In the reaction `H_(2(g)) +l_(2(g)) Leftrightarrow 2HI_((g))`

Options

A. `K_(p)=K_(c)`
B. `K_(p) ne K_(c)`
C. `K_(p) gt K_(c)`
D. `K_(p) lt K_(c)`

Answer

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

\[ H_2(g) + I_2(g) \leftrightarrow 2 HI(g) \]

we will follow these steps:

### Step 1: Identify the reaction and its components
The given reaction involves hydrogen gas (\( H_2 \)), iodine gas (\( I_2 \)), and hydrogen iodide gas (\( HI \)).

### Step 2: Write the expression for \( K_c \) and \( K_p \)
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

The equilibrium constant \( K_p \) is given by:

\[ K_p = \frac{P_{HI}^2}{P_{H_2} P_{I_2}} \]

where \( P \) represents the partial pressures of the gases.

### Step 3: Determine \( \Delta N_g \)
To relate \( K_p \) and \( K_c \), we need to calculate \( \Delta N_g \), which is defined as:

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

For the given reaction:
- Number of moles of gaseous products (from \( 2 HI \)) = 2
- Number of moles of gaseous reactants (from \( H_2 + I_2 \)) = 1 + 1 = 2

Thus,

\[ \Delta N_g = 2 - 2 = 0 \]

### Step 4: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

Substituting \( \Delta N_g = 0 \):

\[ K_p = K_c (RT)^0 \]

Since any number raised to the power of 0 is 1:

\[ K_p = K_c \cdot 1 \]

Thus, we conclude that:

\[ K_p = K_c \]

### Step 5: Final answer
The relationship between \( K_p \) and \( K_c \) for the reaction \( H_2(g) + I_2(g) \leftrightarrow 2 HI(g) \) is:

\[ K_p = K_c \]