In a catalytic reaction involving the formation of ammonia by Haber's process `N_(2)+3H_(2)rarr 2NH_(3)` the rate of appearance of `NH_(3)` was measured as `2.5xx10^(-4) mol L^(-1)s^(-1)` The rate of disappearance of `H_(2)` will be

Options

A. `2.50 xx10^(-4) mol L^(-1)s^(-1)`
B. `1.25xx10^(-4) mol L^(-1)s^(-1)`
C. `3.75xx10^(-4) mol L^(-1)s^(-1)`
D. `5.00xx10^(-4) mol L^(-1)s^(-1)`

Explanation

To find the rate of disappearance of hydrogen gas (H₂) in the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \), we can use the stoichiometry of the reaction along with the given rate of appearance of ammonia (NH₃).

### Step-by-Step Solution:

1. **Identify the Reaction Stoichiometry:**
The balanced chemical equation is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
From this equation, we can see that for every 2 moles of NH₃ produced, 3 moles of H₂ are consumed.

2. **Write the Rate Expressions:**
The rates of disappearance and appearance can be expressed as:
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]
Here, \( \frac{d[NH_3]}{dt} \) is the rate of appearance of ammonia, and \( \frac{d[H_2]}{dt} \) is the rate of disappearance of hydrogen.

3. **Substitute the Given Rate:**
We are given that the rate of appearance of NH₃ is:
\[ \frac{d[NH_3]}{dt} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
Substituting this value into the rate expression gives:
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \times 2.5 \times 10^{-4} \]

4. **Calculate the Rate of Disappearance of H₂:**
Rearranging the equation for \( \frac{d[H_2]}{dt} \):
\[ \frac{d[H_2]}{dt} = -3 \times \left( \frac{1}{2} \times 2.5 \times 10^{-4} \right) \]
Simplifying this:
\[ \frac{d[H_2]}{dt} = -3 \times 1.25 \times 10^{-4} = -3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
The negative sign indicates the disappearance of H₂.

5. **Final Answer:**
The rate of disappearance of hydrogen gas (H₂) is:
\[ 3.75 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]