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 step by step, we will analyze the reaction and the equilibrium conditions provided in the question.

### Step 1: Write the balanced chemical equation.
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Set up the initial moles of reactants.
Initially, we have:
- 1 mole of \( N_2 \)
- 1 mole of \( H_2 \)

### Step 3: Define the change in moles at equilibrium.
Let \( y \) be the change in moles of \( N_2 \) that reacts. According to the stoichiometry of the reaction:
- \( N_2 \) decreases by \( y \)
- \( H_2 \) decreases by \( 3y \)
- \( NH_3 \) increases by \( 2y \)

### Step 4: Write the equilibrium moles.
At equilibrium, the moles of each substance will be:
- Moles of \( N_2 \) at equilibrium: \( 1 - y \)
- Moles of \( H_2 \) at equilibrium: \( 1 - 3y \)
- Moles of \( NH_3 \) at equilibrium: \( 2y \)

### Step 5: Use the information given in the problem.
According to the problem, at equilibrium, there are \( x \) moles of \( H_2 \):
\[ 1 - 3y = x \]

### Step 6: Solve for \( y \).
Rearranging the equation:
\[ 3y = 1 - x \]
\[ y = \frac{1 - x}{3} \]

### Step 7: Find the number of moles of \( NH_3 \).
The number of moles of \( NH_3 \) formed at equilibrium is given by:
\[ 2y = 2 \left( \frac{1 - x}{3} \right) = \frac{2(1 - x)}{3} \]

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