One mole of A and 2 moles of B are allowed to react in a 0.5 lit flask. What is the value of K if at equilibirum, 0.4 moles of C is formed in the reaction `A+2B Leftrightarrow C+2D`

Options

A. `4//9`
B. `9//4`
C. `8//27`
D. `27//8`

Explanation

To solve the problem step-by-step, we will analyze the given reaction and the information provided.

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ A + 2B \leftrightarrow C + 2D \]

### Step 2: Set up the initial moles
Initially, we have:
- Moles of A = 1 mole
- Moles of B = 2 moles
- Moles of C = 0 moles (since it hasn't formed yet)
- Moles of D = 0 moles (since it hasn't formed yet)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the number of moles of C formed at equilibrium. According to the problem, at equilibrium, 0.4 moles of C is formed, so:
\[ x = 0.4 \]

### Step 4: Calculate the change in moles for A and B
From the balanced equation:
- For every 1 mole of C formed, 1 mole of A is consumed and 2 moles of B are consumed.
- Therefore, at equilibrium:
- Moles of A = Initial moles of A - moles of A consumed = \( 1 - x = 1 - 0.4 = 0.6 \)
- Moles of B = Initial moles of B - moles of B consumed = \( 2 - 2x = 2 - 2(0.4) = 2 - 0.8 = 1.2 \)

### Step 5: Calculate the moles of D formed
From the balanced equation, for every mole of C formed, 2 moles of D are formed:
- Moles of D = \( 2x = 2(0.4) = 0.8 \)

### Step 6: Summarize the equilibrium moles
At equilibrium, we have:
- Moles of A = 0.6
- Moles of B = 1.2
- Moles of C = 0.4
- Moles of D = 0.8

### Step 7: Calculate the concentrations
We need to calculate the concentrations of each species in a 0.5 L flask:
- Concentration of A = \( \frac{\text{Moles of A}}{\text{Volume}} = \frac{0.6}{0.5} = 1.2 \, \text{M} \)
- Concentration of B = \( \frac{\text{Moles of B}}{\text{Volume}} = \frac{1.2}{0.5} = 2.4 \, \text{M} \)
- Concentration of C = \( \frac{\text{Moles of C}}{\text{Volume}} = \frac{0.4}{0.5} = 0.8 \, \text{M} \)
- Concentration of D = \( \frac{\text{Moles of D}}{\text{Volume}} = \frac{0.8}{0.5} = 1.6 \, \text{M} \)

### Step 8: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{C}]^1 [\text{D}]^2}{[\text{A}]^1 [\text{B}]^2} \]

### Step 9: Substitute the concentrations into the \( K_c \) expression
Substituting the calculated concentrations:
\[ K_c = \frac{(0.8)^1 (1.6)^2}{(1.2)^1 (2.4)^2} \]

### Step 10: Calculate \( K_c \)
Calculating the values:
- \( (1.6)^2 = 2.56 \)
- \( (2.4)^2 = 5.76 \)

Now substituting these values:
\[ K_c = \frac{0.8 \times 2.56}{1.2 \times 5.76} = \frac{2.048}{6.912} \]

Calculating the final value:
\[ K_c \approx 0.2963 \approx \frac{8}{27} \]

### Final Answer
The value of \( K_c \) is approximately \( \frac{8}{27} \).

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