Question

The temperature at which the reaction $$ \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) $$ is at equilibrium is \(\ldots \ldots \ldots\) Given \(\Delta \mathrm{H}=30 \mathrm{~kJ}\) mol and \(\Delta S=0.066 \mathrm{~kJ} \mathrm{k}^{\prime} \mathrm{mol}^{\prime}\). (1) \(462.12 \mathrm{~K}\) (2) \(362.12 \mathrm{~K}\) (3) \(262.12 \mathrm{~K}\) (4) \(562.12 \mathrm{~K}\)

Short answer

462.12 K

Step-by-step solution

- Understand the given data

Identify the given values: \( \Delta \mathrm{H}=30 \mathrm{~kJ} / \mathrm{mol} \) and \( \Delta S=0.066 \mathrm{~kJ} / (\mathrm{mol} \cdot \mathrm{K}) \)

- Recall the condition for equilibrium

For a reaction to be at equilibrium, the Gibbs free energy change, \( \Delta G \), must be zero. The relationship is given by: \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \( \Delta G = 0 \) which implies \( \Delta H = T \Delta S \)

- Solve for temperature (T)

Using the equation \[ \Delta H = T \Delta S \], solve for \( T \): \[ T = \frac{ \Delta H}{ \Delta S } \] Substitute the given values: \[ T = \frac{30 \mathrm{~kJ} / \mathrm{mol}}{0.066 \mathrm{~kJ} / (\mathrm{mol} \cdot \mathrm{K})} \]

- Calculate the value of T

Perform the division: \[ T = \frac{30}{0.066} = 454.55 \]Round to one of the given options closest to 454.55, which is 462.12 \( K \)

Key concepts

Gibbs Free Energy

Gibbs free energy is a thermodynamic potential used to predict whether a process will occur spontaneously at constant pressure and temperature. It is defined as: \[ \text{Gibbs free energy} (\text{G}) = \text{enthalpy} (\text{H}) - \text{temperature} (\text{T}) \times \text{entropy} (\text{S}) \]
In this equation:

• \textbf{Enthalpy (H)} represents the total energy of the system.
• \textbf{Temperature (T)} is in Kelvins and must always be positive.
• \textbf{Entropy (S)} measures the disorder or randomness of the system.

When \

Enthalpy Change (ΔH)

Enthalpy (\text{ΔH}) refers to the heat content of a system and is a measure of the total energy in the system. It accounts for energy stored in chemical bonds and during phase changes.

• Enthalpy change is given by \( \text{ΔH} = \text{H}_{\text{products}} - \text{H}_{\text{reactants}} \).
  If \( \text{ΔH} > 0 \), the process is endothermic (heat is absorbed).
  If \( \text{ΔH} < 0 \), the process is exothermic (heat is released).

In the given problem, the enthalpy change for the decomposition of \

Entropy Change (ΔS)

Entropy (\text{ΔS}) is a measure of randomness or disorder in a system. In a reaction, the entropy change can help predict the feasibility of the process.

• A positive entropy change (\text{ΔS} > 0) indicates an increase in disorder.
• A negative entropy change (\text{ΔS} < 0) indicates a decrease in disorder.

In the given exercise, the entropy change, \( \text{ΔS} = 0.066 \text{kJ/(mol⋅K)} \), measures how the disorder or randomness of the system changes during the decomposition reaction of \