In the gaseous phase reaction
`C_(2)H_(4) +H_(2) hArr C_(2)H_(6),` the equilibrium constant can be expressed in the units to :

Options

A. `"litre"^(-1) " mol"^(-1)`
B. `"mol"^(2) " litre" ^(-2)`
C. `"litre " "mol"^(-1)`
D. `"mol " " litre"`

Explanation

To find the units of the equilibrium constant for the reaction:

\[ C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g) \]

we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (Kc)
The equilibrium constant \( K_c \) for a reaction is defined in terms of the concentrations of the products and reactants at equilibrium. For the given reaction, the expression is:

\[ K_c = \frac{[C_2H_6]}{[C_2H_4][H_2]} \]

### Step 2: Identify the concentrations
In the expression, \([C_2H_6]\), \([C_2H_4]\), and \([H_2]\) represent the molar concentrations (in moles per liter, mol/L) of the respective gases at equilibrium.

### Step 3: Substitute the units into the expression
Now, we can substitute the units into the equilibrium constant expression:

\[ K_c = \frac{[\text{C}_2\text{H}_6]}{[\text{C}_2\text{H}_4][\text{H}_2]} = \frac{\text{mol/L}}{(\text{mol/L})(\text{mol/L})} \]

### Step 4: Simplify the units
This simplifies to:

\[ K_c = \frac{\text{mol/L}}{\text{mol}^2/\text{L}^2} = \frac{\text{mol/L}}{\text{mol}^2} \cdot \text{L}^2 = \frac{\text{L}}{\text{mol}} \]

### Step 5: Final units of the equilibrium constant
Thus, the units of the equilibrium constant \( K_c \) for this reaction are:

\[ K_c = \text{L mol}^{-1} \]

### Summary of the solution:
The units of the equilibrium constant \( K_c \) for the reaction \( C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g) \) are \( \text{L mol}^{-1} \).

---