Problem 89 - Chemistry: Principles and Reactions

Question

Consider the following reaction with its thermodynamic data: \(2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{AB}(\mathrm{g}) \quad \Delta H^{\circ}<0 ; \Delta S^{\circ}<0 ; \Delta G^{\circ}\) at \(60^{\circ} \mathrm{C}=+10 \mathrm{~kJ}\) Which statements about the reaction are true? (a) When \(\Delta G=1\), the reaction is at equilibrium. (b) When \(Q=1, \Delta G=\Delta G^{\circ}\). (c) At \(75^{\circ} \mathrm{C}\), the reaction is definitely nonspontaneous. (d) At \(100^{\circ} \mathrm{C}\), the reaction has a positive entropy change. (e) If \(\mathrm{A}\) and \(\mathrm{B}_{2}\) are elements in their stable states, \(S^{\circ}\) for \(\mathrm{A}\) and \(\mathrm{B}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(0 .\) (f) \(K\) for the reaction at \(60^{\circ} \mathrm{C}\) is less than 1 .

Step-by-Step Solution

Answer

Answer: Statements (b), (e), and (f) are true.

Step 1: Statement (a)

When ΔG = 1, the reaction is at equilibrium. This statement is incorrect. A reaction is at equilibrium when ΔG = 0 (not 1). So, we can discard this statement.

Step 2: Statement (b)

When Q = 1, ΔG = ΔG°. This statement is correct. We can use the formula \(ΔG = ΔG° + RT\ln{Q}\), where R is the ideal gas constant and T is the temperature in Kelvin. When Q = 1, the natural logarithm ln(Q) is equal to 0. Thus, the formula becomes \(ΔG = ΔG°\), confirming the statement.

Step 3: Statement (c)

At 75°C, the reaction is definitely nonspontaneous. We cannot definitively say if this statement is correct or incorrect without knowing the specific values of ΔH° and ΔS°. We can only infer that ΔG° is positive at 60°C, and since ΔH° is negative and ΔS° is negative, the change in temperature could make ΔG positive or negative, depending on the magnitudes of ΔH° and ΔS°. Further information would be required to confirm or deny this statement.

Step 4: Statement (d)

At 100°C, the reaction has a positive entropy change. This statement is incorrect. The problem states that ΔS° < 0, meaning the entropy change is negative. This does not change depending on the temperature.

Step 5: Statement (e)

If A and B₂ are elements in their stable states, S° for A and B₂ at 25°C is 0. This statement is correct. By definition, the standard molar entropy (S°) of an element in its stable state at 25°C is equal to 0.

Step 6: Statement (f)

K for the reaction at 60°C is less than 1. This statement is correct. The equation that relates ΔG° and the equilibrium constant K is: \(ΔG° = -RT\ln{K}\), where R is the ideal gas constant and T is the temperature in Kelvin. At 60°C (333K), the equation becomes \(10\,\text{kJ} = -8.314\,\text{J/mol-K} × 333\,\text{K} × ln K\) To solve for K, we need to isolate K: \(ln K = \frac{10,000\,\text{J}}{-8.314\,\text{J/mol-K} × 333\,\text{K}}\) After calculating the value of ln K, we find that K is less than 1, making the statement correct. In conclusion, statements (b), (e), and (f) are true.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \(\Delta G\), is a critical concept in understanding chemical reactions and their spontaneity. The equation \(\Delta G = \Delta H - T\Delta S\) links enthalpy (\(\Delta H\)), temperature (T), and entropy (\(\Delta S\)) to predict the energy available to do work at constant pressure and temperature. When \(\Delta G\) is negative, a reaction is spontaneous, meaning it can proceed without any additional energy input. If \(\Delta G\) equals zero, the reaction is at equilibrium, and there is no net change in the system. A positive \(\Delta G\) indicates that the reaction is non-spontaneous.In the context of our problem, \(\Delta G^\circ\) at \(60^\circ C\) is positive (+10 kJ), suggesting that the reaction requires energy input to proceed under standard conditions. This helps evaluate whether reactions will occur naturally or need external energy, vital for chemists in various fields.

Reaction Equilibrium

Reaction equilibrium occurs when the forward and reverse rates of a chemical reaction are equal, resulting in no net change in the concentration of reactants and products. At this point, the reaction has reached a state of balance, often referred to as chemical equilibrium.This concept is tightly connected to Gibbs Free Energy through the condition \(\Delta G = 0\). Equilibrium is also related to the equilibrium constant, \(K\), which quantifies the ratio of product concentrations to reactant concentrations at equilibrium. In the presented problem, the reaction is at equilibrium when \(\Delta G = 0\). Statement that \(\Delta G = 1\) indicates equilibrium is incorrect. A correct interpretation implies understanding the relationship between \(\Delta G^\circ\), the reaction quotient \(Q\), and \(K\). When \(Q = 1\), \(\Delta G = \Delta G^\circ\), meaning the reaction quotient equals the equilibrium constant.

Entropy Change

Entropy change (\(\Delta S\)) is an integral part of thermodynamics, describing the amount of disorder or randomness introduced in a system during a process. In a reaction, \(\Delta S\) plays a crucial role in determining the spontaneity and direction of the reaction.For the provided chemical reaction, \(\Delta S^\circ\) is negative, suggesting that the system's disorder decreases as the reaction proceeds. A negative \(\Delta S\) often implies that products are more ordered than the reactants.Understanding \(\Delta S\) is essential for insights into how temperature affects reaction spontaneity. In situations with negative \(\Delta S\), increased temperature could make the reaction less favorable if \(\Delta H\) is also negative, which is a critical component of our problem where the reaction becomes less spontaneous and potentially nonspontaneous at higher temperatures.

Standard Molar Entropy

Standard Molar Entropy, denoted as \(S^\circ\), refers to the entropy content of a substance under standardized conditions (1 atm pressure and a specific temperature, usually \(25^\circ C\)). For elements in their most stable form under standard conditions, \(S^\circ\) is often set to zero as a reference point.This baseline helps chemists calculate entropy changes in reactions, as knowing the \(S^\circ\) values of reactants and products allows for precise computation of \(\Delta S^\circ\).In the specified exercise, the standard molar entropy \(S^\circ\) for elements such as \(A\) and \(B_2\) in their stable states is zero at \(25^\circ C\). This simplification is crucial in thermochemical calculations, allowing for straightforward assessments of reaction feasibility and equilibrium state predictions. Understanding \(S^\circ\) assists in predicting how variations in temperature can impact \(\Delta G\) and ultimately influence reaction pathways.