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 determine the rate of change of the concentration of hydrogen gas (H₂) based on the given rate of change of ammonia (NH₃) concentration. We will use the stoichiometry of the reaction to relate the rates.

### Step-by-Step Solution:

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

2. **Understand the Rate Expression:**
The rate of the reaction can be expressed in terms of the change in concentrations of the reactants and products. The rate of formation of NH₃ is given by:
\[ \text{Rate} = \frac{\Delta [NH_3]}{\Delta t} = \frac{1}{2} \left(-\frac{\Delta [N_2]}{\Delta t}\right) = \frac{1}{3} \left(-\frac{\Delta [H_2]}{\Delta t}\right) \]
Here, the stoichiometric coefficients (2 for NH₃, 3 for H₂) are used to relate the rates.

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

4. **Relate the Rate of H₂ to NH₃:**
Using the stoichiometry of the reaction, we can relate the rate of change of H₂ to NH₃:
\[ -\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} \cdot \frac{\Delta [NH_3]}{\Delta t} \]

5. **Substitute the Given Rate:**
Now, substitute the given value of \(\frac{\Delta [NH_3]}{\Delta t}\):
\[ -\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} \cdot (4 \times 10^{-4}) \, \text{mol L}^{-1} \text{s}^{-1} \]

6. **Calculate the Rate of Change of H₂:**
\[ -\frac{\Delta [H_2]}{\Delta t} = \frac{3 \times 4}{2} \times 10^{-4} = 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

7. **Final Answer:**
Therefore, the value of \(-\frac{\Delta [H_2]}{\Delta t}\) is:
\[ 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]