Equivalent weight of `N_(2)` in the change
`N_(2) rarr NH_(3)` is

Options

A. 28/6
B. 28
C. 28/2
D. 28/3

Answer

To find the equivalent weight of \( N_2 \) in the reaction \( N_2 \rightarrow NH_3 \), we will follow these steps:

### Step 1: Determine the oxidation states
- In \( N_2 \) (elemental nitrogen), the oxidation state of nitrogen is 0.
- In \( NH_3 \) (ammonia), the oxidation state of nitrogen is -3 (since hydrogen is +1, the nitrogen must be -3 to balance the charge).

### Step 2: Calculate the change in oxidation state
- The change in oxidation state for nitrogen from \( N_2 \) to \( NH_3 \) is:
\[ \text{Change} = \text{Final Oxidation State} - \text{Initial Oxidation State} = -3 - 0 = -3 \]

### Step 3: Identify the number of nitrogen atoms involved
- In the reaction, there are 2 nitrogen atoms in \( N_2 \).

### Step 4: Calculate the valence factor
- The valence factor is calculated as:
\[ \text{Valence Factor} = \text{Number of atoms} \times \text{Change in oxidation state} \]
\[ \text{Valence Factor} = 2 \times (-3) = 6 \]

### Step 5: Determine the molecular weight of \( N_2 \)
- The molecular weight of \( N_2 \) is calculated as:
\[ \text{Molecular Weight} = 14 \, \text{g/mol} \times 2 = 28 \, \text{g/mol} \]

### Step 6: Calculate the equivalent weight
- The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valence Factor}} \]
\[ \text{Equivalent Weight} = \frac{28 \, \text{g/mol}}{6} = \frac{28}{6} \, \text{g} \]

### Final Answer
The equivalent weight of \( N_2 \) in the change \( N_2 \rightarrow NH_3 \) is \( \frac{28}{6} \, \text{g} \).

---