For the reaction : `N_(2) +3H_(2) rarr 2NH_(3)," If "(Delta [NH_(3)])/(Delta t)=2xx10^(-4)" mol L"^(-1) s^(-1)`, the value of `-(Delta [H_(2)])/(Delta t)` would be :

Options

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

Explanation

To solve the problem, we need to analyze the given chemical reaction and the relationship between the rates of change of the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Understand the Rate of Reaction**: The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For the reaction:
\[ -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t} \]
Here, the coefficients from the balanced equation are used to relate the rates of change of the concentrations.

3. **Given Information**: We are given that:
\[ \frac{\Delta [NH_3]}{\Delta t} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Substituting the Known Value**: Substitute the value of \(\frac{\Delta [NH_3]}{\Delta t}\) into the rate equation:
\[ -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} (2 \times 10^{-4}) \]

5. **Calculate the Right Side**: Calculate the right side:
\[ \frac{1}{2} (2 \times 10^{-4}) = 1 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

6. **Rearranging the Equation**: Now, rearranging the equation to solve for \(-\frac{\Delta [H_2]}{\Delta t}\):
\[ -\frac{\Delta [H_2]}{\Delta t} = 3 \times (1 \times 10^{-4}) = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

7. **Final Answer**: Therefore, the value of \(-\frac{\Delta [H_2]}{\Delta t}\) is:
\[ 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]