For the reaction `N_(2)(g) + 2 H_(2) (g) to 2 NH_(3) (g)` under certain conditions of temperature and partial pressure of the reactants , the rate of formation of `NH_(3)` is `0.001 kg h^(-1)` . The rate of conversion of `H_(2)` under the same conditions is

Options

A. `1.82 xx 10^(-4)`kg/hr
B. `0.0015` kg/hr
C. `1.52 xx 10^(4)` kg/hr
D. `1.82 xx 10^(-14)` kg/hr

Explanation

To solve the problem, we need to analyze the given reaction and the rates of formation and conversion of the reactants and products.

### Given Reaction:
\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \]

### Given Data:
- Rate of formation of \( NH_3 \) = \( 0.001 \, \text{kg/h} \)

### Step 1: Convert the rate of formation of \( NH_3 \) to moles.
To convert the mass of \( NH_3 \) to moles, we use the molar mass of ammonia (\( NH_3 \)):
- Molar mass of \( NH_3 \) = 14 (N) + 3 × 1 (H) = 17 g/mol = 0.017 kg/mol

Now, we can convert the rate of formation of \( NH_3 \) from kg/h to mol/h:
\[ \text{Rate of formation of } NH_3 = \frac{0.001 \, \text{kg/h}}{0.017 \, \text{kg/mol}} \approx 0.05882 \, \text{mol/h} \]

### Step 2: Relate the rate of formation of \( NH_3 \) to the rate of conversion of \( H_2 \).
From the stoichiometry of the reaction:
- 2 moles of \( NH_3 \) are produced from 3 moles of \( H_2 \).

Using the stoichiometric coefficients, we can relate the rate of disappearance of \( H_2 \) to the rate of formation of \( NH_3 \):
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

### Step 3: Solve for the rate of conversion of \( H_2 \).
Rearranging the equation gives:
\[ \frac{d[H_2]}{dt} = -\frac{3}{2} \frac{d[NH_3]}{dt} \]

Substituting the rate of formation of \( NH_3 \):
\[ \frac{d[H_2]}{dt} = -\frac{3}{2} \times 0.05882 \, \text{mol/h} \]
\[ \frac{d[H_2]}{dt} = -0.08823 \, \text{mol/h} \]

### Step 4: Convert the rate of conversion of \( H_2 \) back to kg/h.
Using the molar mass of \( H_2 \) (approximately 2 g/mol or 0.002 kg/mol):
\[ \text{Rate of conversion of } H_2 = -0.08823 \, \text{mol/h} \times 0.002 \, \text{kg/mol} = -0.00017646 \, \text{kg/h} \]

### Final Answer:
The rate of conversion of \( H_2 \) is approximately:
\[ \text{Rate of conversion of } H_2 \approx 0.000176 \, \text{kg/h} \]