For the reaction `N_(2) + 3H_(2) to 2NH_(3)` if
`(Delta[NH_(3)])/(Deltat) = 2 xx 10^(-4) mol L^(-1)s^(-1)`, the value of `(-Delta[H_(2)])/(Deltat)` would be

Options

A. `1 xx 10^(-3) mol L^(-1)s^(-1)`
B. `3 xx 10^(-4)molL^(-1)s^(-1)`
C. `4 xx 10^(-4) mol L^(-1)s^(-1)`
D. `6 xx 10^(-4)molL^(-1)s^(-1)`

Explanation

To solve the problem, we need to relate the rate of formation of ammonia (NH₃) to the rate of consumption of hydrogen (H₂) in the given reaction:

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

Given that:

\[ \frac{\Delta[NH_3]}{\Delta t} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

We need to find the value of:

\[ -\frac{\Delta[H_2]}{\Delta t} \]

### Step-by-Step Solution:

1. **Identify the Stoichiometry of the Reaction:**
From the balanced equation, we see that:
- 3 moles of H₂ are consumed for every 2 moles of NH₃ produced.

2. **Set Up the Rate Expressions:**
The rate of the reaction can be expressed in terms of the change in concentration of each species:
\[ \text{Rate} = -\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NH_3]}{\Delta t} \]

3. **Relate the Rates:**
From the stoichiometry, we can set up the following relationship:
\[ -\frac{\Delta[H_2]}{\Delta t} = \frac{3}{2} \frac{\Delta[NH_3]}{\Delta t} \]

4. **Substitute the Given Value:**
Now substitute the given rate of formation of NH₃ into the equation:
\[ -\frac{\Delta[H_2]}{\Delta t} = \frac{3}{2} \left(2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}\right) \]

5. **Calculate the Rate of Consumption of H₂:**
\[ -\frac{\Delta[H_2]}{\Delta t} = \frac{3 \times 2 \times 10^{-4}}{2} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
\[ -\frac{\Delta[H_2]}{\Delta t} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]