Equilivalent amounts of `H_(2)` and `I_(2)` are heated in a closed vessel till equilibrium is obtained. If `80%` of the hydrogen is converted to `HI`, the `K_(c)` at this temperature is

Options

A. 64
B. 16
C. 0.25
D. 4

Explanation

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction between hydrogen (\( H_2 \)) and iodine (\( I_2 \)) to form hydrogen iodide (\( HI \)) given that 80% of the hydrogen is converted to \( HI \).

### Step-by-step Solution:

1. **Identify Initial Conditions**:
- Let the initial amount of \( H_2 \) and \( I_2 \) be \( A \) moles each.
- At the start (time \( t = 0 \)):
- \( [H_2] = A \)
- \( [I_2] = A \)
- \( [HI] = 0 \)

2. **Determine the Change in Concentration**:
- Since 80% of \( H_2 \) is converted to \( HI \), the amount of \( H_2 \) that reacts is \( 0.8A \).
- Therefore, the amount of \( H_2 \) remaining at equilibrium is:
\[ [H_2] = A - 0.8A = 0.2A \]
- The amount of \( I_2 \) that reacts is the same as the amount of \( H_2 \) that reacts (1:1 stoichiometry), so:
\[ [I_2] = A - 0.8A = 0.2A \]
- The amount of \( HI \) formed is:
\[ [HI] = 2 \times 0.8A = 1.6A \]

3. **Write Equilibrium Concentrations**:
- At equilibrium:
- \( [H_2] = 0.2A \)
- \( [I_2] = 0.2A \)
- \( [HI] = 1.6A \)

4. **Write the Expression for \( K_c \)**:
- The equilibrium constant \( K_c \) for the reaction:
\[ H_2 + I_2 \rightleftharpoons 2HI \]
- Is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

5. **Substitute the Equilibrium Concentrations into the \( K_c \) Expression**:
- Plugging in the equilibrium concentrations:
\[ K_c = \frac{(1.6A)^2}{(0.2A)(0.2A)} \]

6. **Simplify the Expression**:
- Calculate \( K_c \):
\[ K_c = \frac{(2.56A^2)}{(0.04A^2)} = \frac{2.56}{0.04} = 64 \]

### Final Answer:
Thus, the equilibrium constant \( K_c \) at this temperature is \( 64 \).

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