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) > K_(c)`
C. `K_(p) = K_(c)`
D. Cannot be predicted

Explanation

To solve the problem regarding 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: Write the expression for \( K_p \) and \( K_c \)

The equilibrium constants \( K_p \) and \( K_c \) are related by the equation:

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

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.

### Step 2: Calculate \( \Delta n_g \)

To find \( \Delta n_g \), we need to determine the number of moles of gaseous products and reactants.

- **Products**: \( C_2H_6(g) \) has 1 mole.
- **Reactants**: \( C_2H_4(g) + H_2(g) \) has 2 moles.

Thus,

\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - 2 = -1 \]

### Step 3: Substitute \( \Delta n_g \) into the equation

Now we can substitute \( \Delta n_g \) into the equation for \( K_p \):

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

This can be rearranged to:

\[ K_p = \frac{K_c}{RT} \]

### Step 4: Analyze the relationship between \( K_p \) and \( K_c \)

Since \( R \) and \( T \) are both positive constants, it follows that:

\[ K_p < K_c \]

This means that \( K_p \) is less than \( K_c \).

### Conclusion

The correct expression between \( K_p \) and \( K_c \) for the given reaction at \( 27^\circ C \) is:

\[ K_p < K_c \]

### Final Answer

Thus, the answer is: **\( K_p < K_c \)**.

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