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 find the rate of reaction and the rate of disappearance of \( H_2 \) given the rate of appearance of \( NH_3 \).

### Step-by-Step Solution:

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

2. **Given Data:**
The rate of appearance of \( NH_3 \) is given as:
\[ \text{Rate of } NH_3 = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

3. **Rate of Reaction:**
The rate of reaction can be expressed in terms of the rate of appearance of products and the rate of disappearance of reactants. For the given reaction, we can write:
\[ R = \frac{1}{2} \frac{d[NH_3]}{dt} \]
where the factor of \( \frac{1}{2} \) comes from the stoichiometry of the reaction (2 moles of \( NH_3 \) are produced).

4. **Calculate the Rate of Reaction:**
Substituting the given rate of appearance of \( NH_3 \):
\[ R = \frac{1}{2} \times (2.5 \times 10^{-4}) = 1.25 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Rate of Disappearance of \( H_2 \):**
The rate of disappearance of \( H_2 \) can be expressed as:
\[ R = -\frac{1}{3} \frac{d[H_2]}{dt} \]
where the factor of \( -\frac{1}{3} \) comes from the stoichiometry of the reaction (3 moles of \( H_2 \) are consumed).

6. **Set Up the Equation:**
From the rate of reaction we calculated:
\[ 1.25 \times 10^{-4} = -\frac{1}{3} \frac{d[H_2]}{dt} \]

7. **Solve for \( \frac{d[H_2]}{dt} \):**
Rearranging gives:
\[ \frac{d[H_2]}{dt} = -3 \times (1.25 \times 10^{-4}) = -3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

8. **Final Answers:**
- Rate of reaction: \( 1.25 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Rate of disappearance of \( H_2 \): \( 3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)