1.50 moles each of hydrogen and iodine were placed in a sealed 10 litre container maintained at 717 K. At equilibrium `1.25` moles each of hydrogen and iodine were left behind. The equilibrium constant, `K_(c)` for the reaction , `H_(2)(g)+I_(2)(g) hArr 2Hl(g)` at 717 K is

Options

A. `0.4`
B. `0.16`
C. 25
D. 50

Explanation

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

\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

we will follow these steps:

### Step 1: Determine Initial Moles
Initially, we have:
- Moles of \( H_2 \) = 1.50 moles
- Moles of \( I_2 \) = 1.50 moles
- Moles of \( HI \) = 0 moles

### Step 2: Determine Moles at Equilibrium
At equilibrium, we are told that:
- Moles of \( H_2 \) = 1.25 moles
- Moles of \( I_2 \) = 1.25 moles

Using this information, we can find the change in moles for \( H_2 \) and \( I_2 \):
- Change in moles of \( H_2 \) = Initial moles - Equilibrium moles = \( 1.50 - 1.25 = 0.25 \) moles
- Change in moles of \( I_2 \) = Initial moles - Equilibrium moles = \( 1.50 - 1.25 = 0.25 \) moles

### Step 3: Calculate Moles of \( HI \) Produced
Since the reaction produces 2 moles of \( HI \) for every mole of \( H_2 \) and \( I_2 \) that reacts, the moles of \( HI \) produced will be:
- Moles of \( HI \) = \( 2 \times \) (Change in moles of \( H_2 \) or \( I_2 \)) = \( 2 \times 0.25 = 0.50 \) moles

### Step 4: Calculate Concentrations at Equilibrium
Next, we need to calculate the concentrations of all species at equilibrium. The volume of the container is 10 liters.

- Concentration of \( H_2 \):
\[ [H_2] = \frac{1.25 \text{ moles}}{10 \text{ L}} = 0.125 \text{ M} \]

- Concentration of \( I_2 \):
\[ [I_2] = \frac{1.25 \text{ moles}}{10 \text{ L}} = 0.125 \text{ M} \]

- Concentration of \( HI \):
\[ [HI] = \frac{0.50 \text{ moles}}{10 \text{ L}} = 0.050 \text{ M} \]

### Step 5: Use the Equilibrium Expression to Calculate \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.050)^2}{(0.125)(0.125)} = \frac{0.0025}{0.015625} \]

Calculating this gives:

\[ K_c = 0.16 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at 717 K is \( 0.16 \).

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