Problem 11 - Chemistry A Guided Inquiry

Question

If \(\Delta H^{\circ}=T \Delta S^{\circ}\), what is the value of \(\Delta G^{\circ}\) ? Predict the value of \(K\) in this case.

Step-by-Step Solution

Answer

The calculated value for \(\Delta G^{\circ}\) is 0 and hence, the predicted value for \(K\) is 1.

Step 1: Substitute the given into the Gibbs-Helmholtz equation

Start from the given formula: \(\Delta H^{\circ}=T \Delta S^{\circ}\). This can be substituted into the Gibbs-Helmholtz equation which is: \(\Delta G^{\circ} = \Delta H^{\circ}-T \Delta S^{\circ}\). The substitution will simplify the equation.

Step 2: Solve for \(\Delta G^{\circ}\)

By replacing the value of \(\Delta H^{\circ}\) into the Gibbs-Helmholtz equation, we get \(\Delta G^{\circ} = T \Delta S^{\circ}-T \Delta S^{\circ}\). Simplifying this we get \(\Delta G^{\circ} = 0\). Thus, the value of \(\Delta G^{\circ}\) is zero.

Step 3: Predict the value of \(K\)

The value of \(K\) can be predicted using the relationship between \(\Delta G^{\circ}\) and \(K\). According to the equation \(\Delta G^{\circ} = -RT \ln K\), where R is the universal gas constant and T is the absolute temperature, when \(\Delta G^{\circ} = 0\), the formula can be rearranged to \(\ln K = -\Delta G^{\circ}/RT\). Since we found that \(\Delta G^{\circ} = 0\) in the previous step, the value of \(K\) will be 1 as \(\ln1 = 0\).

Key Concepts

Gibbs-Helmholtz Equation

The Gibbs-Helmholtz equation is an essential formula in thermodynamics that relates the free energy change of a system to its enthalpy, entropy, and temperature. It is presented as \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). This equation helps to determine how spontaneous a reaction is. A negative \( \Delta G^{\circ} \) result indicates a spontaneous reaction, while a positive one suggests non-spontaneity.

When the exercise states that \( \Delta H^{\circ} = T \Delta S^{\circ} \), substituting this into the Gibbs-Helmholtz equation simplifies it, showing \( \Delta G^{\circ} = 0 \). This means the reaction is at equilibrium and neither favors the products nor the reactants.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H^{\circ} \), measures the heat change at constant pressure during a reaction. It reflects whether a process is endothermic or exothermic:

• Endothermic: Heat is absorbed, and \( \Delta H^{\circ} > 0 \).
• Exothermic: Heat is released, and \( \Delta H^{\circ} < 0 \).

Enthalpy is key to predicting reaction spontaneity when used in combination with entropy. In our original exercise, \( \Delta H^{\circ} = T \Delta S^{\circ} \), implying neither energy gain nor loss as heat since the system is balanced by the entropy term. Consequently, \( \Delta G^{\circ} \) becomes zero, indicating no net change in free energy.

Entropy Change

Entropy, symbolized by \( \Delta S^{\circ} \), quantifies the disorder or randomness in a system. Higher entropy means greater disorder. Changes in entropy can indicate whether a process naturally occurs based on two scenarios:

• Increase in entropy (\( \Delta S^{\circ} > 0 \)): Typically, spontaneous.
• Decrease in entropy (\( \Delta S^{\circ} < 0 \)): Usually non-spontaneous.

In the context of our problem, \( \Delta S^{\circ} \) and \( T \) are such that they balance out \( \Delta H^{\circ} \), ensuring that the system's free energy change remains zero. This means the reaction is ideally at equilibrium.

Thermodynamic Equilibrium Constant

The thermodynamic equilibrium constant, \( K \), represents the ratio of the product concentrations to reactant concentrations at equilibrium. Its value provides insight into the position of equilibrium:

• \( K > 1 \): Products are favored at equilibrium.
• \( K < 1 \): Reactants are favored.
• \( K = 1 \): Reaction is perfectly balanced.

According to the equation \( \Delta G^{\circ} = -RT \ln K \), a \( \Delta G^{\circ} = 0 \) results in \( \ln K = 0 \), translating to \( K = 1 \). Hence, in this situation, both reactants and products are equally favored, denoting a balanced equilibrium where no side is preferred over the other.