For the reaction `N_(2)+3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(dt).=4xx10^(-4)` mol `L^(-1)s^(-1)`, the value of `(-d[H_(2)])/(dt)` would be

Options

A. `3xx10^(-4)" mol l"^(-1)s^(-1)`
B. `4xx10^(-4)"mol l"^(-1)s^(-1)`
C. `6xx10^(-4)"mol l"^(-1)s^(-1)`
D. `6.5xx10^(-4)"mol l"^(-1)s^(-1)`

Answer

To solve the problem, we need to analyze the given reaction and apply the concept of reaction rates based on stoichiometry.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Identify the stoichiometric coefficients:**
- For \(N_2\), the coefficient is 1.
- For \(H_2\), the coefficient is 3.
- For \(NH_3\), the coefficient is 2.

3. **Understand the relationship between the rates of change of concentrations:**
The rate of the reaction can be expressed in terms of the change in concentration over time for each species involved:
\[ -\frac{1}{1} \frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

4. **Given information:**
We know that:
\[ \frac{d[NH_3]}{dt} = 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Relate the rate of formation of \(NH_3\) to the rate of disappearance of \(H_2\):**
From the stoichiometry of the reaction, we can set up the following equation:
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

6. **Substituting the known value of \(\frac{d[NH_3]}{dt}\):**
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \times 4 \times 10^{-4} \]

7. **Calculate \(\frac{d[H_2]}{dt}\):**
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = 2 \times 10^{-4} \]
Now, multiply both sides by -3 to solve for \(\frac{d[H_2]}{dt}\):
\[ \frac{d[H_2]}{dt} = -3 \times 2 \times 10^{-4} = -6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

8. **Final answer:**
The value of \(-\frac{d[H_2]}{dt}\) is:
\[ 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]