For the reaction , `N_2+3H_2rarr2NH_3` if `(d[NH_3])/(dt)=2xx10^(-4)"mol L"^(-1) s^(-1)` , the value of `(-d[H_2])/(dt)` would be

Options

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

Answer

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

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

**Given Rate of Formation of Ammonia:**
\[ \frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

**Step 1: Identify Stoichiometric Coefficients**
From the balanced equation, we see that:
- The stoichiometric coefficient of \( NH_3 \) is 2.
- The stoichiometric coefficient of \( H_2 \) is 3.

**Step 2: Relate the Rates Using Stoichiometry**
According to the stoichiometry of the reaction, the rates of formation and disappearance can be related as follows:
\[ \frac{1}{2} \frac{d[NH_3]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} \]

**Step 3: Substitute the Given Rate**
Substituting the given rate of formation of ammonia into the equation:
\[ \frac{1}{2} (2 \times 10^{-4}) = -\frac{1}{3} \frac{d[H_2]}{dt} \]

**Step 4: Simplify the Left Side**
Calculating the left side:
\[ \frac{1}{2} \times 2 \times 10^{-4} = 1 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

**Step 5: Rearranging the Equation**
Now we can rearrange the equation to solve for \(-\frac{d[H_2]}{dt}\):
\[ 1 \times 10^{-4} = -\frac{1}{3} \frac{d[H_2]}{dt} \]

**Step 6: Solve for \(-\frac{d[H_2]}{dt}\)**
Multiplying both sides by -3 to isolate \(-\frac{d[H_2]}{dt}\):
\[ -\frac{d[H_2]}{dt} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

**Final Answer:**
\[ -\frac{d[H_2]}{dt} = 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

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