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 \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \), we will follow these steps:

### Step 1: Understand the given reaction and equilibrium constant
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 given as 4 at 800 K.

### Step 2: Reverse the given reaction
To find \( K_c \) for the reaction we are interested in, we first need to reverse the original reaction. The reversed reaction will be:
\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons 4NH_3(g) \]
When we reverse a reaction, the equilibrium constant becomes the reciprocal:
\[ K_c' = \frac{1}{K_c} = \frac{1}{4} \]

### Step 3: Adjust the coefficients
Next, we need to multiply the entire reversed reaction by 2 to match the coefficients of the desired reaction:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
When we multiply the coefficients of a reaction by a factor, the equilibrium constant is raised to the power of that factor:
\[ K_c'' = (K_c')^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 4: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot R^{\Delta N_g} \cdot T \]
Where:
- \( R \) is the ideal 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, calculated as the moles of products minus the moles of reactants.

### Step 5: Calculate \( \Delta N_g \)
For the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \):
- Moles of products = 2 (from \( 2NH_3 \))
- Moles of reactants = 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4
Thus,
\[ \Delta N_g = 2 - 4 = -2 \]

### Step 6: Substitute the values into the equation for \( K_p \)
Now we can substitute the values into the equation:
\[ K_p = K_c'' \cdot R^{\Delta N_g} \cdot T = \frac{1}{16} \cdot (0.0821)^{-2} \cdot 800 \]

### Step 7: Calculate \( K_p \)
Calculating \( R^{-2} \):
\[ R^{-2} = \left(0.0821\right)^{-2} \approx 148.76 \]
Now substituting this back:
\[ K_p = \frac{1}{16} \cdot 148.76 \cdot 800 \]
Calculating this gives:
\[ K_p \approx \frac{1}{16} \cdot 119008 \approx 7438 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) is approximately 7438.

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