`N_(2) + 3H_(2) to 2NH_(3)` In this reaction, equivalent weight of `N_(2)` is :

Explanation

To find the equivalent weight of \( N_2 \) in the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \), we can follow these steps:

### Step 1: Understand the Reaction
The balanced chemical equation shows that nitrogen gas (\( N_2 \)) reacts with hydrogen gas (\( H_2 \)) to form ammonia (\( NH_3 \)).

### Step 2: Determine the Oxidation States
In the reaction:
- The oxidation state of nitrogen in \( N_2 \) is 0.
- In ammonia (\( NH_3 \)), the oxidation state of nitrogen can be calculated as follows:
\[ x + 3(1) = 0 \implies x = -3 \]
Thus, the oxidation state of nitrogen changes from 0 in \( N_2 \) to -3 in \( NH_3 \).

### Step 3: Calculate the Change in Oxidation State
Since there are 2 nitrogen atoms in \( N_2 \), the total change in oxidation state for both nitrogen atoms is:
\[ \text{Total change} = 2 \times (0 - (-3)) = 2 \times 3 = 6 \]
This means that 6 electrons are involved in the reduction of \( N_2 \) to \( NH_3 \).

### Step 4: Determine the Molecular Weight of \( N_2 \)
The molecular weight of \( N_2 \) can be calculated as follows:
\[ \text{Molecular weight of } N_2 = 14 \, \text{g/mol} \times 2 = 28 \, \text{g/mol} \]

### Step 5: Calculate the n-factor
The n-factor is defined as the number of moles of electrons transferred per mole of substance. In this case, the n-factor for \( N_2 \) is equal to the total change in oxidation state, which is 6.

### Step 6: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight of } N_2 = \frac{28 \, \text{g/mol}}{6} \approx 4.67 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( N_2 \) is approximately 4.67 g/equiv.

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