`K_(p)=0.04 atm` at `899 K` for the equilibrium shown below. What is the equilibrium concentration of `C_(2)H_(6)` when it is placed in a flask at `4.0 atm` pressure and allowed to come to equilibrium?
`C_(2)H_(6)(g) hArr C_(2)H_(4)(g)+H_(2)(g)`

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial conditions
Initially, we have:
- The pressure of \( C_2H_6 \) is 4.0 atm.
- The pressures of \( C_2H_4 \) and \( H_2 \) are both 0 atm.

### Step 3: Define the change in pressure
Let \( x \) be the amount of \( C_2H_6 \) that dissociates at equilibrium. Therefore, at equilibrium:
- The pressure of \( C_2H_6 \) will be \( 4.0 - x \) atm.
- The pressure of \( C_2H_4 \) will be \( x \) atm.
- The pressure of \( H_2 \) will also be \( x \) atm.

### Step 4: Write the expression for \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{P_{C_2H_4} \cdot P_{H_2}}{P_{C_2H_6}} \]
Substituting the pressures at equilibrium, we get:
\[ K_p = \frac{x \cdot x}{4.0 - x} = \frac{x^2}{4.0 - x} \]

### Step 5: Substitute the value of \( K_p \)
We know that \( K_p = 0.04 \) atm, so we can set up the equation:
\[ \frac{x^2}{4.0 - x} = 0.04 \]

### Step 6: Solve for \( x \)
Cross-multiplying gives us:
\[ x^2 = 0.04(4.0 - x) \]
Expanding this:
\[ x^2 = 0.16 - 0.04x \]
Rearranging gives us:
\[ x^2 + 0.04x - 0.16 = 0 \]

### Step 7: Use the quadratic formula
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 0.04, c = -0.16 \):
\[ x = \frac{-0.04 \pm \sqrt{(0.04)^2 - 4 \cdot 1 \cdot (-0.16)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (0.04)^2 + 0.64 = 0.0016 + 0.64 = 0.6416 \]
Now, substituting back:
\[ x = \frac{-0.04 \pm \sqrt{0.6416}}{2} \]
Calculating \( \sqrt{0.6416} \approx 0.801 \):
\[ x = \frac{-0.04 \pm 0.801}{2} \]

Calculating the two potential values for \( x \):
1. \( x = \frac{0.761}{2} \approx 0.3805 \)
2. \( x = \frac{-0.841}{2} \) (not valid since pressure cannot be negative)

Thus, we take \( x \approx 0.38 \) atm.

### Step 8: Calculate the equilibrium pressure of \( C_2H_6 \)
The equilibrium pressure of \( C_2H_6 \) is:
\[ P_{C_2H_6} = 4.0 - x = 4.0 - 0.38 = 3.62 \text{ atm} \]

### Final Answer
The equilibrium concentration of \( C_2H_6 \) is **3.62 atm**.

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