For this reaction
`N_(2(g))+3H_(2(g))to2NH_(3(g))`
The relation between `(d(NH_(3)))/(dt)` and `-(d(H_(2)))/(dt)` is :-

Options

A. `-(1)/(3) (d[H_(2)])/(dt) = + (1)/(2) (d[NH_(3)])/(dt)`
B. `-(1)/(2) (d[H_(2)])/(dt) = + (1)/(3) (d[NH_(3)])/(dt)`
C. `+(1)/(2) (d[H_(2)])/(dt) = - (1)/(3) (d[NH_(3)])/(dt)`
D. `+(1)/(3) (d[H_(2)])/(dt) = - (1)/(2) (d[NH_(3)])/(dt)`

Answer

To solve the problem, we need to establish the relationship between the rate of change of concentration of ammonia (NH₃) and the rate of change of concentration of hydrogen (H₂) in the given reaction:

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The balanced chemical equation is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

2. **Define the Rate of Reaction:**
The rate of reaction can be expressed in terms of the change in concentration of reactants and products. According to the stoichiometry of the reaction:
- For nitrogen (N₂):
\[ \text{Rate} = -\frac{1}{1} \frac{d[N_2]}{dt} \]
- For hydrogen (H₂):
\[ \text{Rate} = -\frac{1}{3} \frac{d[H_2]}{dt} \]
- For ammonia (NH₃):
\[ \text{Rate} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

3. **Set Up the Equations:**
Since all these expressions represent the same rate of reaction, we can equate them:
\[ -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

4. **Relate \( \frac{d[NH_3]}{dt} \) and \( -\frac{d[H_2]}{dt} \):**
From the expression for hydrogen, we can express \( \frac{d[NH_3]}{dt} \) in terms of \( -\frac{d[H_2]}{dt} \):
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

5. **Rearranging the Equation:**
To find the relationship, we can rearrange the equation:
\[ \frac{d[NH_3]}{dt} = -\frac{3}{2} \frac{d[H_2]}{dt} \]

6. **Final Relation:**
The final relationship between the rates is:
\[ \frac{d[NH_3]}{dt} = -\frac{3}{2} \left(-\frac{d[H_2]}{dt}\right) \]
or
\[ \frac{d[NH_3]}{dt} = \frac{3}{2} \frac{d[H_2]}{dt} \]

### Conclusion:
Thus, the relationship between \( \frac{d[NH_3]}{dt} \) and \( -\frac{d[H_2]}{dt} \) is:
\[ \frac{d[NH_3]}{dt} = \frac{3}{2} \left(-\frac{d[H_2]}{dt}\right) \]