The rate of disappearance of ammonia is `3.4` gm/litre sec. when it dissociates to form nitrogen and hydrogen. The rate of appearance of nitrogen will be :

Options

A. `3.4` gm/litre sec
B. `1.7` gm/litre sec
C. `0.1` gm/litre sec
D. `2.8` gm/litre sec

Answer

To solve the problem, we need to understand the stoichiometry of the reaction involved in the dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂). The balanced chemical equation for the dissociation of ammonia is:

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

### Step 1: Identify the stoichiometric coefficients
From the balanced equation, we can see the following stoichiometric relationships:
- 2 moles of NH₃ produce 1 mole of N₂.
- Therefore, the rate of disappearance of NH₃ is twice the rate of appearance of N₂.

### Step 2: Relate the rates using stoichiometry
Let the rate of disappearance of NH₃ be denoted as \( R_{\text{NH}_3} \) and the rate of appearance of N₂ as \( R_{\text{N}_2} \). According to the stoichiometry of the reaction, we can express this relationship as:

\[ R_{\text{NH}_3} = 2 \times R_{\text{N}_2} \]

### Step 3: Substitute the known values
We know from the problem that the rate of disappearance of ammonia (NH₃) is given as:

\[ R_{\text{NH}_3} = 3.4 \, \text{gm/litre sec} \]

### Step 4: Calculate the rate of appearance of nitrogen
Now, we can rearrange the equation to find \( R_{\text{N}_2} \):

\[ R_{\text{N}_2} = \frac{R_{\text{NH}_3}}{2} \]

Substituting the known value:

\[ R_{\text{N}_2} = \frac{3.4 \, \text{gm/litre sec}}{2} = 1.7 \, \text{gm/litre sec} \]

### Final Answer
The rate of appearance of nitrogen (N₂) is:

\[ \boxed{1.7 \, \text{gm/litre sec}} \]

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