Standrad entropies of `X_(2), Y_(2) and XY_(3)` are 60, 40 and `"50 JK"^(-1)" mol"^(-1)` resepectively. For the reaction
`(1)/(2)X_(2)+(3)/(2)Y_(2)harr XY_(3), DeltaH=-"30 kJ"` to be at equilibrium, the temperature shold be

Options

A. 500 K
B. 750 K
C. 1000 K
D. 1250 K

Answer

To solve the problem, we need to determine the temperature at which the given reaction is at equilibrium using the provided standard entropies and enthalpy change. Here’s a step-by-step solution:

### Step 1: Write down the reaction and the given data
The reaction is:
\[ \frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3 \]
Given standard entropies:
- \( S^\circ(X_2) = 60 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(Y_2) = 40 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(XY_3) = 50 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( \Delta H = -30 \, \text{kJ} = -30000 \, \text{J} \)

### Step 2: Calculate the entropy change (\( \Delta S \)) for the reaction
The change in entropy (\( \Delta S \)) can be calculated using the formula:
\[ \Delta S = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]
Substituting the values:
\[ \Delta S = S^\circ(XY_3) - \left( \frac{1}{2} S^\circ(X_2) + \frac{3}{2} S^\circ(Y_2) \right) \]
\[ \Delta S = 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right) \]
Calculating the reactants' contribution:
\[ \frac{1}{2} \times 60 = 30 \quad \text{and} \quad \frac{3}{2} \times 40 = 60 \]
Thus,
\[ \Delta S = 50 - (30 + 60) = 50 - 90 = -40 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Use the Gibbs free energy equation
At equilibrium, the Gibbs free energy change (\( \Delta G \)) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
This implies:
\[ \Delta H = T \Delta S \]
Rearranging for temperature \( T \):
\[ T = \frac{\Delta H}{\Delta S} \]

### Step 4: Substitute the values
Convert \( \Delta H \) to Joules:
\[ \Delta H = -30000 \, \text{J} \]
Now substituting the values:
\[ T = \frac{-30000 \, \text{J}}{-40 \, \text{J K}^{-1} \text{mol}^{-1}} = \frac{30000}{40} = 750 \, \text{K} \]

### Conclusion
The temperature at which the reaction is at equilibrium is:
\[ \boxed{750 \, \text{K}} \]