For reaction `2NO(g)hArr N_(2)(g)+O_(2)(g)` degree of dissociation of `N_(2) and O_(2)` is x. Initially 1 mole of NO is taken then number of moles of `O_(2)` at equilibrium is

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \]

### Step 2: Define the initial conditions
Initially, we have 1 mole of NO. Therefore, the initial moles of each species are:
- NO: 1 mole
- N2: 0 moles
- O2: 0 moles

### Step 3: Define the degree of dissociation
Let the degree of dissociation of NO be \( x \). This means that \( x \) moles of NO will dissociate.

### Step 4: Determine the change in moles at equilibrium
From the stoichiometry of the reaction:
- For every 2 moles of NO that dissociate, 1 mole of N2 and 1 mole of O2 are produced.

Thus, if \( x \) moles of NO dissociate:
- Moles of NO remaining = \( 1 - x \)
- Moles of N2 produced = \( \frac{x}{2} \)
- Moles of O2 produced = \( \frac{x}{2} \)

### Step 5: Write the equilibrium expression for O2
At equilibrium, the number of moles of O2 will be:
\[ \text{Moles of O2} = \frac{x}{2} \]

### Final Answer
Therefore, the number of moles of O2 at equilibrium is:
\[ \frac{x}{2} \]

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