The bond energies of H - H and I - I bonds are `435 kJ mol^-1 and 150 kJ mol^-1` respectively. If `DeltaH_f^o` for HI is `26.5 kJ mol^-1` then bond enthalpy of H-I bond is

Options

A. 558.5kJ/mol
B. 266kJ/mol
C. 611.5kJ/mol
D. 293kJ/mol

Answer

To find the bond enthalpy of the H-I bond, we can use the given bond energies and the enthalpy of formation of HI. Here’s a step-by-step solution:

### Step 1: Write the formation reaction for HI
The formation of HI can be represented as:
\[ \frac{1}{2} \text{H}_2 + \frac{1}{2} \text{I}_2 \rightarrow \text{HI} \]

### Step 2: Write the expression for the enthalpy change (ΔH)
The enthalpy change (ΔH) for the reaction can be expressed using the bond energies:
\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

### Step 3: Substitute the bond energies
The bond energies given are:
- H-H bond energy = 435 kJ/mol
- I-I bond energy = 150 kJ/mol

Thus, the bond energies for the reactants in the formation of HI are:
\[ \Delta H = \left(\frac{1}{2} \times 435 + \frac{1}{2} \times 150\right) - \text{Bond energy of HI} \]

### Step 4: Calculate the bond energies of the reactants
Calculating the bond energies:
- For H-H: \(\frac{1}{2} \times 435 = 217.5 \, \text{kJ/mol}\)
- For I-I: \(\frac{1}{2} \times 150 = 75 \, \text{kJ/mol}\)

Adding these together gives:
\[ 217.5 + 75 = 292.5 \, \text{kJ/mol} \]

### Step 5: Set up the equation using ΔH_f^o
We know that the standard enthalpy of formation of HI is given as:
\[ \Delta H_f^o = 26.5 \, \text{kJ/mol} \]

Now we can set up the equation:
\[ 26.5 = 292.5 - \text{Bond energy of HI} \]

### Step 6: Solve for the bond energy of HI
Rearranging the equation to find the bond energy of HI:
\[ \text{Bond energy of HI} = 292.5 - 26.5 \]
\[ \text{Bond energy of HI} = 266 \, \text{kJ/mol} \]

### Final Answer
The bond enthalpy of the H-I bond is:
\[ \text{Bond energy of HI} = 266 \, \text{kJ/mol} \]

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