`K_p` is how many times equal to `K_c` for the given reaction ?
`N_2(g) +3H_2 (g) hArr 2NH_3(g)`

Options

A. `1/(R^2T^2)`
B. `R^2T^2`
C. `R/T`
D. `RT`

Answer

To find how many times \( K_p \) is equal to \( K_c \) for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

we can 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 gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas, calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants.

### Step 2: Calculate \( \Delta n_g \)
For the given reaction:

- Moles of gaseous products (NH3): 2 moles
- Moles of gaseous reactants (N2 + 3H2): \( 1 + 3 = 4 \) moles

Now, calculate \( \Delta n_g \):

\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 4 = -2 \]

### Step 3: Substitute \( \Delta n_g \) into the equation
Now substitute \( \Delta n_g \) into the equation for \( K_p \):

\[ K_p = K_c (RT)^{-2} \]

This can also be expressed as:

\[ K_p = \frac{K_c}{(RT)^2} \]

### Step 4: Rearrange to find the relationship between \( K_p \) and \( K_c \)
From the above equation, we can rearrange it to express \( K_c \) in terms of \( K_p \):

\[ K_c = K_p (RT)^2 \]

### Step 5: Conclusion
Thus, we find that:

\[ K_p = K_c \cdot \frac{1}{(RT)^2} \]

This means that \( K_p \) is equal to \( K_c \) divided by \( (RT)^2 \). Therefore, \( K_p \) is \( \frac{1}{(RT)^2} \) times \( K_c \).

### Final Answer
So, \( K_p \) is \( \frac{1}{(RT)^2} \) times equal to \( K_c \).

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