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 relationship between Gibbs free energy (ΔG), enthalpy change (ΔH), and entropy change (ΔS). The relevant equation is:

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

At equilibrium, the change in Gibbs free energy (ΔG) is zero:

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

From this equation, we can rearrange it to solve for the temperature (T):

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

### Step 1: Identify the given values
- ΔH = 40.5 kJ/mol
- ΔS = 0.086 kJ/mol·K

### Step 2: Substitute the values into the equation
Now we substitute the values of ΔH and ΔS into the equation for T:

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

### Step 3: Calculate the temperature
Now we perform the division:

\[ T = \frac{40.5}{0.086} \]

Calculating this gives:

\[ T \approx 470.93 \, \text{K} \]

### Conclusion
The temperature at which the reaction is at equilibrium is approximately **470.93 K**.

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