In a catalytic conversion of `N_(2)` to `NH_(3)` by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is `40 xx 10^(-3) mol L^(-1) s^(-1)`. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol `L^(-1) s^(-1)`)

Options

A. `60 xx 10^(-3)`
B. `20xx 10^(-3)`
C. 1200
D. `10.3 xx 10^(-3) `

Answer

To solve the problem, we need to analyze the reaction and relate the rates of change of the reactants and products based on their stoichiometric coefficients.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction for the Haber process is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Identify the Rate of Reaction**:
The rate of the reaction is given in terms of the change in concentration of ammonia (NH₃):
\[ \text{Rate} = \frac{d[NH_3]}{dt} = 40 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

3. **Relate the Rates of Reactants and Products**:
According to the stoichiometry of the reaction, we can express the rates of disappearance of the reactants and the appearance of the products as follows:
\[ -\frac{d[N_2]}{dt} = \frac{1}{1} \cdot \text{Rate} = \frac{d[NH_3]}{dt} = \frac{2}{3} \cdot \left(-\frac{d[H_2]}{dt}\right) \]

4. **Express the Rate of Hydrogen**:
We need to find the rate of disappearance of hydrogen (H₂) in terms of the rate of formation of ammonia (NH₃). From the stoichiometric coefficients, we have:
\[ -\frac{d[H_2]}{dt} = \frac{3}{2} \cdot \frac{d[NH_3]}{dt} \]

5. **Substitute the Known Rate**:
Now, substituting the given rate of ammonia into the equation:
\[ -\frac{d[H_2]}{dt} = \frac{3}{2} \cdot 40 \times 10^{-3} \]

6. **Calculate the Rate of Hydrogen**:
Performing the multiplication:
\[ -\frac{d[H_2]}{dt} = 60 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

7. **Final Answer**:
Thus, the rate of the reaction expressed in terms of hydrogen is:
\[ \frac{d[H_2]}{dt} = -60 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Summary:
The rate of the reaction in terms of hydrogen is \( 60 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

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