Equivalent 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

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between hydrogen and iodine is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Define initial amounts
Let’s assume we start with 1 mole of \( H_2 \) and 1 mole of \( I_2 \). Initially, the amount of \( HI \) is 0.

- Initial concentrations:
- \( [H_2] = 1 \) mole
- \( [I_2] = 1 \) mole
- \( [HI] = 0 \) mole

### Step 3: Determine the change in amounts
According to the problem, 80% of \( H_2 \) is converted to \( HI \).

- Amount of \( H_2 \) converted:
\[ 0.8 \text{ moles of } H_2 \]

- Remaining \( H_2 \):
\[ 1 - 0.8 = 0.2 \text{ moles of } H_2 \]

- Since the stoichiometry of the reaction shows that 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \), the amount of \( I_2 \) that reacts is also 0.8 moles.

- Remaining \( I_2 \):
\[ 1 - 0.8 = 0.2 \text{ moles of } I_2 \]

- Amount of \( HI \) produced:
\[ 2 \times 0.8 = 1.6 \text{ moles of } HI \]

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [H_2] = 0.2 \) moles
- \( [I_2] = 0.2 \) moles
- \( [HI] = 1.6 \) moles

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

### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(1.6)^2}{(0.2)(0.2)} \]

### Step 7: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{2.56}{0.04} = 64 \]

### Final Answer
Thus, the value of \( K_c \) at this temperature is \( 64 \).

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