For the reaction
`H_(2)+I_(2)hArr2HI`, `DeltaH=12.40 kcal`. The heat of formation `(DeltaH)` of `HI` is

Options

A. 12.4 kcal
B. 6.20 kcal
C. `-6.20` kcal
D. `-12.4` kcal

Explanation

To find the heat of formation of HI from the given reaction \( H_2 + I_2 \rightleftharpoons 2HI \) with \( \Delta H = 12.40 \) kcal, we can follow these steps:

### Step 1: Understand the Reaction
The reaction shows that 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \). The enthalpy change (\( \Delta H \)) for this reaction is given as 12.40 kcal.

### Step 2: Define Heat of Formation
The heat of formation (\( \Delta H_f \)) of a compound is defined as the enthalpy change when 1 mole of the compound is formed from its elements in their standard states.

### Step 3: Relate the Given Reaction to Heat of Formation
Since the reaction produces 2 moles of \( HI \), the heat of formation for 2 moles of \( HI \) is equal to the given \( \Delta H \):
\[ \Delta H_f \text{ for } 2 \text{ moles of } HI = 12.40 \text{ kcal} \]

### Step 4: Calculate Heat of Formation for 1 Mole of HI
To find the heat of formation for 1 mole of \( HI \), we need to divide the total enthalpy change by 2:
\[ \Delta H_f \text{ for } 1 \text{ mole of } HI = \frac{12.40 \text{ kcal}}{2} = 6.20 \text{ kcal} \]

### Step 5: Conclusion
The heat of formation of \( HI \) is \( 6.20 \) kcal.

### Final Answer
The heat of formation of \( HI \) is \( 6.20 \) kcal.

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