`K` for the synthesis of `HI` is `50`. `K` for dissociation of `HI` is

Options

A. `50`
B. `5`
C. `0.2`
D. `0.02`

Answer

To find the equilibrium constant \( K \) for the dissociation of hydrogen iodide (HI), we can follow these steps:

### Step 1: Write the synthesis reaction and its equilibrium expression
The synthesis of HI can be represented as:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]
The equilibrium constant \( K_1 \) for this reaction is given by:
\[ K_1 = \frac{[HI]^2}{[H_2][I_2]} \]
Given that \( K_1 = 50 \).

### Step 2: Write the dissociation reaction and its equilibrium expression
The dissociation of HI can be represented as:
\[ 2 HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
We will denote the equilibrium constant for this reaction as \( K_2 \). The equilibrium expression for this reaction is:
\[ K_2 = \frac{[H_2][I_2]}{[HI]^2} \]

### Step 3: Relate \( K_2 \) to \( K_1 \)
Notice that the dissociation reaction is the reverse of the synthesis reaction. For reactions that are the reverse of each other, the equilibrium constants are reciprocals. Therefore, we have:
\[ K_2 = \frac{1}{K_1} \]

### Step 4: Substitute the value of \( K_1 \)
Now, substituting the given value of \( K_1 \):
\[ K_2 = \frac{1}{50} \]

### Step 5: Calculate \( K_2 \)
Calculating the value:
\[ K_2 = 0.02 \]

### Final Answer
Thus, the equilibrium constant for the dissociation of HI is:
\[ K_2 = 0.02 \]

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