For the homogeneous reaction: `4NH_3+5O_2 hArr 4NO+6H_2O`
The equilibrium constant `K_c` has the units of :

Options

A. `(Conc.)^(-10)`
B. `(Conc.)^(1)`
C. `(Conc.)^(-1)`
D. It is dimensionless

Answer

To determine the units of the equilibrium constant \( K_c \) for the reaction:

\[ 4NH_3 + 5O_2 \rightleftharpoons 4NO + 6H_2O \]

we follow these steps:

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

The equilibrium constant \( K_c \) for a reaction is given by the formula:

\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]

For our reaction, the products are \( NO \) and \( H_2O \), and the reactants are \( NH_3 \) and \( O_2 \).

### Step 2: Substitute the concentrations into the expression

The equilibrium expression for the given reaction is:

\[ K_c = \frac{[NO]^4 \cdot [H_2O]^6}{[NH_3]^4 \cdot [O_2]^5} \]

### Step 3: Determine the units of each concentration

The concentration is typically expressed in moles per liter (mol/L). Therefore, the units for each species in the expression are:

- \([NO]\): mol/L
- \([H_2O]\): mol/L
- \([NH_3]\): mol/L
- \([O_2]\): mol/L

### Step 4: Substitute the units into the equilibrium expression

Now we substitute the units into the \( K_c \) expression:

\[ K_c = \frac{(mol/L)^4 \cdot (mol/L)^6}{(mol/L)^4 \cdot (mol/L)^5} \]

### Step 5: Simplify the expression

Now we simplify the expression:

\[ K_c = \frac{(mol^4/L^4) \cdot (mol^6/L^6)}{(mol^4/L^4) \cdot (mol^5/L^5)} \]

This simplifies to:

\[ K_c = \frac{mol^{4+6}/L^{4+6}}{mol^{4+5}/L^{4+5}} = \frac{mol^{10}/L^{10}}{mol^{9}/L^{9}} = \frac{mol^{10-9}}{L^{10-9}} = \frac{mol^1}{L^1} = \frac{mol}{L} \]

### Step 6: Final units of \( K_c \)

Thus, the units of the equilibrium constant \( K_c \) for this reaction are:

\[ K_c = \text{mol/L} \quad \text{or} \quad \text{M} \]