The following equilibrium are given below,
`A_2 + 3B_2 hArr 2AB_3 ....K_1`
`A_2 + C_2 hArr 2AC ....K_2`
` B_2 + 1/2C_2 hArr B_2C ....K_3`
The equilibrium constant of the reaction
`2AB_3 + 5/2C_2 hArr 2AC + 3B_2C`, in terms of`K_1, K_2`, and `K_3` is

Options

A. `K_1K_2/K_3`
B. `K_1K_3^2/K_2`
C. `K_2K_3^3/K_1`
D. `K_1K_2K_3`

Answer

To find the equilibrium constant for the reaction:

\[ 2AB_3 + \frac{5}{2}C_2 \rightleftharpoons 2AC + 3B_2C \]

in terms of \( K_1, K_2, \) and \( K_3 \), we will manipulate the given equilibrium reactions step by step.

### Step 1: Write the given equilibrium reactions
1. \( A_2 + 3B_2 \rightleftharpoons 2AB_3 \) (Equilibrium constant \( K_1 \))
2. \( A_2 + C_2 \rightleftharpoons 2AC \) (Equilibrium constant \( K_2 \))
3. \( B_2 + \frac{1}{2}C_2 \rightleftharpoons B_2C \) (Equilibrium constant \( K_3 \))

### Step 2: Reverse the first reaction
To obtain \( 2AB_3 \) on the reactant side, we reverse the first reaction:

\[ 2AB_3 \rightleftharpoons A_2 + 3B_2 \]

The equilibrium constant for the reversed reaction is:

\[ K' = \frac{1}{K_1} \]

### Step 3: Keep the second reaction as it is
The second reaction remains unchanged:

\[ A_2 + C_2 \rightleftharpoons 2AC \]

The equilibrium constant remains:

\[ K_2 \]

### Step 4: Multiply the third reaction by 3
To obtain \( 3B_2C \), we multiply the third reaction by 3:

\[ 3(B_2 + \frac{1}{2}C_2 \rightleftharpoons B_2C) \]

This gives:

\[ 3B_2 + \frac{3}{2}C_2 \rightleftharpoons 3B_2C \]

The equilibrium constant for this reaction becomes:

\[ K' = K_3^3 \]

### Step 5: Combine the reactions
Now, we can add the modified reactions together:

1. \( 2AB_3 \rightleftharpoons A_2 + 3B_2 \) (from Step 2)
2. \( A_2 + C_2 \rightleftharpoons 2AC \) (from Step 3)
3. \( 3B_2 + \frac{3}{2}C_2 \rightleftharpoons 3B_2C \) (from Step 4)

When we add these reactions, the \( A_2 \) cancels out, and we get:

\[ 2AB_3 + \frac{5}{2}C_2 \rightleftharpoons 2AC + 3B_2C \]

### Step 6: Combine the equilibrium constants
When adding reactions, the equilibrium constants multiply:

\[ K_{dash} = \frac{1}{K_1} \cdot K_2 \cdot K_3^3 \]

Thus, we can express \( K_{dash} \) as:

\[ K_{dash} = \frac{K_2 \cdot K_3^3}{K_1} \]

### Final Answer
The equilibrium constant for the reaction \( 2AB_3 + \frac{5}{2}C_2 \rightleftharpoons 2AC + 3B_2C \) in terms of \( K_1, K_2, \) and \( K_3 \) is:

\[ K_{dash} = \frac{K_2 \cdot K_3^3}{K_1} \]

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