Calculate the enthalpy change for the reaction `H_(2)(g)+I_(2)(g) rarr 2HI(g)`
Given that the bond energies of H-H,I-I and H-I are433, 151 and 299 `kJmol^(-1)` respectively.

Answer

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

\[ H_2(g) + I_2(g) \rightarrow 2HI(g) \]

we will use the bond energy values provided:

- Bond energy of \( H-H \) = 433 kJ/mol
- Bond energy of \( I-I \) = 151 kJ/mol
- Bond energy of \( H-I \) = 299 kJ/mol

### Step 1: Write the formula for enthalpy change using bond energies.

The enthalpy change for the reaction can be calculated using the following formula:

\[ \Delta H = \text{(Sum of bond energies of reactants)} - \text{(Sum of bond energies of products)} \]

### Step 2: Identify the bonds broken and formed.

In the reaction, we break one \( H-H \) bond and one \( I-I \) bond, and we form two \( H-I \) bonds.

### Step 3: Calculate the total bond energy of the reactants.

The total bond energy of the reactants is the sum of the bond energies of \( H-H \) and \( I-I \):

\[ \text{Bond energy of } H_2 = 433 \text{ kJ/mol} \]
\[ \text{Bond energy of } I_2 = 151 \text{ kJ/mol} \]

Total bond energy of reactants:

\[ \text{Total BE (reactants)} = 433 + 151 = 584 \text{ kJ/mol} \]

### Step 4: Calculate the total bond energy of the products.

The total bond energy of the products is the sum of the bond energies of the two \( H-I \) bonds formed:

\[ \text{Bond energy of } 2HI = 2 \times 299 \text{ kJ/mol} = 598 \text{ kJ/mol} \]

### Step 5: Substitute the values into the formula.

Now, we can substitute the values into the enthalpy change formula:

\[ \Delta H = \text{Total BE (reactants)} - \text{Total BE (products)} \]
\[ \Delta H = 584 \text{ kJ/mol} - 598 \text{ kJ/mol} \]
\[ \Delta H = -14 \text{ kJ/mol} \]

### Final Answer:

The enthalpy change for the reaction is:

\[ \Delta H = -14 \text{ kJ/mol} \]