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 \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), we will follow these steps:

### Step 1: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \([HI]\), \([H_2]\), and \([I_2]\) are the molar concentrations of the respective species at equilibrium.

### Step 2: Convert the volume from mL to L
The volume of the flask is given as 500 mL. To convert this to liters:
\[ \text{Volume in L} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

### Step 3: Calculate the concentrations of each species
Using the number of moles provided and the volume in liters, we can calculate the concentrations:
- For \( HI \):
\[ [HI] = \frac{1.5 \text{ moles}}{0.5 \text{ L}} = 3 \text{ M} \]
- For \( H_2 \):
\[ [H_2] = \frac{3 \text{ moles}}{0.5 \text{ L}} = 6 \text{ M} \]
- For \( I_2 \):
\[ [I_2] = \frac{2 \text{ moles}}{0.5 \text{ L}} = 4 \text{ M} \]

### Step 4: Substitute the concentrations into the \( K_c \) expression
Now we can substitute the concentrations into the \( K_c \) expression:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3)^2}{(6)(4)} \]

### Step 5: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{9}{24} = 0.375 \]

### Final Answer
Thus, the value of \( K_c \) is:
\[ K_c = 0.375 \]