For the reaction:
`N_(2)+3H_(2)hArr2NH_(3)`
If the initial concentration of `N_(2)` and `H_(2)` are a and b moles per litre
and x moles per litre of ammonia is formed at equilibrium,
what is the equilibrium concentration of hydrogen?

Options

A. `(b-3x)`
B. `(3b-3x)`
C. `(b-(3x)/(2))`
D. `(b-(2)/(3)x)`

Explanation

To solve the problem, we need to determine the equilibrium concentration of hydrogen gas (H₂) for the given reaction:

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step-by-Step Solution:

1. **Identify Initial Concentrations**:
- Let the initial concentration of \( N_2 \) be \( a \) moles per litre.
- Let the initial concentration of \( H_2 \) be \( b \) moles per litre.

2. **Define Change in Concentrations**:
- Let \( x \) be the concentration of ammonia (\( NH_3 \)) formed at equilibrium.
- According to the stoichiometry of the reaction, for every 2 moles of \( NH_3 \) produced, 1 mole of \( N_2 \) and 3 moles of \( H_2 \) are consumed.

3. **Express Changes in Concentrations**:
- The change in concentration of \( N_2 \) will be \( -\frac{x}{2} \) (since 2 moles of \( NH_3 \) are produced from 1 mole of \( N_2 \)).
- The change in concentration of \( H_2 \) will be \( -\frac{3x}{2} \) (since 2 moles of \( NH_3 \) are produced from 3 moles of \( H_2 \)).

4. **Write Equilibrium Concentrations**:
- At equilibrium, the concentration of \( N_2 \) will be:
\[ [N_2] = a - \frac{x}{2} \]
- The concentration of \( H_2 \) will be:
\[ [H_2] = b - \frac{3x}{2} \]
- The concentration of \( NH_3 \) will be:
\[ [NH_3] = x \]

5. **Find Equilibrium Concentration of Hydrogen**:
- From the expression for \( [H_2] \):
\[ [H_2] = b - \frac{3x}{2} \]

### Final Answer:
The equilibrium concentration of hydrogen (\( H_2 \)) is:
\[ [H_2] = b - \frac{3x}{2} \]