In the 500 ml flask following reaction takes place `H_(2)(g)+I_(2)(g)hArr 2HI(g)`. At equilibrium, concentration of `H_(2),I_(2) and HI` is 3, 2 and 1.5 mole then the value of `K_(C)` will be

Options

A. 3.75
B. 0.375
C. 3.25
D. 0.187

Answer

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

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

given the concentrations at equilibrium, we follow these steps:

### Step 1: Write the expression for \( K_C \)

The equilibrium constant \( K_C \) is defined as:

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

### Step 2: Determine the concentrations of the reactants and products

We are given the number of moles of each substance at equilibrium and the volume of the flask. The volume of the flask is 500 mL, which is equivalent to 0.5 L.

- For \( H_2 \): 3 moles
- For \( I_2 \): 2 moles
- For \( HI \): 1.5 moles

Now, we can calculate the concentrations:

\[ [H_2] = \frac{3 \text{ moles}}{0.5 \text{ L}} = 6 \text{ M} \]

\[ [I_2] = \frac{2 \text{ moles}}{0.5 \text{ L}} = 4 \text{ M} \]

\[ [HI] = \frac{1.5 \text{ moles}}{0.5 \text{ L}} = 3 \text{ M} \]

### Step 3: Substitute the concentrations into the \( K_C \) expression

Now we substitute the concentrations into the \( K_C \) expression:

\[ K_C = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3)^2}{(6)(4)} \]

### Step 4: Calculate \( K_C \)

Calculating the values:

\[ K_C = \frac{9}{24} = 0.375 \]

### Final Answer

Thus, the value of \( K_C \) is:

\[ K_C = 0.375 \]