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 step by step, we will analyze the given reaction and the conditions provided:

**Given Reaction:**
\[ C \, (Graphite) + CO_2(g) \rightleftharpoons 2CO(g) \]

**Step 1: Initial Setup**
Assume we start with 1 mole of \( C \) (graphite) and 1 mole of \( CO_2 \). At the beginning, the number of moles of each species is:
- Moles of \( C \) = 1
- Moles of \( CO_2 \) = 1
- Moles of \( CO \) = 0

**Step 2: Degree of Dissociation**
Let \( x \) be the degree of dissociation of \( CO_2 \). Thus, at equilibrium:
- Moles of \( CO_2 \) remaining = \( 1 - x \)
- Moles of \( CO \) formed = \( 2x \) (since 1 mole of \( CO_2 \) produces 2 moles of \( CO \))

**Step 3: Total Moles at Equilibrium**
At equilibrium, the total number of moles is:
\[ \text{Total moles} = (1 - x) + 2x = 1 + x \]

**Step 4: Mole Percent of \( CO \)**
The mole percent of \( CO \) at equilibrium is given as 50%. The mole fraction of \( CO \) is:
\[ \text{Mole fraction of } CO = \frac{\text{Moles of } CO}{\text{Total moles}} = \frac{2x}{1 + x} \]

Setting this equal to 50% (or 0.5):
\[ \frac{2x}{1 + x} = 0.5 \]

**Step 5: Solve for \( x \)**
Cross-multiplying gives:
\[ 2x = 0.5(1 + x) \]
\[ 2x = 0.5 + 0.5x \]
Rearranging the equation:
\[ 2x - 0.5x = 0.5 \]
\[ 1.5x = 0.5 \]
\[ x = \frac{0.5}{1.5} = \frac{1}{3} \]

**Step 6: Conclusion**
The degree of dissociation \( x \) is \( \frac{1}{3} \).

Thus, the final answer is:
\[ \text{Degree of dissociation, } x = \frac{1}{3} \]

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