In a catalyst experiment involving the Haber process `N_(2) + 3H_(2) rarr 2NH_(3)`, the rate of reaction was measured as
Rate `= (Delta[NH_(3)])/(Delta t) = 2.0 xx 10^(-4) mol L^(-1) s^(-1)`
What is the rate of reaction expressed in terms of (a) `N_(2)` (b) `H_(2)` ?

Options

A. `2.50 xx 10^(-4) mol L^(-1) s^(-1)`
B. `1.25 xx 10^(-4) mol L^(-1) s^(-1)`
C. `3.75 xx 10^(-4) mol L^(-1) s^(-1)`
D. `5.00 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

To solve the problem, we need to express the rate of reaction in terms of the reactants \(N_2\) and \(H_2\) based on the given rate of formation of \(NH_3\).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Identify the relationship between the rates of change of the reactants and products:**
According to the stoichiometry of the reaction:
- The rate of consumption of \(N_2\) is half the rate of formation of \(NH_3\).
- The rate of consumption of \(H_2\) is one and a half times the rate of formation of \(NH_3\).

Mathematically, this can be expressed as:
\[ -\frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]
\[ -\frac{d[H_2]}{dt} = \frac{1}{3} \frac{d[NH_3]}{dt} \]

3. **Given rate of formation of \(NH_3\):**
\[ \frac{d[NH_3]}{dt} = 2.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Calculate the rate of reaction in terms of \(N_2\):**
Using the relationship for \(N_2\):
\[ -\frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]
Substitute the given rate:
\[ -\frac{d[N_2]}{dt} = \frac{1}{2} \times (2.0 \times 10^{-4}) = 1.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Calculate the rate of reaction in terms of \(H_2\):**
Using the relationship for \(H_2\):
\[ -\frac{d[H_2]}{dt} = \frac{1}{3} \frac{d[NH_3]}{dt} \]
Substitute the given rate:
\[ -\frac{d[H_2]}{dt} = \frac{1}{3} \times (2.0 \times 10^{-4}) = \frac{2.0 \times 10^{-4}}{3} = \frac{2.0}{3} \times 10^{-4} \approx 6.67 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answers:
- (a) Rate of reaction in terms of \(N_2\): \(1.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}\)
- (b) Rate of reaction in terms of \(H_2\): \(6.67 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}\)