At a certain temperature , only 50% HI is dissociated at equilibrium in the following reaction:
`2HI(g)hArrH_(2)(g)+I_(2)(g)`
the equilibrium constant for this reaction is:

Options

A. `0.25`
B. `1.0`
C. `3.0`
D. `0.5`

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

given that 50% of HI is dissociated at equilibrium, we can follow these steps:

### Step 1: Define the initial concentrations
Let's assume the initial concentration of HI is \( A \) (in moles). At time \( t = 0 \):
- \([ \text{HI} ] = A\)
- \([ \text{H}_2 ] = 0\)
- \([ \text{I}_2 ] = 0\)

### Step 2: Determine the change in concentrations at equilibrium
Since 50% of HI is dissociated, this means that at equilibrium, 50% of \( A \) will have reacted. Therefore, the amount of HI that dissociates is \( \alpha = 0.5 \). The change in concentrations at equilibrium will be:
- \([ \text{HI} ] = A - 2 \times \frac{A \alpha}{2} = A - A \cdot 0.5 = A \cdot 0.5\)
- \([ \text{H}_2 ] = \frac{A \alpha}{2} = \frac{A \cdot 0.5}{2} = \frac{A}{4}\)
- \([ \text{I}_2 ] = \frac{A \alpha}{2} = \frac{A \cdot 0.5}{2} = \frac{A}{4}\)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

Substituting the equilibrium concentrations we found:

\[ K_c = \frac{\left(\frac{A}{4}\right) \left(\frac{A}{4}\right)}{\left(0.5A\right)^2} \]

### Step 4: Simplify the expression
Now, substituting the values:

\[ K_c = \frac{\frac{A^2}{16}}{(0.5A)^2} \]

Calculating \( (0.5A)^2 \):

\[ (0.5A)^2 = 0.25A^2 \]

Now substituting this back into the equation for \( K_c \):

\[ K_c = \frac{\frac{A^2}{16}}{0.25A^2} = \frac{A^2/16}{A^2/4} \]

### Step 5: Further simplify
The \( A^2 \) cancels out:

\[ K_c = \frac{1/16}{1/4} = \frac{1}{16} \times \frac{4}{1} = \frac{4}{16} = \frac{1}{4} \]

### Final Result
Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 4 \]