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 will follow these steps:

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

Also, the enthalpy change for the reaction is:
\[ \Delta H = -30 \, \text{kJ} = -30000 \, \text{J} \]

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

### Step 3: Use Gibbs Free Energy Equation
At equilibrium, the Gibbs free energy change (\( \Delta G \)) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
Rearranging gives:
\[ \Delta H = T \Delta S \]
Substituting the values:
\[ -30000 = T \times (-40) \]

### Step 4: Solve for Temperature (T)
Rearranging the equation:
\[ T = \frac{-30000}{-40} = 750 \, \text{K} \]

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