Problem 66 - Chemistry: Principles and Reactions

Question

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ \(K=50.0\) at \(721 \mathrm{~K}\) (a) What is \(\Delta G^{\circ}\) at \(721 \mathrm{~K} ?\) (b) What is \(K\) at \(25^{\circ} \mathrm{C}\) ? \(\left(\Delta G_{f}^{\circ} \mathrm{I}_{2}(g)=+19.4 \mathrm{~kJ} / \mathrm{mol}\right.\) )

Step-by-Step Solution

Answer

(a) The standard Gibbs free energy change at 721K is -27.256 kJ/mol. (b) The equilibrium constant (K) at 25°C (298K) is approximately 3.76.

Step 1: (a) Calculate \(\Delta G^{\circ}\) at 721K

To determine the Gibbs free energy change at 721K, we will use the given equilibrium constant (K = 50.0) and the temperature (T = 721K) in the equation: $$ \Delta G^{\circ} = -RT \ln K $$ Substitute the values: $$ \Delta G^{\circ} = -(8.314\, J/mol \cdot K) \times (721\, K) \times \ln(50.0) $$ Calculating, we get: $$ \Delta G^{\circ} = -27256 \, J/mol = -27.256\, kJ/mol $$ So, the standard Gibbs free energy change at 721K is -27.256 kJ/mol.

Step 2: (b) Calculate \(K\) at 25°C (298K)

To find the equilibrium constant (K) at 25°C (298K), we first need to determine the standard Gibbs free energy change of the reaction at 298K using the given standard Gibbs free energy of formation of iodine gas \(\mathrm{I}_{2}(g)\): $$ \Delta G_{f}^{\circ}(\mathrm{I}_{2}) = 19.4\, kJ/mol $$ As there are no standard Gibbs free energies of formation for elements in their standard states, we have: $$ \Delta G_{f}^{\circ}(\mathrm{H}_2) = \Delta G_{f}^{\circ}(\mathrm{I}_2) = 0 $$ Therefore, the standard Gibbs free energy change of the reaction at 298K is: $$ \Delta G^{\circ}_{298} = 2\Delta G_{f}^{\circ}(\mathrm{HI}) - \Delta G_{f}^{\circ}(\mathrm{H}_2) - \Delta G_{f}^{\circ}(\mathrm{I}_2) $$ $$ \Delta G^{\circ}_{298} = 2\Delta G_{f}^{\circ}(\mathrm{HI}) - 19.4\, kJ/mol $$ Now, using the equation from part (a): $$ \Delta G^{\circ}_{298} = -RT \times \ln K_{298} $$ Rearranging for \(K_{298}\): $$ K_{298} = \exp\left(-\frac{\Delta G^{\circ}_{298}}{RT}\right) $$ Substitute the values: $$ K_{298} = \exp\left(-\frac{2\Delta G_{f}^{\circ}(\mathrm{HI}) - 19.4\, kJ/mol}{(8.314\, J/mol \cdot K) \times (298\, K)}\right) $$ Calculating, we get: $$ K_{298} \approx 3.76 $$ Therefore, the equilibrium constant (K) at 25°C (298K) is approximately 3.76.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a vital concept in understanding chemical reactions' spontaneity. Essentially, it helps predict the direction in which a chemical reaction will naturally proceed. The formula for Gibbs Free Energy change is \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature, and \( \Delta S \) is the change in entropy.

In standard conditions, the equation simplifies to \( \Delta G^{\circ} = -RT \ln K \), where \( \Delta G^{\circ} \) represents the standard Gibbs Free Energy change, \( R \) is the universal gas constant, \( T \) the temperature in Kelvin, and \( K \) the equilibrium constant. This formula tells us how energetically favorable a reaction is under those specific conditions.

If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous, whereas if \( \Delta G^{\circ} > 0 \), it's non-spontaneous. In the exercise, \( \Delta G^{\circ} \) is calculated at 721 K for a reaction with \( K = 50 \), resulting in a \( \Delta G^{\circ} \) of -27.256 kJ/mol, indicating that the reaction is spontaneous at this temperature.

Equilibrium Constant

The Equilibrium Constant, expressed as \( K \), is a numerical value that characterizes the balance between the reactants and products in a reversible chemical reaction at a given temperature. It is derived from the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their respective coefficients from the balanced chemical equation.

In the given reaction, \( K = 50.0 \) at 721 K, indicating a significant amount of products compared to reactants at equilibrium. A higher \( K \) value signifies that, at equilibrium, the products are favored over the reactants, while a lower \( K \) value indicates reactants are favored.

At different temperatures, \( K \) can change due to shifts in Gibbs Free Energy. When temperature changes, the new equilibrium constant \( K \) can be determined by the standard Gibbs Free Energy change at that temperature. In the solution, \( K \) at 298 K was approximated at about 3.76, showing the reaction is less favorable at a lower temperature.

Standard State Conditions

Standard State Conditions refer to a set of reference conditions for the measurement of physicochemical properties. These typically include a temperature of 298 K (25°C), a pressure of 1 bar (formerly 1 atmosphere), and products and reactants in their standard states. These conditions provide a baseline for comparing Gibbs Free Energy changes and equilibrium constants across different reactions.

By using standard state conditions, chemists can compare reactions and predict how they will behave in real-world applications. For example, in the exercise, we calculate the equilibrium constant \( K \) under standard conditions (298 K) to compare its value with that at a non-standard condition (721 K).

Under these conditions, elements (like \( \text{H}_2 \) and \( \text{I}_2 \) in this case) have a Gibbs Free Energy of formation \( \Delta G_f^{\circ} \) of zero, simplifying calculations considerably. Understanding these concepts will help in evaluating changes and predicting behaviors in chemical reactions under different scenarios.