`NH_(3)` gas can be prepared by Haber’s process as, `N_(2)(g) + 3H_(2) (g) rarr 2NH_(3)(g)`. At a particular moment concentration of all the species is 2 moles, calculate the concentration of `N_(2)` and `H_(2)` taken initially

Answer

To solve the problem, we need to determine the initial concentrations of nitrogen (N₂) and hydrogen (H₂) in the reaction given by the Haber process:

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step-by-Step Solution:

1. **Define Initial Concentrations**:
Let the initial concentration of nitrogen (N₂) be \( A \) moles and the initial concentration of hydrogen (H₂) be \( B \) moles. Initially, the concentration of ammonia (NH₃) is 0 moles since the reaction has not started.

2. **Set Up the Change in Concentration**:
As the reaction proceeds, let \( x \) be the amount of N₂ that reacts. According to the stoichiometry of the reaction:
- The concentration of N₂ after time \( t \) will be \( A - x \).
- The concentration of H₂ will decrease by \( 3x \), so it will be \( B - 3x \).
- The concentration of NH₃ produced will be \( 2x \).

3. **Equate Concentrations at a Particular Moment**:
We are given that at a particular moment, the concentration of all species (N₂, H₂, and NH₃) is 2 moles. Therefore, we can set up the following equations:
- \( A - x = 2 \) (for N₂)
- \( B - 3x = 2 \) (for H₂)
- \( 2x = 2 \) (for NH₃)

4. **Solve for \( x \)**:
From the equation \( 2x = 2 \):
\[ x = 1 \]

5. **Substitute \( x \) Back into the Equations**:
Now substitute \( x = 1 \) into the equations for \( A \) and \( B \):
- For N₂:
\[ A - 1 = 2 \implies A = 2 + 1 = 3 \]
- For H₂:
\[ B - 3(1) = 2 \implies B - 3 = 2 \implies B = 2 + 3 = 5 \]

6. **Final Answer**:
The initial concentrations are:
- Concentration of N₂ (A) = 3 moles
- Concentration of H₂ (B) = 5 moles

### Summary:
- Initial concentration of N₂ = 3 moles
- Initial concentration of H₂ = 5 moles