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. `mol^(2)dm^(-3)`
B. `dm^(3)mol^(-1)`
C. `dm^(3)mol^(-1)`
D. `mol.dm^(-3)`

Answer

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

\[ C_2H_4 + H_2 \rightleftharpoons C_2H_6, \]

we can 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 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 reaction:

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

### Step 2: Identify the stoichiometric coefficients

In this reaction, the stoichiometric coefficients for all species are 1. Therefore, we can simplify the expression:

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

### Step 3: Determine the units of concentration

The concentration of each species is typically expressed in moles per cubic decimeter (mol/dm³). Therefore, we can write:

- \([C_2H_6]\) has units of mol/dm³
- \([C_2H_4]\) has units of mol/dm³
- \([H_2]\) has units of mol/dm³

### Step 4: Substitute the units into the \( K_c \) expression

Now substituting the units into the \( K_c \) expression:

\[ K_c = \frac{[C_2H_6]}{[C_2H_4] \cdot [H_2]} = \frac{\text{mol/dm}^3}{(\text{mol/dm}^3) \cdot (\text{mol/dm}^3)} \]

### Step 5: Simplify the units

This simplifies to:

\[ K_c = \frac{\text{mol/dm}^3}{\text{mol}^2/\text{dm}^6} = \frac{\text{mol/dm}^3 \cdot \text{dm}^6}{\text{mol}^2} = \frac{\text{dm}^3}{\text{mol}} \]

### Final Result

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

\[ K_c = \text{dm}^3 \cdot \text{mol}^{-1} \]