`Delta H` for the reaction, `I_((g))+I_((g))rarr I_(2(g))` will be :-

Options

A. Zero
B. `-ve`
C. `+ve`
D. `oo`

Answer

To determine the change in enthalpy (ΔH) for the reaction \( I_{(g)} + I_{(g)} \rightarrow I_2_{(g)} \), we can follow these steps:

### Step 1: Identify the reactants and products
In this reaction, the reactants are two iodine atoms (\( I_{(g)} \)) and the product is one molecule of iodine gas (\( I_2_{(g)} \)).

### Step 2: Understand the concept of ΔH
The change in enthalpy (ΔH) for a reaction is calculated using the formula:
\[ \Delta H = \text{Enthalpy of products} - \text{Enthalpy of reactants} \]

### Step 3: Analyze the enthalpy values
- The enthalpy of the reactants consists of two iodine atoms. Since each iodine atom has a certain enthalpy, the total enthalpy for the reactants will be higher because there are two separate iodine atoms.
- The product, \( I_2 \), has a lower enthalpy compared to the sum of the enthalpies of the two separate iodine atoms because forming a bond in \( I_2 \) releases energy.

### Step 4: Compare the enthalpy values
Since the enthalpy of the reactants (two separate iodine atoms) is greater than the enthalpy of the product (one \( I_2 \) molecule), we can conclude that:
\[ \text{Enthalpy of products} < \text{Enthalpy of reactants} \]

### Step 5: Calculate ΔH
Using the formula:
\[ \Delta H = \text{Enthalpy of products} - \text{Enthalpy of reactants} \]
Since the enthalpy of the reactants is greater, ΔH will be negative:
\[ \Delta H < 0 \]

### Conclusion
Thus, the change in enthalpy (ΔH) for the reaction \( I_{(g)} + I_{(g)} \rightarrow I_2_{(g)} \) is negative.

**Final Answer:**
\[ \Delta H < 0 \]

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