Calculate the temperature, at which the reaction given below is at equilibrium,
`Ag_(2)O(s) to 1/2 O_(2)(g)`
Given, `DeltaH = 30.5 kJ mol^(-1)` and `DeltaS = 0.066 kJK^(-1)"mol"^(-1)`

Explanation

To calculate the temperature at which the reaction \( \text{Ag}_2\text{O}(s) \rightleftharpoons \frac{1}{2} \text{O}_2(g) \) is at equilibrium, we can use the relationship between Gibbs free energy (\( \Delta G \)), enthalpy (\( \Delta H \)), and entropy (\( \Delta S \)). At equilibrium, the Gibbs free energy change is zero (\( \Delta G = 0 \)).

### Step-by-step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The equation relating Gibbs free energy, enthalpy, and entropy is:
\[
\Delta G = \Delta H - T \Delta S
...