For the reaction `C_(2)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g)`
`K_(p)` is `5xx10^(-2)` atm. Calculate the mole per cent of `C_(2)H_(6)(g)` at equilibruium if pure `C_(2)H_(6)` at `1` atm is passed over a suitable catalyt at `900 K` :

Options

A. `20`
B. `33.33`
C. `66.66`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ C_2H_6(g) \rightleftharpoons C_2H_4(g) + H_2(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Pressure of \( C_2H_6 = 1 \, \text{atm} \)
- Pressure of \( C_2H_4 = 0 \, \text{atm} \)
- Pressure of \( H_2 = 0 \, \text{atm} \)

### Step 3: Define changes at equilibrium
Let \( x \) be the change in pressure of \( C_2H_6 \) that reacts. At equilibrium, the pressures will be:
- Pressure of \( C_2H_6 = 1 - x \)
- Pressure of \( C_2H_4 = x \)
- Pressure of \( H_2 = x \)

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{C_2H_4} \cdot P_{H_2}}{P_{C_2H_6}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{x \cdot x}{1 - x} = \frac{x^2}{1 - x} \]

### Step 5: Substitute the given value of \( K_p \)
We know that \( K_p = 5 \times 10^{-2} \):
\[ 5 \times 10^{-2} = \frac{x^2}{1 - x} \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ x^2 = 5 \times 10^{-2} (1 - x) \]
\[ x^2 + 5 \times 10^{-2} x - 5 \times 10^{-2} = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 5 \times 10^{-2} \), and \( c = -5 \times 10^{-2} \).

Calculating the discriminant:
\[ b^2 - 4ac = (5 \times 10^{-2})^2 - 4(1)(-5 \times 10^{-2}) = 2.5 \times 10^{-3} + 0.2 = 0.2025 \]

Now, substituting into the quadratic formula:
\[ x = \frac{-5 \times 10^{-2} \pm \sqrt{0.2025}}{2} \]
Calculating \( \sqrt{0.2025} \approx 0.45 \):
\[ x = \frac{-0.05 \pm 0.45}{2} \]
Taking the positive root:
\[ x = \frac{0.4}{2} = 0.2 \, \text{atm} \]

### Step 8: Calculate the total pressure at equilibrium
Total pressure at equilibrium:
\[ P_{total} = P_{C_2H_6} + P_{C_2H_4} + P_{H_2} = (1 - x) + x + x = 1 + x = 1 + 0.2 = 1.2 \, \text{atm} \]

### Step 9: Calculate the mole fraction of \( C_2H_6 \)
The mole fraction of \( C_2H_6 \):
\[ \text{Mole fraction of } C_2H_6 = \frac{P_{C_2H_6}}{P_{total}} = \frac{1 - x}{1.2} = \frac{1 - 0.2}{1.2} = \frac{0.8}{1.2} = \frac{2}{3} \]

### Step 10: Convert mole fraction to mole percent
Mole percent of \( C_2H_6 \):
\[ \text{Mole percent} = \left( \frac{2}{3} \right) \times 100 \approx 66.67\% \]

### Final Answer
The mole percent of \( C_2H_6 \) at equilibrium is approximately **66.67%**.

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