For homogeneous gas reaction `4NH_(3) + 5O_(2)hArr4NO + 6H_(2)O`. The equilibrium constant `K_(c)` has the unit of

Options

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

Answer

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

**Step 1: Write the balanced chemical equation.**
The reaction is:
\[ 4NH_3 + 5O_2 \rightleftharpoons 4NO + 6H_2O \]

**Step 2: Write the expression for the equilibrium constant \( K_c \).**
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[NO]^4 \cdot [H_2O]^6}{[NH_3]^4 \cdot [O_2]^5} \]
where \([X]\) denotes the molar concentration of species \( X \).

**Step 3: Identify the units of concentration.**
The concentration of each species is measured in moles per liter (mol/L), which can also be represented as M (molarity).

**Step 4: Substitute the units into the \( K_c \) expression.**
Substituting the units into the expression for \( K_c \):
\[ K_c = \frac{(M)^4 \cdot (M)^6}{(M)^4 \cdot (M)^5} \]

**Step 5: Simplify the expression.**
Now simplify the expression:
\[ K_c = \frac{M^{4+6}}{M^{4+5}} = \frac{M^{10}}{M^{9}} = M^{10-9} = M^1 \]

**Step 6: Conclusion.**
Thus, the unit of the equilibrium constant \( K_c \) is:
\[ \text{Unit of } K_c = M \text{ (molarity)} \]

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