Consider the following reaction `N_(2(g))+3H_(2(g))to2NH_(3(g))`. The rate of the reaction in terms in terms of `N_(2)` at `T(k)` is `-(d[N_(2)])/(dt)=0.02 ` mole `"lit"^(-1) "sec"^(-1)`. What is the value of `-(d[H_(2)])/(dt)` (in mole `"lit"^(-1)"sec"^(-1)`) at the same temperature?

Options

A. 0.02
B. 50
C. 0.06
D. 0.04

Answer

To solve the problem, we need to analyze the given reaction and the rates of disappearance of the reactants involved.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The balanced chemical reaction is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

2. **Identify the Rate of Reaction:**
The rate of the reaction in terms of \(N_2\) is given as:
\[ -\frac{d[N_2]}{dt} = 0.02 \, \text{mole} \, \text{lit}^{-1} \, \text{sec}^{-1} \]

3. **Relate the Rates Using Stoichiometry:**
From the balanced equation, we see that:
- For every 1 mole of \(N_2\) that reacts, 3 moles of \(H_2\) react.
- Therefore, the relationship between the rates of disappearance of \(N_2\) and \(H_2\) can be expressed as:
\[ -\frac{d[N_2]}{dt} = \frac{1}{3} \left(-\frac{d[H_2]}{dt}\right) \]

4. **Substituting the Known Rate:**
We can substitute the known rate of \(N_2\) into the equation:
\[ 0.02 = \frac{1}{3} \left(-\frac{d[H_2]}{dt}\right) \]

5. **Solving for the Rate of Disappearance of \(H_2\):**
Rearranging the equation gives:
\[ -\frac{d[H_2]}{dt} = 0.02 \times 3 \]
\[ -\frac{d[H_2]}{dt} = 0.06 \, \text{mole} \, \text{lit}^{-1} \, \text{sec}^{-1} \]

### Final Answer:
Thus, the value of \(-\frac{d[H_2]}{dt}\) at the same temperature is:
\[ 0.06 \, \text{mole} \, \text{lit}^{-1} \, \text{sec}^{-1} \]