`N_(2) + 3 H_(2) rarr 2NH_(3)`
Molecular weight of `NH_(3)` and `N_(2)` and `x_(1)` and `x_(2)`, respectively. Their equivalent weights are `y_(1)` and `y_(2)`, respectively. Then `(y_(1) - y_(2))`

Options

A. `((2x_(1) - x_(2))/(6))`
B. `(x_(1) - x_(2))`
C. `(3x_(1) - x_(2))`
D. `(x_(1) - 3x_(2))`

Explanation

To solve the problem, we need to calculate the difference between the equivalent weights of ammonia (NH₃) and nitrogen (N₂) based on the given chemical reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The balanced chemical equation is:
\[ N_2 + 3 H_2 \rightarrow 2 NH_3 \]

2. **Determine Molecular Weights**:
Let:
- \( x_1 \) = Molecular weight of \( NH_3 \)
- \( x_2 \) = Molecular weight of \( N_2 \)

3. **Calculate Equivalent Weights**:
The equivalent weight is defined as the molecular weight divided by the number of moles of reactive units (in this case, hydrogen atoms) involved in the reaction.

- For \( N_2 \):
- In the reaction, 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \), which involves 6 moles of hydrogen (since \( 3 H_2 \) gives 6 H atoms).
- Therefore, the equivalent weight \( y_2 \) of \( N_2 \) is:
\[ y_2 = \frac{x_2}{6} \]

- For \( NH_3 \):
- In the reaction, 2 moles of \( NH_3 \) are produced from the reaction, which also involves 6 moles of hydrogen.
- Therefore, the equivalent weight \( y_1 \) of \( NH_3 \) is:
\[ y_1 = \frac{2x_1}{6} = \frac{x_1}{3} \]

4. **Calculate \( y_1 - y_2 \)**:
Now, we can find the difference between the equivalent weights:
\[ y_1 - y_2 = \frac{x_1}{3} - \frac{x_2}{6} \]

To combine these fractions, we need a common denominator:
\[ y_1 - y_2 = \frac{2x_1}{6} - \frac{x_2}{6} = \frac{2x_1 - x_2}{6} \]

5. **Final Result**:
Thus, the expression for the difference in equivalent weights is:
\[ y_1 - y_2 = \frac{2x_1 - x_2}{6} \]