For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` the value of `Delta H=30.56 KJ mol^(_1)` and `Delta S = 66 JK^(-1)mol^(-1)`. The temperature at which the free energy change for the reaction will be zero is :-

Options

A. 373k
B. 413k
C. 463k
D. 493k

Answer

To find the temperature at which the Gibbs free energy change (ΔG) for the reaction becomes zero, we can use the Gibbs free energy equation:

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

Where:
- ΔG = Gibbs free energy change
- ΔH = Enthalpy change
- T = Temperature in Kelvin
- ΔS = Entropy change

Given:
- ΔH = 30.56 kJ/mol
- ΔS = 66 J/K·mol

### Step 1: Convert ΔH from kJ to J
Since ΔH is given in kJ, we need to convert it to J for consistency with ΔS, which is in J/K·mol.

\[ \Delta H = 30.56 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 30560 \, \text{J/mol} \]

### Step 2: Set ΔG to zero
We are looking for the temperature at which ΔG = 0.

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

### Step 3: Rearrange the equation to solve for T
Rearranging the equation gives us:

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

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

### Step 4: Substitute the values
Now we can substitute the values of ΔH and ΔS into the equation:

\[ T = \frac{30560 \, \text{J/mol}}{66 \, \text{J/K·mol}} \]

### Step 5: Calculate T
Perform the division:

\[ T = \frac{30560}{66} \approx 463.64 \, \text{K} \]

### Step 6: Round to appropriate significant figures
Since we are dealing with significant figures, we can round this to:

\[ T \approx 463 \, \text{K} \]

### Final Answer
The temperature at which the free energy change for the reaction will be zero is approximately **463 K**.

---