The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6, and 180.7 J `mol^(-1)K^(-1)` respectively. Using the `DeltaG^(@)` given below, calculate the bond energy of HI.
`HI(g)rarrH(g)+I(g)," "DeltaG^(@)=271.8 kJ`

Options

A. `282.4 kJ "mol"^(-1)`
B. `298.3 kJ "mol"^(-1)`
C. `290.1 kJ "mol"^(-1)`
D. `315.4 kJ "mol"^(-1)`

Explanation

To calculate the bond energy of HI using the given data, we will follow these steps:

### Step 1: Calculate ΔS° (Change in Entropy)
The change in entropy (ΔS°) for the reaction can be calculated using the formula:
\[ \Delta S° = S_{\text{products}} - S_{\text{reactants}} \]
For the reaction:
\[ \text{HI(g)} \rightarrow \text{H(g)} + \text{I(g)} \]
The molar entropies are given as:
- \(S_{\text{HI}} = 206.5 \, \text{J mol}^{-1} \text{K}^{-1}\)
- \(S_{\text{H}} = 114.6 \, \text{J mol}^{-1} \text{K}^{-1}\)
- \(S_{\text{I}} = 180.7 \, \text{J mol}^{-1} \text{K}^{-1}\)

Now, substituting the values:
\[ \Delta S° = (S_{\text{H}} + S_{\text{I}}) - S_{\text{HI}} = (114.6 + 180.7) - 206.5 \]
Calculating this gives:
\[ \Delta S° = 295.3 - 206.5 = 88.8 \, \text{J mol}^{-1} \text{K}^{-1} \]

### Step 2: Use the Gibbs Free Energy Equation
We know from thermodynamics that:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Rearranging this gives us:
\[ \Delta H° = \Delta G° + T \Delta S° \]

### Step 3: Convert ΔG° to Joules
The given ΔG° is in kJ, so we convert it to J:
\[ \Delta G° = 271.8 \, \text{kJ} = 271800 \, \text{J} \]

### Step 4: Substitute Values into the Equation
Now we can substitute the values into the equation. The temperature \(T\) is given as 298 K:
\[ \Delta H° = 271800 \, \text{J} + (298 \, \text{K} \times 88.8 \, \text{J mol}^{-1} \text{K}^{-1}) \]
Calculating \(T \Delta S°\):
\[ T \Delta S° = 298 \times 88.8 = 26478.4 \, \text{J} \]

### Step 5: Calculate ΔH°
Now substituting this back into the equation for ΔH°:
\[ \Delta H° = 271800 + 26478.4 = 298278.4 \, \text{J} \]
Converting back to kJ:
\[ \Delta H° = 298.3 \, \text{kJ mol}^{-1} \]

### Conclusion
The bond energy of HI is approximately:
\[ \Delta H° \approx 298.3 \, \text{kJ mol}^{-1} \]

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