For the reaction, `N_(2)+3H_(2)hArr2NH_(3)`, the units of `K_(c) "and" K_(p)` respectively are:

Options

A. `mol^(-2)L^(2) "and" b ar^(-2)`
B. `mol^(-2)L^(2) "and" B ar^(-1)`
C. `mol^(-1)L "and" ba r^(-2)`
D. `mol^(-1)L^(-1) "and" b ar^(-1)`

Answer

To determine the units of \( K_c \) and \( K_p \) for the reaction:

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

we will follow these steps:

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined in terms of the concentrations of the reactants and products at equilibrium. For the given reaction, the expression for \( K_c \) is:

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

### Step 2: Determine the units for concentration

The unit of concentration is typically expressed in moles per liter (mol/L). Therefore, we can write:

- \([NH_3]\) has units of mol/L
- \([N_2]\) has units of mol/L
- \([H_2]\) has units of mol/L

### Step 3: Substitute the units into the \( K_c \) expression

Substituting the units into the \( K_c \) expression gives:

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

### Step 4: Simplify the units for \( K_c \)

Now, simplify the expression:

\[ K_c = \frac{(mol)^2}{(mol)^4} \cdot \frac{1}{(L)^2} = mol^{-2} \cdot L^2 \]

Thus, the units for \( K_c \) are:

\[ K_c \text{ units} = mol^{-2} \cdot L^2 \]

### Step 5: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the reactants and products. The expression for \( K_p \) for the same reaction is:

\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

### Step 6: Determine the units for pressure

Assuming we are using bar as the unit of pressure, we have:

- \( P_{NH_3} \) has units of bar
- \( P_{N_2} \) has units of bar
- \( P_{H_2} \) has units of bar

### Step 7: Substitute the units into the \( K_p \) expression

Substituting the units into the \( K_p \) expression gives:

\[ K_p = \frac{(bar)^2}{(bar)(bar)^3} = \frac{(bar)^2}{(bar)^4} \]

### Step 8: Simplify the units for \( K_p \)

Now, simplify the expression:

\[ K_p = \frac{(bar)^2}{(bar)^4} = bar^{-2} \]

Thus, the units for \( K_p \) are:

\[ K_p \text{ units} = bar^{-2} \]

### Final Answer

The units of \( K_c \) and \( K_p \) respectively are:

\[ K_c: mol^{-2} \cdot L^2 \quad \text{and} \quad K_p: bar^{-2} \]