Bond energies of H - H bond is 80 kJ/mol, I - I bond is 100 kJ/mol and for H - I bond is 200 kJ/mol, the enthalpy of the reaction : `H_(2)(g) + I_(2)(g) rarr 2HI(g)` is

Options

A. `-120` kJ
B. `-220` kJ
C. `+100` kJ
D. `+120` kJ

Answer

To calculate the enthalpy change (ΔH) for the reaction:

\[ \text{H}_2(g) + \text{I}_2(g) \rightarrow 2 \text{HI}(g) \]

we will use the bond energies provided:

- Bond energy of H-H = 80 kJ/mol
- Bond energy of I-I = 100 kJ/mol
- Bond energy of H-I = 200 kJ/mol

### Step-by-Step Solution:

1. **Identify the bonds broken and formed:**
- In the reactants, we have:
- 1 H-H bond (in H₂)
- 1 I-I bond (in I₂)
- In the products, we have:
- 2 H-I bonds (in 2 HI)

2. **Calculate the total bond energy of the reactants:**
- The total bond energy of the reactants is the sum of the bond energies of the bonds that are broken.
- For H₂: 1 H-H bond = 80 kJ/mol
- For I₂: 1 I-I bond = 100 kJ/mol
- Total bond energy of reactants = 80 kJ + 100 kJ = 180 kJ

3. **Calculate the total bond energy of the products:**
- The total bond energy of the products is the sum of the bond energies of the bonds that are formed.
- For 2 HI: 2 H-I bonds = 2 × 200 kJ/mol = 400 kJ

4. **Apply the bond energy formula to find ΔH:**
- The enthalpy change (ΔH) for the reaction can be calculated using the formula:
\[ \Delta H = \text{(Total bond energy of reactants)} - \text{(Total bond energy of products)} \]
- Substituting the values:
\[ \Delta H = 180 \text{ kJ} - 400 \text{ kJ} = -220 \text{ kJ} \]

5. **Conclusion:**
- The enthalpy change for the reaction \( \text{H}_2(g) + \text{I}_2(g) \rightarrow 2 \text{HI}(g) \) is:
\[ \Delta H = -220 \text{ kJ} \]

### Final Answer:
The enthalpy of the reaction is \(-220 \text{ kJ}\).

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