For the reaction : `C "(Graphite)"+CO_(2)(g)hArr 2CO(g)`. At equilibrium the mole precent of CO is equal to `50%` in gas phase. What is the degree of dissociation, x. (Assume equal number of moles of C(graphite) and `CO_(2)` at the start)

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `(1)/(5)`

Answer

To solve the problem, we need to determine the degree of dissociation (x) for the reaction:

\[ C (graphite) + CO_2 (g) \rightleftharpoons 2 CO (g) \]

Given that at equilibrium, the mole percent of CO is 50%, we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( CO_2 \) and no \( CO \) initially. Since \( C \) is a solid, we do not consider its moles in the equilibrium expression.

- Initial moles of \( CO_2 = 1 \)
- Initial moles of \( CO = 0 \)

### Step 2: Define Change in Moles
Let \( x \) be the degree of dissociation of \( CO_2 \). This means:

- Moles of \( CO_2 \) dissociated = \( x \)
- Moles of \( CO \) produced = \( 2x \) (because of the stoichiometry of the reaction)

### Step 3: Calculate Equilibrium Moles
At equilibrium, the moles of each species will be:

- Moles of \( CO_2 \) at equilibrium = \( 1 - x \)
- Moles of \( CO \) at equilibrium = \( 2x \)

### Step 4: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:

\[ \text{Total moles} = (1 - x) + (2x) = 1 + x \]

### Step 5: Calculate Mole Fraction of CO
The mole fraction of \( CO \) is given by:

\[ \text{Mole fraction of } CO = \frac{\text{Moles of } CO}{\text{Total moles}} = \frac{2x}{1 + x} \]

### Step 6: Set Up the Equation
We know from the problem that the mole fraction of \( CO \) is 0.5 (or 50%). Therefore, we can set up the equation:

\[ \frac{2x}{1 + x} = 0.5 \]

### Step 7: Solve for x
To eliminate the fraction, we can cross-multiply:

\[ 2x = 0.5(1 + x) \]

Expanding the right side:

\[ 2x = 0.5 + 0.5x \]

Rearranging gives:

\[ 2x - 0.5x = 0.5 \]

\[ 1.5x = 0.5 \]

Now, solving for \( x \):

\[ x = \frac{0.5}{1.5} = \frac{1}{3} \]

### Conclusion
The degree of dissociation \( x \) is \( \frac{1}{3} \).

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