If the equilibrium constant for the reaction `0.125`.
`P_(4(g))+6Cl_(2(g))hArr4PCl_(3(g))`
The value of equilibrium constant for this reaction
`4PCl_(3(g))hArrP_(4(g))+6Cl_(2(g))`

Options

A. `0.25`
B. `8`
C. `0.125`
D. `6`

Answer

To find the equilibrium constant for the reaction \(4PCl_3(g) \rightleftharpoons P_4(g) + 6Cl_2(g)\) given that the equilibrium constant for the reaction \(P_4(g) + 6Cl_2(g) \rightleftharpoons 4PCl_3(g)\) is \(K_{c1} = 0.125\), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Given Reaction and Its Equilibrium Constant**:
The first reaction is:
\[ P_4(g) + 6Cl_2(g) \rightleftharpoons 4PCl_3(g) \]
The equilibrium constant for this reaction is given as:
\[ K_{c1} = 0.125 \]

2. **Write the Reverse Reaction**:
The reverse of the given reaction is:
\[ 4PCl_3(g) \rightleftharpoons P_4(g) + 6Cl_2(g) \]

3. **Relate the Equilibrium Constants**:
For any reaction, if you reverse the reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Therefore:
\[ K_{c2} = \frac{1}{K_{c1}} \]

4. **Substitute the Value of \(K_{c1}\)**:
Now, substituting the value of \(K_{c1}\):
\[ K_{c2} = \frac{1}{0.125} \]

5. **Calculate \(K_{c2}\)**:
Performing the calculation:
\[ K_{c2} = 8 \]

### Final Answer:
The equilibrium constant for the reaction \(4PCl_3(g) \rightleftharpoons P_4(g) + 6Cl_2(g)\) is:
\[ K_{c2} = 8 \]

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