Problem 4 - Chemistry A Guided Inquiry

Question

Recall that \(\Delta G^{\circ}\) can be written as a function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ} .\) Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not temperature dependent and answer each of the following: a) Derive an expression relating \(\ln K, \Delta H^{\circ}\), and \(\Delta S^{\circ} .\) That is, derive an expression that looks like \(\ln K=\) some function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\). The temperature, \(T\), should appear only once in this equation. b) How are the equilibrium constants for reactions with \(\Delta H^{\circ}>0\) affected by an increase in temperature? c) How are equilibrium constants for reactions with \(\Delta H^{\circ}<0\) affected by an increase in temperature?

Step-by-Step Solution

Answer

The derived relationship between the equilibrium constant and the enthalpy, entropy, and temperature is \(\ln(K) = \Delta S - \Delta H / T\). For reactions with positive enthalpy (\(\Delta H^{\circ}>0\)), an increase in temperature will lead to an increase in the equilibrium constant. For reactions with negative enthalpy (\(\Delta H^{\circ}<0\)), an increase in temperature will result in a decrease in equilibrium constant.

Step 1: Obtaining the Gibbs Free Energy Equation

The formula for Gibbs free energy is defined as follows: \[\Delta G = \Delta H - T\Delta S\]where \(\Delta G\) = Gibbs free energy change, \(\Delta H\) = Enthalpy change, \(\Delta S\) = Entropy change,and \(T\) = Temperature in Kelvin.

Step 2: Deriving an expression relating lnK, ΔHo, and ΔSo

Recall that at equilibrium, \(\Delta G = 0\), yielding:\[0 = \Delta H - T\Delta S\]Solving for \(\Delta S\) gives us \(\Delta S = \Delta H / T\).We also know that \(\Delta G = -RT\ln(K)\), where \(R\) is the ideal gas constant and \(K\) is the equilibrium constant. At equilibrium, this becomes \(0 = -RT\ln(K)\), yielding \(\ln(K) = -\Delta H / (RT)\), substituting \(\Delta S\) into the equation we get \(\ln(K) = -\Delta S\).So, the expression relating \(\ln K\), \(\Delta H^{\circ}\), and \(\Delta S^{\circ}\) is \(\ln(K) = \Delta S - \Delta H / T\).

Step 3: Determine the effect of temperature on lnK when ΔH > 0

In the derived expression, \(\ln(K) = \Delta S - \Delta H / T\), when \(\Delta H\) is greater than 0 (endothermic reaction) and the temperature increases, the term \(-\Delta H / T\) decreases. Therefore, \(\ln K\) will increase, meaning that the equilibrium constant increases when the temperature is increased for reactions with a positive enthalpy.

Step 4: Determine the effect of temperature on lnK when ΔH < 0

In the derived expression, \(\ln(K) = \Delta S - \Delta H / T\), when \(\Delta H\) is less than 0 (exothermic reaction) and the temperature increases, the term \(-\Delta H / T\) increases, leading to a decrease in \(\ln K\). Therefore, the equilibrium constant decreases when the temperature is increased for reactions with a negative enthalpy.

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \(\Delta H\), is a fundamental concept in thermodynamics that represents the heat absorbed or released during a chemical reaction at constant pressure. It provides insight into whether a reaction is exothermic or endothermic. \

• In an exothermic reaction, heat is released, resulting in \(\Delta H < 0\). This means the products have less energy than the reactants.
• Conversely, in an endothermic reaction, heat is absorbed, leading to \(\Delta H > 0\). Here, the products have more energy than the reactants.

For students, it's crucial to understand that \(\Delta H\) is temperature-independent under many circumstances. When Gibbs Free Energy \(\Delta G\) is related to \(\Delta H\) in equations, it helps predict the spontaneity of reactions. The interaction of enthalpy with other variables such as temperature and entropy dictates the overall feasibility of reactions. \ Understanding enthalpy change allows you to predict how energy flows during reactions, which is essential for comprehending reaction mechanisms and the design of experimental procedures.

Entropy Change

Entropy can be seen as a measure of disorder or randomness in a system. Entropy change, represented by \(\Delta S\), is crucial for determining how the energy is spread out in a process. It gives an indication of the number of ways the energy can be distributed.

• If \(\Delta S > 0\), the system's disorder increases, which is usually favorable for spontaneity.
• If \(\Delta S < 0\), the system becomes more ordered, which may not favor spontaneity unless compensated by enthalpy changes.

In the context of Gibbs Free Energy, \(\Delta G = \Delta H - T\Delta S\), entropy plays a key role. It's this balance between enthalpy and entropy that determines whether a reaction will proceed without external work. Entropy not only impacts reactions by itself, but it also influences how \(\Delta G\) changes as temperature changes.

Equilibrium Constant

The equilibrium constant \(K\) is a vital concept in chemical equilibrium, representing the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients. It indicates the extent to which a reaction progresses. \ Incorporating the expression \(\ln(K) = \Delta S - \frac{\Delta H}{T}\), we see the connection between equilibrium and thermodynamic quantities. This equation shows how \(\Delta H\) and \(\Delta S\) dictate the position of the equilibrium.

• For endothermic reactions (\