The unit for the equilibrium constant of the reaction

Options

A. `"[mole/lit]"^(-)`
B. `"[mole/lit]"^(-2)`
C. `"Mole/lit"`
D. `"[mole/lit]"^(2)`

Answer

To determine the unit for the equilibrium constant (K) of the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \), we can follow these steps:

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

\[ K = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 2: Identify the concentrations
In this expression:
- \([NH_3]\) is the concentration of ammonia in moles per liter (mol/L).
- \([N_2]\) is the concentration of nitrogen in moles per liter (mol/L).
- \([H_2]\) is the concentration of hydrogen in moles per liter (mol/L).

### Step 3: Substitute the concentrations into the K expression
Substituting the units into the expression for K, we get:

\[ K = \frac{(mol/L)^2}{(mol/L)(mol/L)^3} = \frac{(mol/L)^2}{(mol/L)^4} \]

### Step 4: Simplify the expression
Now, simplifying the expression gives:

\[ K = \frac{(mol/L)^2}{(mol^4/L^4)} = (mol/L)^{2-4} = (mol/L)^{-2} \]

### Step 5: Final unit of the equilibrium constant
Thus, the unit for the equilibrium constant \( K \) for the given reaction is:

\[ K = mol/L^{-2} \]

### Conclusion
The correct unit for the equilibrium constant of the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) is \( mol/L^{-2} \).

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