For the reaction `N_2(g) + 3H_2(g) to 2NH_3(g) " if " (Delta[NH_3])/(Deltat) = 4xx 10^(-4) mol.l^(-1) s^(-1)`, the value of `(-Delta[H_2])/(Deltat)` would be

Options

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

Answer

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

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

We are provided with the rate of formation of ammonia (\(NH_3\)):

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

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

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

### Step 1: Write the Rate Expressions
From the stoichiometry of the reaction, we can express the rates of change of concentrations as follows:

\[ \text{Rate} = -\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NH_3]}{\Delta t} \]

### Step 2: Substitute the Known Value
We know the rate of formation of \(NH_3\):

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

Substituting this value into the rate expression for \(H_2\):

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

### Step 3: Calculate the Rate of Disappearance of \(H_2\)
Now we can calculate the right-hand side:

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

Now, we can relate this to the rate of disappearance of \(H_2\):

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

### Step 4: Solve for \(-\frac{\Delta[H_2]}{\Delta t}\)
To find \(-\frac{\Delta[H_2]}{\Delta t}\), we can rearrange the equation:

\[ \frac{\Delta[H_2]}{\Delta t} = -3 \times (2 \times 10^{-4}) \]

Calculating this gives:

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

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

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

### Final Answer
The value of \(-\frac{\Delta[H_2]}{\Delta t}\) is:

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