The rate of production of `NH_(3)" in "N_(2)+3H_(2) rarr 2NH_(3)" is "3.4" kg min"^(-1)`. The rate of consumption of `H_(2)` is :

Options

A. `5.1" kg min"^(-1)`
B. `0.01" kg sec"^(-1)`
C. `0.6" kg hr"^(-1)`
D. None of these

Answer

To find the rate of consumption of \( H_2 \) in the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \), we can use the stoichiometry of the reaction and the given rate of production of \( NH_3 \).

### Step-by-Step Solution:

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

2. **Determine the Rate of Production of \( NH_3 \)**:
The rate of production of \( NH_3 \) is given as \( 3.4 \, \text{kg/min} \).

3. **Convert the Rate of Production to Moles**:
To relate the rate of production to the rate of consumption, we need to convert the mass of \( NH_3 \) produced to moles. The molar mass of \( NH_3 \) (ammonia) is approximately \( 17 \, \text{g/mol} \) or \( 0.017 \, \text{kg/mol} \).
\[ \text{Moles of } NH_3 = \frac{3.4 \, \text{kg/min}}{0.017 \, \text{kg/mol}} \approx 200 \, \text{mol/min} \]

4. **Use Stoichiometry to Find the Rate of Consumption of \( H_2 \)**:
From the balanced equation, the ratio of \( NH_3 \) produced to \( H_2 \) consumed is:
\[ \frac{2 \, \text{mol } NH_3}{3 \, \text{mol } H_2} \]
Therefore, if \( 200 \, \text{mol/min} \) of \( NH_3 \) is produced, the rate of consumption of \( H_2 \) can be calculated as follows:
\[ \text{Rate of consumption of } H_2 = \frac{3}{2} \times \text{Rate of production of } NH_3 = \frac{3}{2} \times 200 \, \text{mol/min} = 300 \, \text{mol/min} \]

5. **Convert Moles of \( H_2 \) Back to Mass**:
The molar mass of \( H_2 \) (hydrogen gas) is approximately \( 2 \, \text{g/mol} \) or \( 0.002 \, \text{kg/mol} \).
\[ \text{Rate of consumption of } H_2 = 300 \, \text{mol/min} \times 0.002 \, \text{kg/mol} = 0.6 \, \text{kg/min} \]

### Final Answer:
The rate of consumption of \( H_2 \) is \( 0.6 \, \text{kg/min} \).