For the gaseous reaction
`C_2H_4 + H_2 The equilibrium constant, has the units

Options

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

Answer

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

\[ C_2H_4 + H_2 \implies C_2H_6 \]

we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (K)
The equilibrium constant (K) for a reaction is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced equation.

For the given reaction:

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

### Step 2: Identify the coefficients
In this reaction:
- The coefficient of \(C_2H_6\) (product) is 1.
- The coefficient of \(C_2H_4\) (reactant) is 1.
- The coefficient of \(H_2\) (reactant) is also 1.

### Step 3: Substitute the concentrations
The concentrations are expressed in moles per liter (mol/L). Therefore, we can substitute the concentrations into the equilibrium expression:

\[ K = \frac{(mol/L)^1}{(mol/L)^1 \times (mol/L)^1} \]

### Step 4: Simplify the expression
Now, simplify the expression:

\[ K = \frac{mol/L}{(mol/L) \times (mol/L)} = \frac{mol/L}{(mol^2/L^2)} \]

This simplifies to:

\[ K = \frac{mol}{L} \times \frac{L^2}{mol^2} = \frac{L}{mol} \]

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

\[ K = \frac{L}{mol} \]

### Conclusion
The final answer is that the units of the equilibrium constant \(K\) for the reaction \(C_2H_4 + H_2 \implies C_2H_6\) are \(L \cdot mol^{-1}\).

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