The equilibrium constant for the reaction
`I_2(g) +Cl_2(g) hArr 2ICl(g)`
at 740 K is 640. What is the equilibrium constant for the reaction ?
`2ICl(g) hArr I_2(g)+Cl_2(g)` at the same temperature .

Explanation

To find the equilibrium constant for the reaction

\[ 2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g) \]

given that the equilibrium constant for the reaction

\[ I_2(g) + Cl_2(g) \rightleftharpoons 2ICl(g) \]

is 640 at 740 K, we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:

\[ I_2(g) + Cl_2(g) \rightleftharpoons 2ICl(g) \]

The equilibrium constant \( K_1 \) for this reaction is given as:

\[ K_1 = 640 \]

### Step 2: Write the reverse of the given reaction
To find the equilibrium constant for the reverse reaction:

\[ 2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g) \]

### Step 3: Relate the equilibrium constants
When a reaction is reversed, the equilibrium constant for the new reaction \( K_2 \) is the reciprocal of the original equilibrium constant \( K_1 \). Therefore:

\[ K_2 = \frac{1}{K_1} \]

### Step 4: Substitute the value of \( K_1 \)
Now, substituting the value of \( K_1 \):

\[ K_2 = \frac{1}{640} \]

### Step 5: Calculate \( K_2 \)
Calculating \( K_2 \):

\[ K_2 = 0.0015625 \]

Thus, the equilibrium constant for the reaction

\[ 2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g) \]

at 740 K is

\[ K_2 = 0.0015625 \]

### Final Answer:
The equilibrium constant for the reaction \( 2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g) \) at 740 K is \( 0.0015625 \).

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