At a certain temp. `2HI hArrH_(2) + I_(2)` . Only `50%` HI is dissociated at equilibrium. The equilibrium constant is

Options

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

Answer

To find the equilibrium constant for the reaction \( 2HI \rightleftharpoons H_2 + I_2 \) where 50% of HI is dissociated at equilibrium, we can follow these steps:

### Step 1: Set up the initial conditions
Assume we start with 1 mole of HI. Initially, we have:
- Moles of HI = 1
- Moles of H2 = 0
- Moles of I2 = 0

### Step 2: Determine the change at equilibrium
Since 50% of HI is dissociated, this means:
- Moles of HI dissociated = 50% of 1 = 0.5 moles
- Moles of HI remaining = 1 - 0.5 = 0.5 moles

### Step 3: Calculate the moles of products formed
According to the stoichiometry of the reaction:
- For every 2 moles of HI that dissociate, 1 mole of H2 and 1 mole of I2 are formed.
- Therefore, if 0.5 moles of HI dissociate, the moles of H2 formed = \( \frac{0.5}{2} = 0.25 \) moles.
- Similarly, moles of I2 formed = \( \frac{0.5}{2} = 0.25 \) moles.

### Step 4: Write the equilibrium concentrations
At equilibrium, we have:
- Moles of HI = 0.5
- Moles of H2 = 0.25
- Moles of I2 = 0.25

Assuming the reaction occurs in a container of volume \( V \):
- Concentration of HI = \( \frac{0.5}{V} \)
- Concentration of H2 = \( \frac{0.25}{V} \)
- Concentration of I2 = \( \frac{0.25}{V} \)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
Substituting the concentrations:
\[ K_c = \frac{\left(\frac{0.25}{V}\right) \left(\frac{0.25}{V}\right)}{\left(\frac{0.5}{V}\right)^2} \]

### Step 6: Simplify the expression
\[ K_c = \frac{\frac{0.25 \times 0.25}{V^2}}{\frac{0.25}{V^2}} = \frac{0.0625}{0.25} = 0.25 \]

### Final Answer
The equilibrium constant \( K_c \) is \( 0.25 \).

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