`K_(p)` and `K_(c)` are inter related as
`K_(p)=K_(c)(RT)^(Deltan)`
Answer the following questions:
The unit of equilibrium constant for
`H_(2)(g)+I_(2)(g) hArr 2HI(g)`

Options

A. mol `L^(-2)`
B. `mol^(2) L^(-2)`
C. `L mol^(-2)`
D. None of these

Answer

To find the unit of the equilibrium constant for the reaction:

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

we will follow these steps:

### Step 1: Identify the reaction and count moles
The reaction involves:
- Reactants: \( H_2 \) and \( I_2 \) (both are gases)
- Products: \( 2HI \) (gas)

### Step 2: Calculate the change in moles (Δn)
The change in moles (Δn) is calculated as:
\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

From the reaction:
- Moles of gaseous products = 2 (from \( 2HI \))
- Moles of gaseous reactants = 1 (from \( H_2 \)) + 1 (from \( I_2 \)) = 2

Thus,
\[ \Delta n = 2 - 2 = 0 \]

### Step 3: Determine the unit of \( K_c \)
The unit of the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]

For our reaction, the unit of \( K_c \) can be expressed as:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

The concentration units for each component are in moles per liter (mol/L). Therefore, substituting the units:
\[ K_c = \frac{(mol/L)^2}{(mol/L)(mol/L)} = \frac{(mol/L)^2}{(mol/L)^2} = 1 \]

### Step 4: Conclusion
Since \( \Delta n = 0 \), we find that:
\[ K_c = (mol/L)^0 = 1 \]

Thus, the unit of the equilibrium constant \( K_c \) for the given reaction is dimensionless (no units).