Consider the imaginary equilibrium `4A(g)+5B(g)hArr 4X(g)+6Y(g)`, The unit of equilibrium constant `K_(C)` is

Options

A. `"Mole"^(2) "litre"^(-2)`
B. `"Litre mole"^(-1)`
C. `"Mole litre"^(-1)`
D. `"Litre"^(2)" mole"^(-2)`

Answer

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

\[ 4A(g) + 5B(g) \rightleftharpoons 4X(g) + 6Y(g) \]

we will follow these steps:

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

The equilibrium constant \( K_c \) is defined as the ratio of the concentrations of the products raised to the power of their coefficients to the concentrations of the reactants raised to the power of their coefficients. For the given reaction, the expression for \( K_c \) is:

\[ K_c = \frac{[X]^4[Y]^6}{[A]^4[B]^5} \]

### Step 2: Identify the units of concentration

Concentration is typically expressed in moles per liter (mol/L). Therefore, the units for the concentrations of the species in the reaction are:

- \([X]\) has units of mol/L
- \([Y]\) has units of mol/L
- \([A]\) has units of mol/L
- \([B]\) has units of mol/L

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

Now, substituting the units of concentration into the \( K_c \) expression:

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

### Step 4: Simplify the expression

Now we simplify the expression:

\[ K_c = \frac{(mol^4/L^4) \cdot (mol^6/L^6)}{(mol^4/L^4) \cdot (mol^5/L^5)} = \frac{mol^{4+6}/L^{4+6}}{mol^{4+5}/L^{4+5}} = \frac{mol^{10}/L^{10}}{mol^{9}/L^{9}} \]

This simplifies to:

\[ K_c = \frac{mol^{10-9}}{L^{10-9}} = \frac{mol^1}{L^1} = \frac{mol}{L} \]

### Conclusion

Thus, the unit of the equilibrium constant \( K_c \) for the given reaction is:

\[ K_c = \text{mol/L} \]