Standard entropies of `x_(2), y_(2) and xy_(3)` are 70, 50 and `"60 j K"^(-1)" mol"^(-1)` respectively. For the reaction
`(1)/(2)x_(2)+(3)/(2)y_(2)hArr xy_(3). DeltaH=-30kJ` to be at equilibrium, the temperature should be

Options

A. 450 K
B. 600 K
C. 1200 K
D. 300 K

Answer

To solve the problem, we need to determine the temperature at which the given reaction is at equilibrium. The reaction is:

\[ \frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightleftharpoons XY_3 \]

Given data:
- Standard entropies:
- \( S(X_2) = 70 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S(Y_2) = 50 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S(XY_3) = 60 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Enthalpy change \( \Delta H = -30 \, \text{kJ} = -30 \times 10^3 \, \text{J} \)

### Step 1: Calculate the change in entropy (\( \Delta S \)) for the reaction.

The change in entropy can be calculated using the formula:

\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

For the reaction, we have:

\[ \Delta S = S(XY_3) - \left( \frac{1}{2} S(X_2) + \frac{3}{2} S(Y_2) \right) \]

Substituting the values:

\[ \Delta S = 60 - \left( \frac{1}{2} \times 70 + \frac{3}{2} \times 50 \right) \]

Calculating the reactants:

\[ \Delta S = 60 - \left( 35 + 75 \right) = 60 - 110 = -50 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 2: Use the Gibbs free energy equation at equilibrium.

At equilibrium, the Gibbs free energy change (\( \Delta G \)) is zero:

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

Rearranging gives:

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

### Step 3: Substitute known values into the equation.

We know:

\[ \Delta H = -30 \times 10^3 \, \text{J} \]
\[ \Delta S = -50 \, \text{J K}^{-1} \text{mol}^{-1} \]

Substituting these into the equation:

\[ -30 \times 10^3 = T \times (-50) \]

### Step 4: Solve for \( T \).

Rearranging gives:

\[ T = \frac{-30 \times 10^3}{-50} = \frac{30 \times 10^3}{50} = 600 \, \text{K} \]

### Conclusion

The temperature at which the reaction is at equilibrium is:

\[ \boxed{600 \, \text{K}} \]