For the reaction, `N_(2) + 3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(d t) = 2 xx 10^(-4) "mol L"^(-1) s^(-1)`, the value of `(-d[H_(2)])/(d t)` would be:

Options

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

Explanation

To solve the problem, we need to relate the rate of formation of ammonia (NH₃) to the rate of disappearance of hydrogen (H₂) using the stoichiometry of the reaction:

**Given Reaction:**
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

**Step 1: Write the rate expressions based on stoichiometry.**
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products:
- For nitrogen (N₂):
\[ -\frac{d[N_2]}{dt} \]
- For hydrogen (H₂):
\[ -\frac{1}{3} \frac{d[H_2]}{dt} \]
- For ammonia (NH₃):
\[ +\frac{1}{2} \frac{d[NH_3]}{dt} \]

From the stoichiometry of the balanced equation, we can relate these rates as follows:
\[ -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

**Step 2: Substitute the known rate of formation of NH₃.**
We are given:
\[ \frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

**Step 3: Calculate the rate of disappearance of H₂.**
Using the relationship:
\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

We can rearrange this to find \(-\frac{d[H_2]}{dt}\):
\[ \frac{d[H_2]}{dt} = -3 \times \frac{1}{2} \frac{d[NH_3]}{dt} \]

Substituting the value of \(\frac{d[NH_3]}{dt}\):
\[ -\frac{d[H_2]}{dt} = 3 \times \frac{1}{2} \times (2 \times 10^{-4}) \]

**Step 4: Simplify the expression.**
Calculating the right-hand side:
\[ -\frac{d[H_2]}{dt} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Thus, the value of \(-\frac{d[H_2]}{dt}\) is:
\[ \boxed{3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}} \]

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