For the reaction N_(2)(g)+3H_(2)(g) rarr 2NH_(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 same rate of converison of hydrogen under the same condition is................kg h^(-1).

Explanation

To solve the problem, we need to determine the rate of conversion of hydrogen (H₂) based on the given rate of formation of ammonia (NH₃) for the reaction:

\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]

### Step-by-Step Solution:

1. **Identify the Stoichiometry of the Reaction:**
The balanced chemical equation shows that:
- 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
- This gives us the stoichiometric ratios:
- For every 2 moles of NH₃ produced, 3 moles of H₂ are consumed.

2. **Write the Rate Expressions:**
The rate of the reaction can be expressed in terms of the change in concentration (or mass) of the reactants and products:
\[ -\frac{1}{3} \frac{d[\text{H}_2]}{dt} = \frac{1}{2} \frac{d[\text{NH}_3]}{dt} \]
Here, the negative sign indicates the consumption of H₂, while the positive sign indicates the formation of NH₃.

3. **Convert the Rate of Formation of NH₃ to Moles:**
We are given the rate of formation of NH₃ as 0.001 kg/h. To convert this to moles, we use the molar mass of NH₃:
- Molar mass of NH₃ = 14 (N) + 3 × 1 (H) = 17 g/mol = 0.017 kg/mol.
\[ \text{Rate of formation of NH}_3 = \frac{0.001 \text{ kg/h}}{0.017 \text{ kg/mol}} \approx 0.05882 \text{ mol/h} \]

4. **Calculate the Rate of Consumption of H₂:**
Using the stoichiometric ratio from the balanced equation:
\[ -\frac{1}{3} \frac{d[\text{H}_2]}{dt} = \frac{1}{2} \frac{d[\text{NH}_3]}{dt} \]
Rearranging gives:
\[ \frac{d[\text{H}_2]}{dt} = -\frac{3}{2} \frac{d[\text{NH}_3]}{dt} \]
Substituting the rate of formation of NH₃:
\[ \frac{d[\text{H}_2]}{dt} = -\frac{3}{2} \times 0.05882 \text{ mol/h} \approx -0.08823 \text{ mol/h} \]

5. **Convert the Rate of Consumption of H₂ to kg/h:**
Now, we convert the rate of consumption of H₂ back to kg/h using its molar mass:
- Molar mass of H₂ = 2 g/mol = 0.002 kg/mol.
\[ \text{Rate of consumption of H}_2 = 0.08823 \text{ mol/h} \times 0.002 \text{ kg/mol} \approx 0.00017646 \text{ kg/h} \]

6. **Final Answer:**
The same rate of conversion of hydrogen under the same conditions is approximately:
\[ \text{Rate of conversion of H}_2 \approx 0.00017646 \text{ kg/h} \]

### Summary:
The rate of conversion of hydrogen (H₂) is approximately 0.00017646 kg/h.