Problem 56 - Chemistry The Molecular Science

Question

Predict whether this reaction is product-favored at \(25^{\circ} \mathrm{C}\) by calculating the change in standard Gibbs free energy from the entropy and enthalpy changes. $$ \begin{aligned} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) & \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \\ \Delta_{\mathrm{r}} H^{\circ} &=52.96 \mathrm{~kJ} / \mathrm{mol} \quad \Delta_{\mathrm{r}} S^{\circ}=21.81 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$

Step-by-Step Solution

Answer

The reaction is not product-favored at \(25^{\circ} \text{C}\).

Step 1: Convert Units

Ensure consistency by converting the enthalpy change from kJ to J. \[ \Delta_{\mathrm{r}} H^{\circ} = 52.96 \times 1000 = 52960 \, \text{J/mol} \]

Step 2: Recall Gibbs Free Energy Equation

The formula for calculating the change in standard Gibbs free energy is: \[ \Delta_{\mathrm{r}} G^{\circ} = \Delta_{\mathrm{r}} H^{\circ} - T \Delta_{\mathrm{r}} S^{\circ} \]

Step 3: Calculate Temperature in Kelvin

Convert the temperature from Celsius to Kelvin. \[ T = 25 + 273.15 = 298.15 \, \text{K} \]

Step 4: Calculate Gibbs Free Energy Change

Substitute the known values into the Gibbs free energy equation: \[ \Delta_{\mathrm{r}} G^{\circ} = 52960 - (298.15 \times 21.81) \] Simplifying further: \[ \Delta_{\mathrm{r}} G^{\circ} = 52960 - 6501.11 \approx 46458.89 \, \text{J/mol} \]

Step 5: Analyze the Result

Since \( \Delta_{\mathrm{r}} G^{\circ} > 0 \), the reaction is not product-favored at \( 25^{\circ} \text{C} \). Product-favored reactions have a negative Gibbs free energy change.

Key Concepts

Enthalpy Change

In chemical reactions, enthalpy change is crucial for understanding how much heat energy is absorbed or released during a reaction. It is represented by the symbol \( \Delta H \) and is measured in joules or kilojoules per mole. To grasp this concept, imagine it as the energy required to break bonds in reactants minus the energy released when new bonds are formed in the products.

• **Exothermic reactions**: In these reactions, energy is released to the surroundings, resulting in a negative \( \Delta H \). This means that products have lower energy than reactants.
• **Endothermic reactions**: These reactions absorb energy, leading to a positive \( \Delta H \), indicating that products are higher in energy than reactants.

In the case of the reaction between hydrogen gas \( \text{H}_2\) and iodine gas \( \text{I}_2\), forming hydrogen iodide \( \text{HI}\), the enthalpy change \( \Delta_{\text{r}} H^{\circ} = 52.96 \text{kJ/mol} \) tells us that this reaction absorbs energy, hence it is endothermic.

Entropy Change

Entropy change, denoted as \( \Delta S \), measures the disorder or randomness in a system. It's a reflection of how much the positional freedom and energy dispersal increase as a result of the reaction. When the products of a reaction have greater disorder compared to reactants, \( \Delta S \) is positive. Conversely, if the products are more ordered, \( \Delta S \) is negative.

Let's look at the reaction \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \). The standard entropy change \( \Delta_{\text{r}} S^{\circ} = 21.81 \text{J/K mol} \) suggests that the products have slightly more disorder compared to reactants. A positive \( \Delta S \) indicates an increase in randomness. In general, gas-phase reactions that increase the number of molecules often see an increase in entropy. However, it's not just the quantity but the nature and distribution of molecular motion contributing to entropy.

Product-Favored Reactions

Understanding whether a reaction is product-favored involves observing its Gibbs free energy change \( \Delta G \). This value helps predict the direction and spontaneity of a chemical reaction under standard conditions.

The equation \( \Delta G = \Delta H - T \Delta S \) allows us to determine whether a reaction will proceed spontaneously to form products:

• When \( \Delta G < 0 \), the reaction is product-favored, indicating that products are more stable and the reaction is spontaneous.
• When \( \Delta G > 0 \), the reaction is reactant-favored, meaning reactants are more stable and the reaction is non-spontaneous at the given temperature.

In the given example involving hydrogen and iodine gases to form hydrogen iodide, the calculated Gibbs free energy change was positive \( \Delta_{\text{r}} G^{\circ} \approx 46458.89 \text{J/mol} \). This tells us that the reaction is not spontaneous or product-favored at this temperature. Indeed, for it to become product-favored, conditions such as temperature, pressure, or concentrations might need to change to shift the balance favorably towards products.