When 1 mole of `N_2` and 1 mole of `H_2` is enclosed in 3L vessel and the reaction is allowed to attain equilibrium , it is found that at equilibrium there is 'x' mole of `H_2` . The number of moles of `NH_3` formed would be

Options

A. `(2x)/3`
B. `(2(1+x))/3`
C. `(2(1-x))/3`
D. `((1-x))/2`

Answer

To solve the problem, we need to analyze the given reaction and the information provided about the initial and equilibrium conditions.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction between nitrogen and hydrogen to form ammonia is given by:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

2. **Set Up Initial Conditions**:
Initially, we have:
- Moles of \(N_2 = 1\)
- Moles of \(H_2 = 1\)
- Moles of \(NH_3 = 0\)

3. **Define Changes at Equilibrium**:
Let \(y\) be the number of moles of \(N_2\) that react at equilibrium. According to the stoichiometry of the reaction:
- For every 1 mole of \(N_2\) that reacts, 3 moles of \(H_2\) react and 2 moles of \(NH_3\) are formed.
- Therefore, at equilibrium:
- Moles of \(N_2 = 1 - y\)
- Moles of \(H_2 = 1 - 3y\)
- Moles of \(NH_3 = 2y\)

4. **Use the Given Information**:
We know that at equilibrium there are \(x\) moles of \(H_2\):
\[ H_2 \text{ at equilibrium} = 1 - 3y = x \]
Rearranging this gives:
\[ 3y = 1 - x \implies y = \frac{1 - x}{3} \]

5. **Calculate Moles of \(NH_3\) Formed**:
The number of moles of \(NH_3\) formed at equilibrium is given by:
\[ NH_3 \text{ at equilibrium} = 2y = 2 \left(\frac{1 - x}{3}\right) = \frac{2(1 - x)}{3} \]

### Final Answer:
The number of moles of \(NH_3\) formed at equilibrium is:
\[ \frac{2(1 - x)}{3} \]