Question

Standard entropy of \(\mathrm{X}_{2}, \mathrm{Y}_{2}\) and \(\mathrm{XY}_{3}\) are 60,40 and 50 \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. For the reaction: \(1 / 2 \mathrm{X}_{2}+3 / 2 \mathrm{Y}_{2} \longrightarrow \mathrm{XY}_{3}, \Delta \mathrm{H}=-30 \mathrm{~kJ}\), to be at equilibrium, the temperature will be: (a) \(1250 \mathrm{~K}\) (b) \(500 \mathrm{~K}\) (c) \(750 \mathrm{~K}\) (d) \(1000 \mathrm{~K}\)

Short answer

The temperature for equilibrium is 750 K (option c).

Step-by-step solution

Understand the Reaction and Entropy Change

The given reaction is \(\frac{1}{2} \mathrm{X}_{2} + \frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3}\). We need to calculate the change in entropy, \(\Delta S^\circ\), for this reaction using the standard entropy values. The formula for the entropy change of a reaction is given by \(\Delta S^\circ = S^\circ_{products} - S^\circ_{reactants}\).

Calculate Standard Entropy Change (\(\Delta S^\circ\))

Calculate \(\Delta S^\circ\) by subtracting the sum of the entropies of the reactants from the entropy of the products: \(\Delta S^\circ = S^\circ(\mathrm{XY}_{3}) - \left(\frac{1}{2}S^\circ(\mathrm{X}_{2}) + \frac{3}{2}S^\circ(\mathrm{Y}_{2})\right)\).Substitute the given entropies:\(\Delta S^\circ = 50 - \left(\frac{1}{2} \times 60 + \frac{3}{2} \times 40\right) = 50 - (30 + 60) = 50 - 90 = -40 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1}\).

Use the Gibbs Free Energy Equation at Equilibrium

The Gibbs free energy change at equilibrium is given by the equation \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0\). This implies that \(T = \frac{\Delta H^\circ}{\Delta S^\circ}\) when the reaction is at equilibrium.

Insert Values and Solve for Temperature

Substitute the values for enthalpy change, \(\Delta H^\circ = -30 \, \mathrm{kJ/mol} = -30000 \, \mathrm{J/mol}\) and entropy change, \(\Delta S^\circ = -40 \, \mathrm{J/K/mol}\) into the equation for temperature:\[T = \frac{-30000}{-40} = 750 \, \mathrm{K}\].

Identify the Correct Option

Compare the calculated temperature (750 K) with the given options: (a) 1250 K, (b) 500 K, (c) 750 K, (d) 1000 K. The correct answer is option (c) 750 K.

Key concepts

Standard Entropy

Standard entropy, denoted as \(S^\circ\), reflects the amount of disorder or randomness in a system and is a crucial concept in thermodynamics. For chemical reactions, it's essential to calculate the entropy change \( \Delta S^\circ \) to understand how entropy influences the reaction. This is calculated using the formula: \( \Delta S^\circ = S^\circ_{products} - S^\circ_{reactants} \). Here, each substance involved in the reaction has an associated standard entropy value, which is usually measured in \( \mathrm{J}\, \mathrm{K}^{-1} \mathrm{mol}^{-1} \).

• Example: In the reaction \( \frac{1}{2} \mathrm{X}_2 + \frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3 \), we subtract the sum of the reactants' entropies from the product's entropy to find \( \Delta S^\circ \).
• The negative \( \Delta S^\circ \) value in our example indicates that the reaction leads to a decrease in disorder, which is often unfavorable for spontaneity at higher temperatures.

Understanding standard entropy enables students to better predict how changes in entropy influence the direction and feasibility of chemical reactions.

Gibbs Free Energy

Gibbs free energy, \( \Delta G^\circ \), is pivotal in determining a reaction's spontaneity. It combines enthalpy, temperature, and entropy to provide a criterion for spontaneity:

• A negative \( \Delta G^\circ \) indicates a spontaneous reaction.
• At equilibrium, however, \( \Delta G^\circ \) is zero, signifying no net change in the system.

The relationship is expressed through the equation: \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \). This formula helps chemists understand how energy and matter flow in reactions. By setting \( \Delta G^\circ = 0 \) for equilibrium, we can find the temperature where both reactants and products are stable. This knowledge allows us to control reactions effectively in industrial and laboratory settings.

Equilibrium Temperature

Equilibrium temperature in a chemical reaction is where the free energy change \( \Delta G^\circ \) becomes zero. This temperature indicates the point where the forward and reverse reaction rates are equal, meaning no net change in the concentrations of reactants and products.
In practical terms, equilibrium is achieved when the following holds true:

• \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0 \)
• Solving for temperature gives \( T = \frac{\Delta H^\circ}{\Delta S^\circ} \).
• In our example, substituting the given values leads to an equilibrium temperature of 750 K.

Understanding equilibrium temperature is essential for controlling and optimizing reactions, especially in industrial processes where precise temperature regulation can lead to more efficient and cost-effective production.

Enthalpy Change

Enthalpy change \(\Delta H^\circ\) is the heat absorbed or released under constant pressure during a reaction. This measure is vital for understanding whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).
Key points to consider:

• A negative \(\Delta H^\circ\) indicates an exothermic reaction, where heat is released.
• A positive \(\Delta H^\circ\) indicates an endothermic reaction, where heat is absorbed.

In the provided example, \(\Delta H^\circ = -30 \mathrm{kJ/mol}\) indicates that heat is released, making the reaction exothermic. Comprehending enthalpy change helps in predicting how a reaction will behave in different conditions, such as varying temperatures and pressures. It's a fundamental aspect of thermodynamics and pivotal for designing chemical processes.