The equilibrium constant for
`A(g)+B_(2)(g) hArr AB_(2)(g)" "k_(p)=100` at `522K`.
Structure of `AB_(2)` is like `H_(2)O` . If bond energy of `A-B` bond is `200 kJ//mol` and that of `B-B` bond is `100 kJ//mol,` the find `Delta S^(@)` of the above reaction `:`

Options

A. `-0.53J //mol-K`
B. `-536 J //mol-K`
C. `-550J//mol-K`
D. `-5.36J//mol-K`

Explanation

To solve the problem, we need to find the standard entropy change (ΔS°) for the reaction:

\[ A(g) + B_2(g) \rightleftharpoons AB_2(g) \]

Given data:
- Equilibrium constant \( K_p = 100 \) at \( 522 \, K \)
- Bond energy of \( A-B \) bond = \( 200 \, kJ/mol \)
- Bond energy of \( B-B \) bond = \( 100 \, kJ/mol \)

### Step 1: Write the reaction and identify the bonds
The reaction is:
\[ A(g) + B_2(g) \rightarrow AB_2(g) \]

In this reaction:
- We break one \( B-B \) bond in \( B_2 \).
- We form two \( A-B \) bonds in \( AB_2 \).

### Step 2: Calculate the change in enthalpy (ΔH°)
Using the bond energies, we can calculate the change in enthalpy (ΔH°) using the formula:
\[ \Delta H° = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

For the reactants:
- Breaking one \( B-B \) bond: \( +100 \, kJ/mol \)

For the products:
- Forming two \( A-B \) bonds: \( -2 \times 200 \, kJ/mol = -400 \, kJ/mol \)

Thus, we have:
\[ \Delta H° = 100 - 400 = -300 \, kJ/mol \]

### Step 3: Calculate the change in Gibbs free energy (ΔG°)
Using the equilibrium constant \( K_p \), we can calculate ΔG° using the formula:
\[ \Delta G° = -RT \ln K_p \]

Where:
- \( R = 8.314 \, J/(mol \cdot K) = 0.008314 \, kJ/(mol \cdot K) \)
- \( T = 522 \, K \)
- \( K_p = 100 \)

Calculating:
\[ \Delta G° = - (0.008314 \, kJ/(mol \cdot K)) \times (522 \, K) \times \ln(100) \]
Using \( \ln(100) = 4.605 \):
\[ \Delta G° = - (0.008314 \times 522 \times 4.605) \approx -19.99 \, kJ/mol \]

### Step 4: Calculate the standard entropy change (ΔS°)
Using the relationship between ΔG°, ΔH°, and ΔS°:
\[ \Delta G° = \Delta H° - T \Delta S° \]

Rearranging gives:
\[ \Delta S° = \frac{\Delta H° - \Delta G°}{T} \]

Substituting the values:
\[ \Delta S° = \frac{-300 \, kJ/mol - (-19.99 \, kJ/mol)}{522 \, K} \]
\[ \Delta S° = \frac{-300 + 19.99}{522} = \frac{-280.01}{522} \approx -0.536 \, kJ/(mol \cdot K) \]

Converting to Joules:
\[ \Delta S° \approx -536 \, J/(mol \cdot K) \]

### Final Answer:
The standard entropy change (ΔS°) for the reaction is approximately:
\[ \Delta S° = -536 \, J/(mol \cdot K) \]