For the reaction : `N_2+3H_2 to 2NH_3`. If the rate of disappearance of hydrogen is `1.8 times 10^3 mol//l -sec`. What is the rate of formation of ammonia?

Options

A. `1.8 times 10^3 mol//l-sec`
B. `1.2 times 10^3 mol//l-sec`
C. `2.7 times 10^3 mol//l-sec`
D. `0.9 times 10^3 mol//l-sec`

Answer

To solve the problem, we need to find the rate of formation of ammonia (NH₃) given the rate of disappearance of hydrogen (H₂). The balanced chemical equation for the reaction is:

\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step-by-Step Solution:

1. **Identify the Stoichiometric Ratios**:
From the balanced equation, we can see the stoichiometric coefficients:
- For H₂: 3
- For NH₃: 2

This means that for every 3 moles of H₂ that disappear, 2 moles of NH₃ are formed.

2. **Write the Rate Expressions**:
The rate of disappearance of H₂ and the rate of formation of NH₃ can be expressed as:
\[ -\frac{d[H_2]}{dt} = \frac{1}{3} \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

3. **Given Rate of Disappearance of H₂**:
We are given that the rate of disappearance of H₂ is:
\[ -\frac{d[H_2]}{dt} = 1.8 \times 10^3 \, \text{mol/L/s} \]

4. **Relate the Rate of Formation of NH₃ to H₂**:
From the rate expressions, we can relate the rate of formation of NH₃ to the rate of disappearance of H₂:
\[ \frac{d[NH_3]}{dt} = \frac{2}{3} \left(-\frac{d[H_2]}{dt}\right) \]

5. **Substitute the Given Value**:
Now, substituting the value of \(-\frac{d[H_2]}{dt}\):
\[ \frac{d[NH_3]}{dt} = \frac{2}{3} \times 1.8 \times 10^3 \]

6. **Calculate the Rate of Formation of NH₃**:
\[ \frac{d[NH_3]}{dt} = \frac{2 \times 1.8 \times 10^3}{3} = \frac{3.6 \times 10^3}{3} = 1.2 \times 10^3 \, \text{mol/L/s} \]

### Final Answer:
The rate of formation of ammonia (NH₃) is:
\[ 1.2 \times 10^3 \, \text{mol/L/s} \]