The unit of `K_(c)` from the reaction `N_(2)+H_(2)hArr2NH_(3)+Q` is

Options

A. `lit^(2)mol^(-2)`
B. `mollit^(-1)`
C. `mol^(2)lit^(2)`
D. `lit mol^(-2)`

Explanation

To find the unit of \( K_c \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 + Q \), we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced equation for the reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is expressed in terms of the concentrations of the products and reactants:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 3: Assign Units to Concentrations
The concentration of a substance is measured in moles per liter (mol/L). Therefore, we can assign the following units:
- \([NH_3]\) has units of \( \text{mol/L} \)
- \([N_2]\) has units of \( \text{mol/L} \)
- \([H_2]\) has units of \( \text{mol/L} \)

### Step 4: Substitute Units into the \( K_c \) Expression
Substituting the units into the \( K_c \) expression gives:
\[ K_c = \frac{(\text{mol/L})^2}{(\text{mol/L})(\text{mol/L})^3} \]

### Step 5: Simplify the Units
Now, simplify the expression:
\[ K_c = \frac{(\text{mol}^2/\text{L}^2)}{(\text{mol}^1/\text{L}^1)(\text{mol}^3/\text{L}^3)} = \frac{\text{mol}^2/\text{L}^2}{\text{mol}^4/\text{L}^4} \]
This can be simplified further:
\[ K_c = \frac{\text{mol}^2 \cdot \text{L}^4}{\text{mol}^4 \cdot \text{L}^2} = \frac{\text{L}^2}{\text{mol}^2} \]

### Step 6: Final Units of \( K_c \)
Thus, the final units of \( K_c \) are:
\[ K_c = \frac{\text{L}^2}{\text{mol}^2} \]

### Conclusion
The unit of \( K_c \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 + Q \) is \( \text{L}^2/\text{mol}^2 \).

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