RT \ln K_p \]where \( R = 8.314 \frac{J}{mol \cdot K} \) and \( T = 350^{\circ}C = 623.15 \ K \).

Step 5: Calculate ΔfG° using Kp

Substitute the values of \( R \), \( T \), and \( K_p \) into the ΔfG° expression:\[ \Delta_{f}G^{\circ} = -8.314 \cdot 623.15 \cdot \ln(66.57) \]Compute the logarithm and perform the multiplication to find:\[ \Delta_{f}G^{\circ} = -8.314 \cdot 623.15 \cdot 4.1989 \approx -21778 \ J/mol \] which is approximately \(-21.8 \ kJ/mol\).

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, represented as \( K_p \) for gas-phase reactions, is a crucial factor. It calculates the ratio of partial pressures of products to reactants at equilibrium. Specifically, for the reaction \( \frac{1}{2} \mathrm{H}_2(\mathrm{g}) + \frac{1}{2} \mathrm{I}_2(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g}) \), the equilibrium constant is given by:

• \( K_p = \frac{{(P_{\mathrm{HI}})^2}}{{P_{\mathrm{H}_2} \cdot P_{\mathrm{I}_2}}} \)

It reflects the balance point of the reaction at a given temperature. Equilibrium does not mean equal concentrations of reactants and products, but rather a steady state where their ratios remain constant.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. Each gas in a mixture acts independently regarding its contribution to the total pressure. For our reaction, the given partial pressures are:

• \( P(\mathrm{H}_2) = 0.132 \) bar
• \( P(\mathrm{I}_2) = 0.295 \) bar
• \( P(\mathrm{HI}) = 1.61 \) bar

Partial pressures are essential in the calculation of \( K_p \), providing insight into the balance of reactants and products under equilibrium conditions. It's critical to express these pressures accurately to solve for the equilibrium constant and eventual Gibbs free energy changes.

Thermodynamics

Thermodynamics explores the principles governing energetic processes, such as chemical reactions and phase changes. Central to thermodynamics is the concept of Gibbs free energy (\( \Delta_{f}G^{\circ} \)). It indicates whether a reaction is spontaneous under constant pressure and temperature. The relationship to the equilibrium constant is given by:

• \( \Delta_{f}G^{\circ} = -RT \ln K_p \)

Depicting this relationship allows us to predict the flow of a reaction:

• If \( \Delta_{f}G^{\circ} < 0 \), the reaction is spontaneous
• If \( \Delta_{f}G^{\circ} > 0 \), the reaction is non-spontaneous

In this scenario, the calculation shows a negative value for \( \Delta_{f}G^{\circ} \), signifying a spontaneous forward reaction at the given conditions.

Chemical Equilibrium

Chemical equilibrium occurs when a reversible chemical reaction proceeds at an equal rate in both directions, maintaining the concentrations of reactants and products constant over time. At equilibrium, the forward reaction rate equals the reverse reaction rate. For the reaction \( \frac{1}{2} \mathrm{H}_2(\mathrm{g}) + \frac{1}{2} \mathrm{I}_2(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g}) \), equilibrium is achieved when the partial pressures match the established ratio of \( K_p \).The state of chemical equilibrium is dynamic, meaning molecules continue to form and revert at the same rate but without a net change in concentrations. This concept is vital to understanding reaction behavior in real-world and laboratory settings.