The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm. Calculate `DeltaG^(@)` for the formation of `1` mole `Ag_(2)O(S)` from metal and oxygen at this temperature. (`"log" 207=2.315`)

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the formation of 1 mole of silver oxide (Ag₂O) from silver (Ag) and oxygen (O₂) at 445 °C, we will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of silver oxide can be represented as:
\[ \text{Ag}_2\text{O}(s) \rightleftharpoons 2 \text{Ag}(s) + \frac{1}{2} \text{O}_2(g) \]

### Step 2: Convert the temperature to Kelvin
The temperature in Celsius is given as 445 °C. To convert this to Kelvin:
\[ T(K) = 445 + 273 = 718 \, K \]

### Step 3: Identify the dissociation pressure (Kp)
The dissociation pressure of silver oxide at this temperature is given as 207 atm. The equilibrium constant \( K_p \) for the dissociation reaction can be expressed as:
\[ K_p = \frac{P_{\text{O}_2}^{1/2}}{1} = \frac{1}{(P_{\text{O}_2})^{1/2}} \]
Thus, we have:
\[ K_p = \frac{1}{207^{1/2}} \]

### Step 4: Use the relationship between ΔG° and Kp
At equilibrium, the relationship between ΔG° and Kp is given by:
\[ \Delta G° = -RT \ln K_p \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 718 \, K \)

### Step 5: Calculate Kp using the logarithm
Using the logarithmic property:
\[ \ln K_p = \ln \left(\frac{1}{207^{1/2}}\right) = -\frac{1}{2} \ln(207) \]
Given \( \log(207) = 2.315 \), we can convert it to natural logarithm:
\[ \ln(207) = 2.303 \times 2.315 \approx 5.319 \]
Thus,
\[ \ln K_p = -\frac{1}{2} \times 5.319 = -2.6595 \]

### Step 6: Substitute values into the ΔG° equation
Now substituting the values into the ΔG° equation:
\[ \Delta G° = - (8.314 \, J/(mol \cdot K)) \times (718 \, K) \times (-2.6595) \]

### Step 7: Calculate ΔG°
Calculating the above expression:
\[ \Delta G° = 8.314 \times 718 \times 2.6595 \]
\[ \Delta G° \approx 8.314 \times 718 \times 2.6595 \approx 13,747.65 \, J \]

### Step 8: Convert to kilocalories
To convert joules to kilocalories:
\[ \Delta G° \approx 13,747.65 \, J \times 0.000239 \, \text{kcal/J} \approx 3.80 \, \text{kcal} \]

### Final Answer
Thus, the ΔG° for the formation of 1 mole of Ag₂O from metal and oxygen at 445 °C is approximately:
\[ \Delta G° \approx 3.80 \, \text{kcal} \]

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