Standard entropy of `X _(2) , Y _(2) and X Y _(3)` are 60, 40 and 50 J `K ^(-1) mol ^(-1)` , respectively. For the reaction, ` 1 / 2 X _(2) + 3 / 2 Y _ (2) to X Y _(3) , DeltaH = -3 0 K J` , to be at equilibrium, the temperature will be to be at equilibrium, the temperature will be

Options

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

Explanation

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

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

We are given the standard entropies of the substances involved and the enthalpy change for the reaction.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Standard entropy of \( X_2 \) = 60 J/K/mol
- Standard entropy of \( Y_2 \) = 40 J/K/mol
- Standard entropy of \( XY_3 \) = 50 J/K/mol
- Enthalpy change (\( \Delta H \)) = -30 kJ = -30000 J

2. **Write the Gibbs Free Energy Equation:**
At equilibrium, the change in Gibbs free energy (\( \Delta G \)) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
Rearranging gives:
\[ \Delta H = T \Delta S \]

3. **Calculate the Change in Entropy (\( \Delta S \)):**
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]
For the reaction:
\[ \Delta S = S_{XY_3} - \left( \frac{1}{2} S_{X_2} + \frac{3}{2} S_{Y_2} \right) \]
Substituting the values:
\[ \Delta S = 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right) \]
Calculate the reactants' entropy:
\[ \frac{1}{2} \times 60 = 30 \quad \text{and} \quad \frac{3}{2} \times 40 = 60 \]
Therefore:
\[ \Delta S = 50 - (30 + 60) = 50 - 90 = -40 \, \text{J/K/mol} \]

4. **Substitute Values into the Gibbs Equation:**
Now substitute \( \Delta H \) and \( \Delta S \) into the equation:
\[ -30000 = T \times (-40) \]
Rearranging gives:
\[ T = \frac{30000}{40} \]

5. **Calculate Temperature:**
\[ T = 750 \, \text{K} \]

Thus, the temperature at which the reaction is at equilibrium is **750 K**.