The equilibrium constant at 717 K for the reaction: `H_(2(g))+I_(2(g))lArr2HI_((g))` is 50.
The equilibrium constant for the reaction:
`2HI_(2(g))lArrH_(2(g))+I_(2(g))` is

Options

A. 0.5
B. `2xx10^(-2)`
C. `4.0`
D. `1xx10^(-1)`

Answer

To find the equilibrium constant for the reaction \( 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \), we can use the relationship between the equilibrium constants of forward and reverse reactions.

### Step-by-Step Solution:

1. **Identify the given reaction and its equilibrium constant**:
The equilibrium constant \( K \) for the reaction:
\[ H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)} \]
is given as \( K = 50 \) at 717 K.

2. **Write the expression for the equilibrium constant**:
For the forward reaction, the equilibrium constant \( K \) can be expressed as:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the equilibrium concentrations of hydrogen iodide, hydrogen, and iodine, respectively.

3. **Determine the equilibrium constant for the reverse reaction**:
The reverse reaction is:
\[ 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \]
The equilibrium constant for the reverse reaction, which we will denote as \( K' \), is related to \( K \) by the following relationship:
\[ K' = \frac{1}{K} \]

4. **Calculate \( K' \)**:
Substituting the value of \( K \):
\[ K' = \frac{1}{50} = 0.02 \]

5. **Final Answer**:
The equilibrium constant for the reaction \( 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \) is:
\[ K' = 0.02 \]