For the reaction , `N_2+3H_2rarr2NH_3` if `(d[NH_3])/(dt)=2xx10^(-4)"mol L"^(-1) s^(-1)` , the value of `(-d[H_2])/(dt)` would be

Options

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

Answer

To solve the problem, we need to analyze the reaction and the rates of change of the concentrations of the reactants and products. The reaction given is:

\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

We are given that the rate of formation of ammonia (\(NH_3\)) is:

\[ \frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

We need to find the value of \(-\frac{d[H_2]}{dt}\).

### Step-by-Step Solution:

1. **Write the Rate Expressions:**
From the balanced chemical equation, we can write the rate expressions for the reactants and products:
- For \(NH_3\):
\[ \frac{d[NH_3]}{dt} = \frac{1}{2} \frac{d[H_2]}{dt} \]
- For \(H_2\):
\[ -\frac{d[H_2]}{dt} = \frac{1}{3} \frac{d[NH_3]}{dt} \]

2. **Relate the Rates:**
From the stoichiometry of the reaction, we can relate the rates of change of the concentrations:
\[ -\frac{d[H_2]}{dt} = \frac{3}{2} \frac{d[NH_3]}{dt} \]

3. **Substitute the Given Rate:**
Substitute the given rate of formation of \(NH_3\) into the equation:
\[ -\frac{d[H_2]}{dt} = \frac{3}{2} \times 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Calculate the Rate of Change of \(H_2\):**
\[ -\frac{d[H_2]}{dt} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Final Result:**
Therefore, the value of \(-\frac{d[H_2]}{dt}\) is:
\[ 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]