Standard entropy of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:

Options

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

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Write the reaction and identify the given data
The reaction is:
\[ \frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightarrow XY_3 \]
The standard entropies are given as:
- \( S^\circ(X_2) = 60 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(Y_2) = 40 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(XY_3) = 50 \, \text{J K}^{-1} \text{mol}^{-1} \)

The change in enthalpy for the reaction is:
\[ \Delta H = -30 \, \text{kJ} = -30 \times 10^3 \, \text{J} \]

### Step 2: Calculate the change in entropy (\(\Delta S\))
Using the formula for the change in entropy:
\[ \Delta S = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]
For our reaction:
\[ \Delta S = S^\circ(XY_3) - \left( \frac{1}{2} S^\circ(X_2) + \frac{3}{2} S^\circ(Y_2) \right) \]
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 the 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 we have:
\[ -30 \times 10^3 = T \times (-40) \]

### Step 4: Solve for temperature (T)
Rearranging the equation:
\[ T = \frac{-30 \times 10^3}{-40} = \frac{30 \times 10^3}{40} \]
Calculating:
\[ T = 750 \, \text{K} \]

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