For the equilibrium `CH_3CH_2CH_2CH_(3(g))hArr CH_3-underset("iso-butane")underset(CH_3)underset(|)(CH)-CH_(3(g))`
If the value of `K_c` is 3.0 the percentage by mass of iso-butane in the equilibrium mixture would be

Options

A. 0.75
B. 0.9
C. 0.3
D. 0.6

Answer

To solve the problem, we need to find the percentage by mass of iso-butane in the equilibrium mixture given the equilibrium constant \( K_c = 3.0 \) for the reaction:

\[ CH_3CH_2CH_2CH_3 (g) \rightleftharpoons CH_3CH(CH_3)CH_3 (g) \]

### Step-by-Step Solution:

1. **Define Initial Concentrations:**
- Let the initial concentration of n-butane (\( CH_3CH_2CH_2CH_3 \)) be 1 (in arbitrary units).
- The initial concentration of iso-butane (\( CH_3CH(CH_3)CH_3 \)) is 0.

2. **Set Up the Change in Concentration:**
- Let \( \alpha \) be the amount of n-butane that converts to iso-butane at equilibrium.
- Therefore, at equilibrium:
- Concentration of n-butane = \( 1 - \alpha \)
- Concentration of iso-butane = \( \alpha \)

3. **Write the Expression for \( K_c \):**
- The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{iso-butane}]}{[\text{n-butane}]} = \frac{\alpha}{1 - \alpha} \]
- We know \( K_c = 3.0 \), so we can set up the equation:
\[ 3.0 = \frac{\alpha}{1 - \alpha} \]

4. **Solve for \( \alpha \):**
- Cross-multiply to solve for \( \alpha \):
\[ 3.0(1 - \alpha) = \alpha \]
- This simplifies to:
\[ 3.0 - 3.0\alpha = \alpha \]
- Rearranging gives:
\[ 3.0 = 4.0\alpha \]
- Thus, we find:
\[ \alpha = \frac{3.0}{4.0} = 0.75 \]

5. **Calculate the Percentage by Mass of Iso-butane:**
- The total concentration at equilibrium is:
\[ \text{Total concentration} = [\text{n-butane}] + [\text{iso-butane}] = (1 - \alpha) + \alpha = 1 \]
- The concentration of iso-butane is \( \alpha = 0.75 \).
- The percentage by mass of iso-butane in the mixture is given by:
\[ \text{Percentage by mass} = \left( \frac{\text{mass of iso-butane}}{\text{total mass}} \right) \times 100 \]
- Since the mass is directly proportional to concentration (assuming constant volume and molecular weight), we can directly use the concentration:
\[ \text{Percentage by mass of iso-butane} = 0.75 \times 100 = 75\% \]

### Final Answer:
The percentage by mass of iso-butane in the equilibrium mixture is **75%**.