For the reaction `N_(2)+3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(dt).=4xx10^(-4)` mol `L^(-1)s^(-1)`, the value of `(-d[H_(2)])/(dt)` would be

Options

A. `1 xx 10^(-4)` mol `L^(-1)s^(-1)`
B. `3xx10^(-4) mol L^(-1) s^(-1)`
C. `4 xx 10^(-4) mol L^(-1) s^(-1)`
D. `6xx10^(-4) mol L^(-1) s^(-1)`

Explanation

To solve the problem, we need to relate the rates of change of the concentrations of the reactants and products in the given reaction:

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

We are given that the rate of formation of ammonia \((NH_3)\) is:

\[ \frac{d[NH_3]}{dt} = 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

We need to find the rate of decrease of hydrogen \((H_2)\), which is represented as:

\[ -\frac{d[H_2]}{dt} \]

### Step 1: Write the rate expressions based on stoichiometry

From the balanced chemical equation, we can express the rates of change of the concentrations as follows:

\[ \frac{1}{2} \frac{d[NH_3]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} \]

### Step 2: Substitute the known value

We know that:

\[ \frac{d[NH_3]}{dt} = 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Substituting this into the rate expression gives:

\[ \frac{1}{2} (4 \times 10^{-4}) = -\frac{1}{3} \frac{d[H_2]}{dt} \]

### Step 3: Calculate the left-hand side

Calculating the left-hand side:

\[ \frac{1}{2} (4 \times 10^{-4}) = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 4: Rearranging to find \(-\frac{d[H_2]}{dt}\)

Now we can rearrange the equation to find \(-\frac{d[H_2]}{dt}\):

\[ 2 \times 10^{-4} = -\frac{1}{3} \frac{d[H_2]}{dt} \]

Multiplying both sides by -3 gives:

\[ -3 \times 2 \times 10^{-4} = \frac{d[H_2]}{dt} \]

\[ \frac{d[H_2]}{dt} = -6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Final answer

Thus, the value of \(-\frac{d[H_2]}{dt}\) is:

\[ -\frac{d[H_2]}{dt} = 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Summary

The final answer is:

\[ \boxed{6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}} \]