The temperature at which the given reaction is at equilibrium `Ag_(2)O_(s) rightarrow 2Ag(s) + 1/2 O_(2)(g)`
`Delta H = 40.5 kJ mol^(-1)` and `DeltaS= 0.086 k J mol^(-1) K^(-1)`

Options

A. 470.93 K
B. 362.12 K
C. 262.12 K
D. 562.12 K

Answer

To find the temperature at which the given reaction is at equilibrium, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

At equilibrium, the change in Gibbs free energy (\(\Delta G\)) is zero. Therefore, we can set up the equation as follows:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this equation to solve for temperature \(T\):

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Identify the values of \(\Delta H\) and \(\Delta S\)

From the question, we have:
- \(\Delta H = 40.5 \, \text{kJ/mol}\)
- \(\Delta S = 0.086 \, \text{kJ/mol K}\)

### Step 2: Substitute the values into the equation

Now, we can substitute the values into the equation we derived for \(T\):

\[ T = \frac{40.5 \, \text{kJ/mol}}{0.086 \, \text{kJ/mol K}} \]

### Step 3: Perform the calculation

Calculating the temperature:

\[ T = \frac{40.5}{0.086} \approx 470.93 \, \text{K} \]

### Conclusion

The temperature at which the reaction is at equilibrium is approximately \(470.93 \, \text{K}\).

---