Question

For reversible reaction : \(X_{(g)}+3 Y_{(g)} \rightleftharpoons 2 Z_{(g)} ;\) \(\Delta H=-40 \mathrm{~kJ}\) Standard entropies of \(X, Y\) and \(Z\) are 60,40 and \(50 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The temperature at which the above reaction is in equilibrium is (a) \(273 \mathrm{~K}\) (b) \(600 \mathrm{~K}\) (c) \(500 \mathrm{~K}\) (d) \(400 \mathrm{~K}\)

Short answer

The temperature at which the reaction is in equilibrium is 500 K.

Step-by-step solution

Write the expression for the standard Gibbs free energy change

The standard Gibbs free energy change (\(\Delta G^\circ\)) for a reaction at equilibrium is zero. The relationship between the standard Gibbs free energy change, the standard enthalpy change (\(\Delta H^\circ\)) and the standard entropy change (\(\Delta S^\circ\)) is given by the equation \(\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\), where \(T\) is the absolute temperature in Kelvins.

Calculate the standard entropy change of the reaction

The standard entropy change of the reaction, \(\Delta S^\circ\), is calculated by taking the sum of the standard entropies of the products multiplied by their stoichiometric coefficients minus the sum of the standard entropies of the reactants multiplied by their stoichiometric coefficients. Hence, \(\Delta S^\circ = 2(50 \mathrm{~J/K\cdot mol}) - (60 \mathrm{~J/K\cdot mol} + 3 \times 40 \mathrm{~J/K\cdot mol}) = 100 \mathrm{~J/K\cdot mol} - (60 + 120) \mathrm{~J/K\cdot mol} = -80 \mathrm{~J/K\cdot mol}\).

Find the equilibrium temperature

Since \(\Delta G^\circ = 0\) at equilibrium, the temperature at which the reaction will be at equilibrium can be found using the equation \(0 = \Delta H^\circ - T \Delta S^\circ\). Solving for the temperature, we get \(T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{-40 \times 10^3 \mathrm{~J/mol}}{-80 \mathrm{~J/K\cdot mol}} = 500 \mathrm{~K}\). Therefore, the reaction is in equilibrium at 500 K.

Key concepts

Gibbs Free Energy

In the realm of chemistry and thermodynamics, Gibbs free energy (\textbf{\(\text{G}\)}) plays a crucial role in understanding the spontaneity and equilibrium of chemical reactions. Defined by the American scientist Josiah Willard Gibbs, this thermodynamic potential can predict whether a process will occur without external intervention.

Gibbs free energy combines the concepts of enthalpy (\textbf{\(\text{H}\)}), entropy (\textbf{\(\text{S}\)}), and temperature (\textbf{\(T\)}) to create a quantity that assesses the 'free' or 'usable' energy in a system. The equation \(ΔG^\text{o} = ΔH^\text{o} - TΔS^\text{o}\) elegantly encapsulates this relationship. At equilibrium, the change in Gibbs free energy is zero, meaning the system has no tendency to change and is at maximum stability.

The importance of this concept in the context of your textbook exercise is in its ability to predict the equilibrium temperature of a reaction by setting the \(ΔG^\text{o}\) to zero and rearranging the equation to solve for \(T\). This fundamental understanding aids in the prediction of reaction behavior under various conditions.

Enthalpy Change

Understanding Enthalpy in Chemical Reactions

The enthalpy change (\textbf{\(ΔH\)}) reflects the heat absorbed or released during a chemical reaction at constant pressure, a key factor in determining whether a reaction is endothermic or exothermic.

In your exercise, the given \(ΔH^\text{o}=-40 \text{kJ}\) signifies an exothermic reaction, where heat is being released into the surroundings. This release of energy can be thought of as the 'driving force' behind a reaction, with exothermic reactions often being naturally more favorable.

Enthalpy change also intertwines with the concept of Gibbs free energy as it is a determinant of its value. Adequate attention to the enthalpy change allows students to better understand the energetics of reactions and the potential impact on the Gibbs free energy.

Entropy Change

The Role of Entropy in Chemical Systems

Entropy (\textbf{\(S\)}) is a measure of the disorder or randomness within a system. An essential point to consider is that nature tends to favor an increase in entropy; in other words, systems will likely evolve towards greater disorder.

When looking at the entropy change (\textbf{\(ΔS\)}) for a reaction, it is calculated by considering the entropy of the products and reactants, taking into account their respective stoichiometry. In your textbook exercise, the calculation of \(ΔS^\text{o}\) gives a negative value, implying that the reaction leads to a decrease in disorder—a less common but feasible situation in chemical processes.

Entropy change gives insight into the 'energy dispersal' within a system. This aspect, combined with enthalpy, helps explain the balance that reactions seek to achieve when reaching equilibrium, as dictated by the second law of thermodynamics.

Equilibrium Temperature

Finding Equilibrium in Chemical Reactions

Equilibrium temperature is the specific temperature at which the forward and reverse reactions occur at the same rate in a closed system, resulting in no net change in the concentration of reactants and products over time.

In the context of the given exercise, you are asked to find the temperature at which the reaction becomes equilibrium. By utilizing the information provided on the enthalpy and entropy changes, and knowing that the Gibbs free energy change is zero at equilibrium, the equilibrium temperature can be deduced by the formula \(T = \frac{ΔH^\text{o}}{ΔS^\text{o}}\). The process demonstrates a practical application of thermodynamic concepts to determine a precise temperature, in this case, 500 K, where the reaction system is balanced.

This concept is fundamental for students to grasp as it interconnects the theoretical aspects of chemistry with real-world scenarios, such as industrial chemical manufacturing and environmental science.