For the reaction `C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g)` , which of the following expressions between `K_(p) "and" K_(c)` is true at `27^(@)C` ?

Options

A. `K_(p)ltK_(c)`
B. `K_(p)ltK_(c)`
C. `K_(p) = K_(c)`
D. Cannot be predicted

Explanation

To determine the relationship between \( K_p \) and \( K_c \) for the reaction

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

at \( 27^\circ C \), we will follow these steps:

### Step 1: Identify the reaction and equilibrium constants
The reaction involves the conversion of ethylene (\( C_2H_4 \)) and hydrogen (\( H_2 \)) gas into ethane (\( C_2H_6 \)). The equilibrium constants are defined as:
- \( K_p \): Equilibrium constant in terms of partial pressures.
- \( K_c \): Equilibrium constant in terms of concentrations.

### Step 2: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ K_p = K_c (RT)^{\Delta n_g} \]

where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas, calculated as:

\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

### Step 3: Calculate \( \Delta n_g \)
For the given reaction:
- Moles of gaseous products: \( 1 \) (from \( C_2H_6 \))
- Moles of gaseous reactants: \( 2 \) (from \( C_2H_4 \) and \( H_2 \))

Thus,

\[ \Delta n_g = 1 - 2 = -1 \]

### Step 4: Substitute \( \Delta n_g \) into the equation
Now substituting \( \Delta n_g \) into the relationship:

\[ K_p = K_c (RT)^{-1} \]

This can be rearranged to:

\[ K_c = K_p (RT) \]

### Step 5: Analyze the temperature factor
At \( 27^\circ C \) (which is \( 300 \, K \)), \( RT \) is always greater than 1 (since \( R \) is a positive constant and \( T \) is positive). Therefore, we can conclude that:

\[ K_c > K_p \]

### Conclusion
The correct relationship is:

\[ K_p < K_c \]