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 \(4A(g) \rightleftharpoons B(g) + 2C(g)\), given the two reactions and their equilibrium constants, we can follow these steps:

### Step 1: Write down the given reactions and their equilibrium constants.
1. \(2A(g) \rightleftharpoons B(g) + Y(g), \quad K_{c1} = 8\)
2. \(C(g) \rightleftharpoons A(g) + \frac{1}{2}Y(g), \quad K_{c2} = \frac{1}{4}\)

### Step 2: Reverse the second reaction to eliminate \(Y\).
To eliminate \(Y\) from the final equation, we reverse the second reaction:
\[ A(g) + \frac{1}{2}Y(g) \rightleftharpoons C(g) \]
When we reverse a reaction, the equilibrium constant is inverted:
\[ K_{c3} = \frac{1}{K_{c2}} = \frac{1}{\frac{1}{4}} = 4 \]

### Step 3: Multiply the reversed second reaction by 2.
Now, we multiply the reversed second reaction by 2 to match the stoichiometry needed for \(4A\):
\[ 2A(g) + Y(g) \rightleftharpoons 2C(g) \]
The new equilibrium constant for this reaction will be:
\[ K_{c4} = K_{c3}^2 = 4^2 = 16 \]

### Step 4: Add the first reaction and the modified second reaction.
Now we add the first reaction and the modified second reaction:
1. \(2A(g) \rightleftharpoons B(g) + Y(g)\)
2. \(2A(g) + Y(g) \rightleftharpoons 2C(g)\)

Adding these reactions gives:
\[ 4A(g) \rightleftharpoons B(g) + 2C(g) \]

### Step 5: Calculate the equilibrium constant for the overall reaction.
The equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions:
\[ K_c = K_{c1} \times K_{c4} = 8 \times 16 = 128 \]

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

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