For the reaction `N_(2)+3H_(2)to2NH_(3)` , the rate `(d[NH_(3)])/(dt)=2xx10^(-4)Ms^(-1)`. Therfore the rate `-(d[N_(2)])/(dt)` is given

Options

A. `10^(-4)M"sec"^(-1)`
B. `10^(4)M"sec^(-1)`
C. `10^(-2)M"sec"^(-1)`
D. `4xx10^(-4)M"sec"^(-1)`

Explanation

To solve the problem, we need to determine the rate of change of concentration of \( N_2 \) based on the given rate of change of concentration of \( NH_3 \) for the reaction:

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

### Step-by-Step Solution:

1. **Write the Rate Expression:**
For the reaction, the rate can be expressed in terms of the change in concentration of the reactants and products. The rate expression is given by:

\[ \text{Rate} = -\frac{1}{1} \frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

2. **Identify the Given Rate:**
We are given that:

\[ \frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, \text{M/s} \]

3. **Relate the Rate of \( NH_3 \) to \( N_2 \):**
From the rate expression, we can relate the rate of formation of \( NH_3 \) to the rate of consumption of \( N_2 \):

\[ -\frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

4. **Substitute the Given Rate:**
Substitute the value of \( \frac{d[NH_3]}{dt} \) into the equation:

\[ -\frac{d[N_2]}{dt} = \frac{1}{2} \times (2 \times 10^{-4}) \, \text{M/s} \]

5. **Calculate the Rate of \( N_2 \):**
Simplifying the right-hand side gives:

\[ -\frac{d[N_2]}{dt} = 10^{-4} \, \text{M/s} \]

6. **Determine the Rate of Change of \( N_2 \):**
To find \( \frac{d[N_2]}{dt} \), we take the negative of the above result:

\[ \frac{d[N_2]}{dt} = -10^{-4} \, \text{M/s} \]

### Final Answer:
The rate of change of concentration of \( N_2 \) is:

\[ -\frac{d[N_2]}{dt} = 10^{-4} \, \text{M/s} \]