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 following steps:

### Step 1: Write the reaction for the formation of HI
The formation of hydrogen iodide (HI) from its elements can be represented by the equation:
\[ \text{H}_2(g) + \text{I}_2(g) \rightarrow 2 \text{HI}(g) \]

### Step 2: Use the given bond energies and enthalpy of formation
We know the bond energies:
- Bond energy of H-H = 435 kJ/mol
- Bond energy of I-I = 150 kJ/mol
- Enthalpy of formation of HI, \( \Delta H_f^o \) = 26.5 kJ/mol

### Step 3: Write the equation for the enthalpy of formation
The enthalpy change for the reaction can be expressed in terms of bond energies:
\[ \Delta H_f^o = \text{(Bond energy of reactants)} - \text{(Bond enthalpy of products)} \]

For the formation of HI, we can express it as:
\[ \Delta H_f^o = \left(\frac{1}{2} \times \text{Bond energy of H-H} + \frac{1}{2} \times \text{Bond energy of I-I}\right) - \text{Bond enthalpy of H-I} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ 26.5 = \left(\frac{1}{2} \times 435 + \frac{1}{2} \times 150\right) - \text{Bond enthalpy of H-I} \]

Calculating the left side:
\[ \frac{1}{2} \times 435 = 217.5 \]
\[ \frac{1}{2} \times 150 = 75 \]
So,
\[ 26.5 = (217.5 + 75) - \text{Bond enthalpy of H-I} \]
\[ 26.5 = 292.5 - \text{Bond enthalpy of H-I} \]

### Step 5: Rearrange the equation to solve for Bond enthalpy of H-I
Rearranging gives:
\[ \text{Bond enthalpy of H-I} = 292.5 - 26.5 \]
\[ \text{Bond enthalpy of H-I} = 266 \text{ kJ/mol} \]

### Final Answer
The bond enthalpy of the H-I bond is **266 kJ/mol**.

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