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 need to analyze the reaction and the degree of dissociation of the reactants and products.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The balanced chemical equation is:
\[ 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \]

2. **Initial Moles:**
We start with 1 mole of NO. Therefore, the initial moles of each species are:
- NO: 1 mole
- N2: 0 moles
- O2: 0 moles

3. **Define Degree of Dissociation:**
Let the degree of dissociation of NO be \( x \). This means that \( x \) fraction of NO will dissociate.

4. **Calculate Moles at Equilibrium:**
When 1 mole of NO dissociates, the changes in moles can be calculated as follows:
- For every 2 moles of NO that dissociate, 1 mole of N2 and 1 mole of O2 are produced.
- Therefore, if \( x \) is the degree of dissociation, the moles at equilibrium will be:
- Moles of NO at equilibrium: \( 1 - x \)
- Moles of N2 at equilibrium: \( \frac{x}{2} \) (since 2 moles of NO produce 1 mole of N2)
- Moles of O2 at equilibrium: \( \frac{x}{2} \) (since 2 moles of NO produce 1 mole of O2)

5. **Find Moles of O2 at Equilibrium:**
From the above calculations, the number of moles of O2 at equilibrium is:
\[ \text{Moles of } O_2 = \frac{x}{2} \]

### Final Answer:
The number of moles of O2 at equilibrium is \( \frac{x}{2} \).

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