Problem 105 - Chemistry The Molecular Science

Question

Suppose that at a certain temperature \(T,\) a chemical reaction is found to have a standard equilibrium constant \(K^{\circ}\) of 1.0 . Indicate whether each statement is true or false and explain why. (a) The enthalpy change for the reaction, \(\Delta_{\mathrm{r}} H^{\circ},\) is zero. (b) The entropy change for the reaction, \(\Delta_{t} S^{\circ},\) is zero. (c) The Gibbs free energy change for the reaction, \(\Delta_{r} G^{\circ},\) is zero. (d) \(\Delta_{\mathrm{r}} H^{\circ}\) and \(\Delta_{\mathrm{r}} \mathrm{S}^{\circ}\) have the same sign. (e) \(\Delta_{\mathrm{r}} H^{\circ} / T=\Delta_{\mathrm{r}} S^{\circ}\) at the temperature \(T\).

Step-by-Step Solution

Answer

(a) False, (b) False, (c) True, (d) True, (e) True.

Step 1: Understanding Equilibrium Constant

The standard equilibrium constant, denoted as \(K^{\circ}\), measures the ratio of products to reactants at equilibrium. A \(K^{\circ}\) value of 1.0 indicates that the reaction is perfectly balanced, with neither reactants nor products being favored.

Step 1: Gibbs Free Energy and Equilibrium Constant

The relationship between Gibbs free energy change and the equilibrium constant at standard temperature is given by the formula \( \Delta_{r} G^{\circ} = -RT \ln K^{\circ} \). Given \( K^{\circ} = 1.0 \), this simplifies to \( \Delta_{r} G^{\circ} = 0 \) because \( \ln(1) = 0 \). Thus, statement (c) is true.

Step 2: Entropy Change and Gibbs Free Energy

The expression for Gibbs free energy is \( \Delta_{r} G^{\circ} = \Delta_{r} H^{\circ} - T\Delta_{r} S^{\circ} \). Given that \( \Delta_{r} G^{\circ} = 0 \), it follows that \( \Delta_{r} H^{\circ} = T\Delta_{r} S^{\circ} \). Therefore, statement (e) is true.

Step 3: Analyze Enthalpy Change

There's no direct correlation between \( K^{\circ} = 1.0 \) and \( \Delta_{r} H^{\circ} = 0 \). It depends on the nature of the reaction, but not true in all cases of \( K^{\circ} = 1.0 \). Thus, statement (a) is false.

Step 4: Analyze Entropy Change

Similarly, \( K^{\circ} = 1.0 \) does not necessarily imply zero entropy change. Hence, the reaction entropy \( \Delta_{r} S^{\circ} \) is not inherently zero. Statement (b) is false.

Step 5: Sign of Enthalpy and Entropy

Since \( \Delta_{r} H^{\circ} - T\Delta_{r} S^{\circ} = 0 \), if \( \Delta_{r} H^{\circ} \) is positive, \( \Delta_{r} S^{\circ} \) must also be positive, and vice versa. Enthalpy and entropy changes have the same sign. Thus, statement (d) is true.

Key Concepts

Equilibrium Constant

In the realm of chemical equilibrium, the equilibrium constant, represented as \( K^{\circ} \), plays a pivotal role. It offers a snapshot of the ratio of concentrations of products to reactants at equilibrium at a specific temperature. When \( K^{\circ} = 1.0 \), it indicates that the concentrations of products and reactants are at a perfect balance. Neither the forward nor the reverse reaction is favored, leading to a situation where the chemical system is deeply stable.
Understanding \( K^{\circ} \) helps chemists predict the direction and extent of a reaction under specific conditions. A \( K^{\circ} \) greater than 1 suggests a preference for products, while less than 1 suggests a preference for reactants.

• A balanced \( K^{\circ} \) (such as 1.0) depicts an ideal equilibrium scenario.
• This balance implies no net change in the concentrations of products and reactants over time at equilibrium.

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta_{r} G^{\circ} \), connects thermodynamics to chemical equilibrium. It reflects the spontaneity of a reaction; a negative value suggests a spontaneous process, while a positive value indicates non-spontaneity.
When the equilibrium constant \( K^{\circ} = 1.0 \), the equation \( \Delta_{r} G^{\circ} = -RT \ln K^{\circ} \) simplifies to \( \Delta_{r} G^{\circ} = 0 \). This zero value signifies that the chemical system is in equilibrium, as there is no inclination for the reaction to proceed in either direction.

• Zero Gibbs Free Energy implies no net work can be extracted because the reaction is already at equilibrium.
• The reaction is in a state of dynamical balance, maintaining constant concentrations of products and reactants.

Enthalpy and Entropy

Enthalpy (\( \Delta_{r} H^{\circ} \)) and entropy (\( \Delta_{r} S^{\circ} \)) are crucial factors in determining a reaction's behavior. Enthalpy reflects the heat change in the system, while entropy measures disorder.
These properties are interconnected by the Gibbs free energy equation: \( \Delta_{r} G^{\circ} = \Delta_{r} H^{\circ} - T\Delta_{r} S^{\circ} \). When \( \Delta_{r} G^{\circ} = 0 \), it follows that \( \Delta_{r} H^{\circ} = T\Delta_{r} S^{\circ} \). This balance means that the enthalpy change is exactly offset by the entropy change adjusted for temperature.

• If \( \Delta_{r} H^{\circ} \) is positive, \( \Delta_{r} S^{\circ} \) is also positive, maintaining equilibrium.
• Enthalpy and entropy changes can help predict temperature effects on reaction direction.

Thermodynamics

Thermodynamics provides the framework for understanding energy changes within chemical reactions. It helps relate macroscopic properties like temperature and pressure to microscopic interactions. At equilibrium, we focus on the first and second laws of thermodynamics to understand how reactions proceed and remain stable.
In our context, thermodynamics relates directly to the balance of enthalpy and entropy, guiding reactions to equilibrium where the free energy is minimized. When \( K^{\circ} = 1.0 \), the thermodynamic system is at a point where no further energy is available to do work.

• Thermodynamics helps explain why systems tend to move towards equilibrium conditions.
• It uses enthalpy, entropy, and Gibbs free energy to predict the feasibility and extent of reactions.