Equilibrium constant for the reaction `4A(g) harr B(g) + 2C (g)` is-
Given : `2A(g) harr B(g) + Y(g), kc_(1) = 8`
`C(g) harr A(g) + 1/2Y(g) , kc_(2) = 1/4`

Options

A. 32
B. 16
C. 2
D. 128

Answer

To find the equilibrium constant for the reaction \(4A(g) \rightleftharpoons B(g) + 2C(g)\), we will use the provided equilibrium constants for the two given reactions.

### Step 1: Write down the given reactions and their equilibrium constants.

1. \(2A(g) \rightleftharpoons B(g) + Y(g)\) with \(K_{c1} = 8\)
2. \(C(g) \rightleftharpoons A(g) + \frac{1}{2}Y(g)\) with \(K_{c2} = \frac{1}{4}\)

### Step 2: Manipulate the second reaction to eliminate \(Y(g)\).

We need to reverse the second reaction to express it in terms of \(C(g)\) on the left side:

\[ A(g) + \frac{1}{2}Y(g) \rightleftharpoons C(g) \]

When we reverse a reaction, the equilibrium constant is inverted:

\[ K_{c2}' = \frac{1}{K_{c2}} = \frac{1}{\frac{1}{4}} = 4 \]

### Step 3: Multiply the reversed second reaction by 2.

To match the stoichiometry needed for our target reaction, we multiply the reversed reaction by 2:

\[ 2A(g) + Y(g) \rightleftharpoons 2C(g) \]

The equilibrium constant for this new reaction will be the square of the equilibrium constant of the reversed reaction:

\[ K_{c2}'' = (K_{c2}')^2 = 4^2 = 16 \]

### Step 4: Add the first reaction and the modified second reaction.

Now we can add the first reaction and the modified second reaction:

1. \(2A(g) \rightleftharpoons B(g) + Y(g)\) with \(K_{c1} = 8\)
2. \(2A(g) + Y(g) \rightleftharpoons 2C(g)\) with \(K_{c2}'' = 16\)

When we add these two reactions, \(Y(g)\) cancels out:

\[ 4A(g) \rightleftharpoons B(g) + 2C(g) \]

### Step 5: Calculate the overall equilibrium constant.

The overall equilibrium constant for the combined reaction is the product of the individual equilibrium constants:

\[ K_c = K_{c1} \times K_{c2}'' = 8 \times 16 = 128 \]

### Final Answer:

Thus, the equilibrium constant for the reaction \(4A(g) \rightleftharpoons B(g) + 2C(g)\) is \(K_c = 128\).