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 determine the percentage by mass of iso-butane in the equilibrium mixture given the equilibrium constant \( K_c \) is 3.0 for the reaction:

\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 (g) \rightleftharpoons \text{CH}_3\text{C}(\text{CH}_3)(g) + \text{CH}_3 \]

### Step 1: Set up the initial concentrations
Assume we start with 1 mole of butane (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \)) and 0 moles of iso-butane (\( \text{CH}_3\text{C}(\text{CH}_3) \)).

- Initial concentration of butane = 1 M
- Initial concentration of iso-butane = 0 M

### Step 2: Define the change in concentration
Let \( x \) be the amount of iso-butane formed at equilibrium. Therefore, at equilibrium:

- Concentration of butane = \( 1 - x \)
- Concentration of iso-butane = \( x \)

### Step 3: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{iso-butane}]}{[\text{butane}]} = \frac{x}{1 - x} \]

Given \( K_c = 3.0 \), we can set up the equation:

\[ 3.0 = \frac{x}{1 - x} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:

\[ 3(1 - x) = x \]

Expanding this, we get:

\[ 3 - 3x = x \]

Rearranging the equation:

\[ 3 = 4x \]

Thus,

\[ x = \frac{3}{4} = 0.75 \]

### Step 5: Calculate the mass percentage of iso-butane
To find the percentage by mass of iso-butane in the equilibrium mixture, we need to calculate the total mass of the mixture at equilibrium.

- Mass of iso-butane = \( 0.75 \) moles
- Mass of butane = \( 1 - 0.75 = 0.25 \) moles

Assuming the molar mass of butane (\( \text{C}_4\text{H}_{10} \)) is approximately 58 g/mol and iso-butane (\( \text{C}_4\text{H}_{10} \)) is also approximately 58 g/mol, we can calculate:

- Mass of iso-butane = \( 0.75 \, \text{mol} \times 58 \, \text{g/mol} = 43.5 \, \text{g} \)
- Mass of butane = \( 0.25 \, \text{mol} \times 58 \, \text{g/mol} = 14.5 \, \text{g} \)

Total mass of the mixture:

\[ \text{Total mass} = 43.5 \, \text{g} + 14.5 \, \text{g} = 58 \, \text{g} \]

Now, calculate the percentage by mass of iso-butane:

\[ \text{Percentage by mass of iso-butane} = \left( \frac{43.5 \, \text{g}}{58 \, \text{g}} \right) \times 100 \approx 75\% \]

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