Problem 95 - Chemistry The Central Science

Question

Consider the following three reactions: $$ \begin{array}{l}{\text { (i) } \operatorname{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{TiCl}_{4}(g)} \\ {\text { (ii) } \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{CCl}_{4}(g)+6 \mathrm{HCl}(g)} \\ {\text { (iii) } \mathrm{BaO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{BaCO}_{3}(s)}\end{array} $$ (a) For each of the reactions, use data in Appendix \(C\) to calculate \(\Delta H^{\circ}, \Delta G^{\circ}, K,\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) . (b) Which of these reactions are spontaneous under standard conditions at \(25^{\circ} \mathrm{C} ?(\mathbf{c})\) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

Step-by-Step Solution

Answer

To find the thermodynamic parameters for each of the given reactions at 25°C, use the following steps: 1. Calculate ∆H° using standard enthalpies of formation values from the appendix and the equation ∆H° = ∑n∆H°(products) - ∑m∆H°(reactants). 2. Calculate ∆S° using standard molar entropy values from the appendix and the equation ∆S° = ∑n∆S°(products) - ∑m∆S°(reactants). 3. Calculate ∆G° using the equation ∆G° = ∆H° - T∆S°, where T = 298 K. 4. Calculate K using the equation K = exp(−∆G°/RT), where R = 8.314 J/(mol K) and T = 298 K. Determine if each reaction is spontaneous at 25°C by checking if ∆G° is negative. Predict how ∆G changes with temperature by analyzing the signs of ∆H° and ∆S° according to the given rules.

Step 1: Calculate ∆H° for each reaction

To calculate the enthalpy change (∆H°) for each reaction, use the following equation: ∆H° = ∑n∆H°(products) - ∑m∆H°(reactants), where n and m are coefficients in the balanced chemical equation, and ∆H° represents the standard enthalpy of formation. You need to find the enthalpy of formation values for each species from the appendix.

Step 2: Calculate ∆S° for each reaction

To calculate the entropy change (∆S°) for each reaction, use the following equation: ∆S° = ∑n∆S°(products) - ∑m∆S°(reactants), where n and m are coefficients in the balanced chemical equation, and ∆S° represents the standard molar entropy. You need to find the molar entropy values for each species from the appendix.

Step 3: Calculate ∆G° for each reaction

Once you have the values for ∆H° and ∆S°, you can calculate the free energy change (∆G°) for each reaction using the following equation: ∆G° = ∆H° - T∆S°, where T is the temperature in Kelvin (298 K in this case)

Step 4: Calculate K for each reaction

To calculate the equilibrium constant (K) for each reaction, use the following equation: K=exp(−∆G°/RT), where R is the gas constant 8.314 J/(mol K), and T is the temperature in Kelvin (298 K in this case)

Step 5: Determine which reactions are spontaneous under standard conditions at 25°C

If ∆G° is negative, the reaction is spontaneous under standard conditions at 25°C; if it is positive, the reaction is non-spontaneous.

Step 6: Predict the behavior of ∆G with an increase in temperature

Use the equation ∆G° = ∆H° - T∆S° to analyze the behavior of ∆G with increasing temperature: - If ∆H° > 0 and ∆S° > 0, the reaction will be spontaneous at high temperatures. - If ∆H° < 0 and ∆S° < 0, the reaction will be spontaneous at low temperatures. - If ∆H° > 0 and ∆S° < 0, the reaction will be non-spontaneous at any temperature. - If ∆H° < 0 and ∆S° > 0, the reaction will be spontaneous at any temperature. Perform these steps for each of the given reactions and analyze the results accordingly.

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It's vital for understanding whether a reaction gives off heat or requires it. To compute \( \Delta H^\circ \) for a reaction, we use enthalpy data for the products and reactants.
This is calculated using the formula: \\[ \Delta H^\circ = \sum n \Delta H^\circ(\text{products}) - \sum m \Delta H^\circ(\text{reactants}) \]\ where \ n\ and \ m\ are the stoichiometric coefficients from the balanced equation. If the result is negative, the reaction is exothermic (releases heat), and if positive, it is endothermic (absorbs heat).

Understanding enthalpy change is crucial because it helps predict if the reaction needs energy to proceed or if it supplies energy. Generally, exothermic reactions are more favorable since they result in releasing energy.

Entropy Change

Entropy change, represented by \( \Delta S \), measures the change in disorder or randomness of a system from reactants to products. It provides insight into the feasibility and spontaneous nature of a chemical process.
Entropy change calculation involves the following equation: \\[ \Delta S^\circ = \sum n \Delta S^\circ(\text{products}) - \sum m \Delta S^\circ(\text{reactants}) \]where \ n \ and \ m \ denote the stoichiometric coefficients.

High positive values of \ \Delta S \ \ indicate an increase in disorder and are usually favorable for the spontaneity of a reaction. It's important to note that not always reactions with positive entropy change are spontaneous because it is part of the Gibbs free energy which includes both entropy and enthalpy changes.

Free Energy Change

Free energy change, known as \( \Delta G \), is a critical quantity that indicates whether a reaction can occur spontaneously under constant temperature and pressure.
The relationship combines enthalpy and entropy changes and is given by the equation: \\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]\where \ T \ is the temperature in Kelvin.

• If \( \Delta G^\circ < 0 \), the reaction is spontaneous and can proceed without external energy input.
• If \( \Delta G^\circ > 0 \), the reaction is non-spontaneous and requires energy to proceed.
• If \( \Delta G^\circ = 0 \), the reaction is at equilibrium.

It's useful in predicting how a reaction is affected by changing conditions, like temperature. The interplay between \( \Delta H \) and \( \Delta S \), and their dependence on temperature, provides a full picture of the reaction's energetic behavior.