In a reaction `N_(2(g))+3H_(2(g))to2NH_(3(g))` rate of appearance of `NH_(3)` is `2.5xx10^(-4)molL^(-1)sec^(-1)` then rate of reaction and rate of disappearance of `H_(2)` respectively is :-

Options

A. `3.75xx10^(-4),1.25xx10^(-4)`
B. `1.25xx10^(-4),2.5xx10^(-4)`
C. `1.25xx10^(-4),3.75xx10^(-4)`
D. `5xx10^(-4),3.75xx10^(-4)`

Explanation

To solve the problem, we need to determine the rate of reaction and the rate of disappearance of \( H_2 \) based on the given rate of appearance of \( NH_3 \).

### Step-by-Step Solution:

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

2. **Understand Rate Relationships:**
The rates of appearance and disappearance can be expressed as:
\[ -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]
Here, the coefficients from the balanced equation indicate how the rates relate to one another.

3. **Given Information:**
The rate of appearance of \( NH_3 \) is given as:
\[ \frac{d[NH_3]}{dt} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \]

4. **Calculate the Rate of Reaction:**
The rate of reaction can be expressed in terms of the rate of appearance of \( NH_3 \):
\[ \text{Rate of Reaction} = \frac{1}{2} \frac{d[NH_3]}{dt} \]
Substituting the value:
\[ \text{Rate of Reaction} = \frac{1}{2} \times 2.5 \times 10^{-4} = 1.25 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \]

5. **Calculate the Rate of Disappearance of \( H_2 \):**
The rate of disappearance of \( H_2 \) can be expressed as:
\[ -\frac{d[H_2]}{dt} = 3 \times \text{Rate of Reaction} \]
Substituting the rate of reaction we calculated:
\[ -\frac{d[H_2]}{dt} = 3 \times 1.25 \times 10^{-4} = 3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \]

### Final Answers:
- Rate of Reaction: \( 1.25 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \)
- Rate of Disappearance of \( H_2 \): \( 3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \)