For reversible reaction : `X_((g))+3Y_((g))hArr 2Z_((g)), DeltaH=-"40 kJ"` Standard entropies of X, Y and Z are 60, 40and `"50 J K"^(-1)",ol"^(-1)` respectively. The temperature at which the above reaction is in equilibrium is

Options

A. 273 K
B. 600 K
C. 500 K
D. 400 K

Explanation

To find the temperature at which the given reaction is in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
The reaction is:
\[ X_{(g)} + 3Y_{(g)} \rightleftharpoons 2Z_{(g)} \]
Given:
- \(\Delta H = -40 \, \text{kJ} = -40000 \, \text{J}\) (since 1 kJ = 1000 J)
- Standard entropies:
- \(S_X = 60 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(S_Y = 40 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(S_Z = 50 \, \text{J K}^{-1} \text{mol}^{-1}\)

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 this reaction:
\[ S_{\text{products}} = 2 \times S_Z = 2 \times 50 = 100 \, \text{J K}^{-1} \]
\[ S_{\text{reactants}} = S_X + 3 \times S_Y = 60 + 3 \times 40 = 60 + 120 = 180 \, \text{J K}^{-1} \]
Therefore,
\[ \Delta S = 100 - 180 = -80 \, \text{J K}^{-1} \]

3. **Use the Gibbs Free Energy Equation**:
At equilibrium, the Gibbs free energy change (\(\Delta G\)) is zero:
\[ \Delta G = \Delta H - T \Delta S \]
Setting \(\Delta G = 0\):
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ T = \frac{\Delta H}{\Delta S} \]

4. **Substituting the Values**:
Substitute \(\Delta H\) and \(\Delta S\) into the equation:
\[ T = \frac{-40000 \, \text{J}}{-80 \, \text{J K}^{-1}} = \frac{40000}{80} = 500 \, \text{K} \]

5. **Final Answer**:
The temperature at which the reaction is in equilibrium is:
\[ T = 500 \, \text{K} \]