Unit of equilibrium constant for the reversible reaction `H_(2) + I_(2)hArr 2HI` is

Options

A. `mol^(-1)` litre
B. `mol^(-2)` litre
C. mol `litre^(-1)`
D. None of these

Explanation

To determine the unit of the equilibrium constant (Kc) for the reversible reaction:

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

we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (Kc)
The equilibrium constant expression for the reaction is given by:

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

where \([HI]\), \([H_2]\), and \([I_2]\) represent the molar concentrations (in moles per liter, M) of the respective species at equilibrium.

### Step 2: Substitute the units into the Kc expression
Now, we will substitute the units into the expression. The concentration units are moles per liter (M), so we can write:

\[ K_c = \frac{(M)^2}{(M)(M)} = \frac{M^2}{M^2} \]

### Step 3: Simplify the expression
When we simplify the expression, we find:

\[ K_c = \frac{M^2}{M^2} = 1 \]

### Step 4: Conclusion about the units
Since the units cancel out, we conclude that the equilibrium constant \(K_c\) is unitless.

### Final Answer
The unit of the equilibrium constant for the reaction \(H_2 + I_2 \rightleftharpoons 2 HI\) is **unitless**.

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