Standard entropies of `X_(2), Y_(2)` and `XY_(3)` are 60, 30 are `50 JK^(-1) mol^(-1)` respectively. For the reaction `(1)/(2)X_(2)+(3)/(2)Y_(2)hArr XY_(3), Delta H = -30 kJ` to be at equilibrium, the temperature should be :

Options

A. 1200 K
B. 1000 K
C. 750 K
D. 500 K

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
We have the standard entropies of the substances involved in the reaction:
- Standard entropy of \(X_2\) = 60 J/K·mol
- Standard entropy of \(Y_2\) = 30 J/K·mol
- Standard entropy of \(XY_3\) = 50 J/K·mol

The reaction is:
\[ \frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3 \]
The change in enthalpy (\(\Delta H\)) for the reaction is given as -30 kJ, which we convert to Joules:
\[ \Delta H = -30 \, \text{kJ} = -30,000 \, \text{J} \]

### Step 2: Calculate the change in entropy (\(\Delta S\))
The change in entropy for the reaction can be calculated using the formula:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

For the products:
- Entropy of \(XY_3\) = 50 J/K·mol

For the reactants:
- For \(X_2\), since we have \(\frac{1}{2}\) mole, the contribution is:
\[ \frac{1}{2} \times 60 = 30 \, \text{J/K·mol} \]
- For \(Y_2\), since we have \(\frac{3}{2}\) moles, the contribution is:
\[ \frac{3}{2} \times 30 = 45 \, \text{J/K·mol} \]

Now, we can calculate the total entropy of the reactants:
\[ S_{\text{reactants}} = 30 + 45 = 75 \, \text{J/K·mol} \]

Now we can find \(\Delta S\):
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} = 50 - 75 = -25 \, \text{J/K·mol} \]

### Step 3: Use the Gibbs free energy equation
At equilibrium, the Gibbs free energy change (\(\Delta G\)) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
From this, we can rearrange to find the temperature:
\[ \Delta H = T \Delta S \]
\[ T = \frac{\Delta H}{\Delta S} \]

### Step 4: Substitute the values
Substituting the values we have:
\[ T = \frac{-30,000 \, \text{J}}{-25 \, \text{J/K}} = \frac{30,000}{25} = 1200 \, \text{K} \]

### Final Answer
The temperature at which the reaction is at equilibrium is:
\[ \boxed{1200 \, \text{K}} \]