For `2NH_(3) overset(Av) to N_(2) + 3H_(2)` rate w.r.t `N_(2)` is `2 xx 10^(-3) M min^(-1)`, then rate w.r.t `N_(2)` after 20 min will be (in `M "min"^(-1)`)

Options

A. `2 xx 10^(-3)`
B. `gt 2 xx 10^(-3)`
C. `10^(-4)`
D. `lt 2 xx 10^(-3)`

Explanation

To solve the problem, we need to determine the rate of the reaction with respect to \( N_2 \) after 20 minutes, given that the initial rate is \( 2 \times 10^{-3} \, M \, min^{-1} \).

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction given is:
\[ 2NH_3 \overset{Av}{\rightarrow} N_2 + 3H_2 \]
This indicates that 2 moles of ammonia (\( NH_3 \)) produce 1 mole of nitrogen (\( N_2 \)) and 3 moles of hydrogen (\( H_2 \)).

2. **Identify the Rate of Reaction**: The rate of formation of \( N_2 \) is given as:
\[ \text{Rate} = \frac{d[N_2]}{dt} = 2 \times 10^{-3} \, M \, min^{-1} \]

3. **Calculate the Change in Concentration of \( N_2 \)**:
- The rate tells us how much \( N_2 \) is produced per minute.
- After 20 minutes, the total amount of \( N_2 \) produced can be calculated as:
\[ \Delta [N_2] = \text{Rate} \times \text{Time} = (2 \times 10^{-3} \, M \, min^{-1}) \times 20 \, min = 4 \times 10^{-2} \, M \]

4. **Determine the Rate After 20 Minutes**:
- The rate of a reaction can change over time depending on the concentrations of the reactants. However, if we assume that the rate remains constant (which is common in introductory kinetics problems unless specified otherwise), the rate after 20 minutes would still be:
\[ \text{Rate after 20 min} = 2 \times 10^{-3} \, M \, min^{-1} \]

5. **Final Answer**: Therefore, the rate with respect to \( N_2 \) after 20 minutes is:
\[ \text{Rate} = 2 \times 10^{-3} \, M \, min^{-1} \]