The relation between `K_p` and `K_x` is

Options

A. `K_p=K_x((P)/(sumn))`
B. `K_p=K_x(P)^(-Deltan)`
C. `K_p=K_x(P)^(Deltan)`
D. `K_p=K_x(RT)^(Deltan)`

Explanation

To find the relation between \( K_p \) and \( K_x \), we can follow these steps:

### Step 1: Write the balanced chemical equation
Consider a general reaction:
\[ aA + bB \rightleftharpoons cC + dD \]
where \( a, b, c, \) and \( d \) are the stoichiometric coefficients of the reactants and products.

### Step 2: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) in terms of partial pressures is given by:
\[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \]
where \( P_A, P_B, P_C, \) and \( P_D \) are the partial pressures of the respective gases.

### Step 3: Relate partial pressures to mole fractions
The partial pressure of a component can be expressed in terms of its mole fraction and the total pressure \( P \):
\[ P_i = y_i \cdot P \]
where \( y_i \) is the mole fraction of component \( i \).

### Step 4: Substitute partial pressures in \( K_p \)
Substituting the expressions for partial pressures into the \( K_p \) expression gives:
\[ K_p = \frac{(y_C P)^c (y_D P)^d}{(y_A P)^a (y_B P)^b} \]
This simplifies to:
\[ K_p = \frac{y_C^c y_D^d P^{c+d}}{y_A^a y_B^b P^{a+b}} \]

### Step 5: Factor out the total pressure
Rearranging the equation, we can factor out \( P^{c+d-a-b} \):
\[ K_p = \frac{y_C^c y_D^d}{y_A^a y_B^b} \cdot P^{(c+d)-(a+b)} \]
Here, \( (c+d)-(a+b) \) is the change in the number of moles, denoted as \( \Delta n \).

### Step 6: Define \( K_x \)
The equilibrium constant \( K_x \) in terms of mole fractions is defined as:
\[ K_x = \frac{y_C^c y_D^d}{y_A^a y_B^b} \]

### Step 7: Relate \( K_p \) and \( K_x \)
Now we can express \( K_p \) in terms of \( K_x \):
\[ K_p = K_x \cdot P^{\Delta n} \]
where \( \Delta n = (c + d) - (a + b) \).

### Conclusion
Thus, the relation between \( K_p \) and \( K_x \) is:
\[ K_p = K_x \cdot P^{\Delta n} \]