Equilibrium constant `K_(C)` for the following reaction at 800 K is, 4 `NH_(3)(g)hArr (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`.
The value of `K_(p)` for the following reaction will be
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`

Options

A. `((800R)/(4))^(2-)`
B. `16xx(800R)^(2)`
C. `[(1)/(4xx800R)]^(2)`
D. `(800R)^(1//2)4`

Answer

To find the value of \( K_p \) for the reaction:

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

given that the equilibrium constant \( K_c \) for the reaction:

\[ 4NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \]

is 4 at 800 K, we will follow these steps:

### Step 1: Write the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

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

### Step 2: Determine \( \Delta n \)

For the reaction \( 4NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \):
- Moles of products = \( \frac{1}{2} + \frac{3}{2} = 2 \)
- Moles of reactants = 4

Thus,

\[ \Delta n = 2 - 4 = -2 \]

### Step 3: Substitute \( K_c \), \( R \), \( T \), and \( \Delta n \) into the equation

Given \( K_c = 4 \), \( R = 0.0821 \, \text{L·atm/(K·mol)} \), and \( T = 800 \, \text{K} \):

\[ K_p = K_c (RT)^{\Delta n} = 4 \cdot (0.0821 \cdot 800)^{-2} \]

### Step 4: Calculate \( RT \)

First, calculate \( RT \):

\[ RT = 0.0821 \cdot 800 = 65.68 \, \text{L·atm/mol} \]

### Step 5: Calculate \( (RT)^{-2} \)

Now calculate \( (RT)^{-2} \):

\[ (RT)^{-2} = \frac{1}{(65.68)^2} \approx \frac{1}{4303.51} \approx 0.000232 \]

### Step 6: Calculate \( K_p \)

Now substitute back to find \( K_p \):

\[ K_p = 4 \cdot 0.000232 \approx 0.000928 \]

### Final Answer

Thus, the value of \( K_p \) for the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) at 800 K is approximately:

\[ K_p \approx 0.000928 \]