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. 373 K
B. 413 K
C. 463 K
D. 493 K

Answer

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

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

Given that ΔG = 0 at equilibrium, we can set up the equation:

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

Rearranging this gives us:

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

From this, we can solve for T:

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

### Step 1: Convert ΔH from kJ to J
The given value of ΔH is 30.56 kJ/mol. To convert this to joules, we multiply by 1000:

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

### Step 2: Use the given value of ΔS
The value of ΔS is given as 66 J/K·mol.

### Step 3: Substitute ΔH and ΔS into the equation for T
Now we can substitute the values of ΔH and ΔS into the equation for T:

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

### Step 4: Calculate T
Now, performing the division:

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

### Step 5: Round to appropriate significant figures
Since we are dealing with significant figures based on the given data, 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**.

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