At `527^(@)C`, the reaction given below has`K_(c)=4`
`NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`
what is the `K_(p)` for the reaction ?
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`

Options

A. `16xx(800R)^(2)`
B. `((800R)/(4))^(-2)`
C. `((1)/(4xx800R))^(2)`
D. none of these

Answer

To solve the problem, we need to find the \( K_p \) for the reaction:

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

given that at \( 527^\circ C \), the equilibrium constant \( K_c \) for the reaction:

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

is \( K_c = 4 \).

### Step-by-Step Solution:

**Step 1: Write the given reaction and its \( K_c \)**

The given reaction is:

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

with \( K_c = 4 \).

**Step 2: Reverse the reaction**

To find the \( K_c \) for the desired reaction, we need to reverse the given reaction. When we reverse a reaction, the equilibrium constant becomes the reciprocal:

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

Thus,

\[ K_c' = \frac{1}{K_c} = \frac{1}{4} \]

**Step 3: Multiply the reversed reaction by 2**

Now, we need to multiply the entire reversed reaction by 2 to match the desired reaction:

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

When we multiply the 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: Relate \( K_c \) to \( K_p \)**

To find \( K_p \), we use the relation between \( K_c \) and \( K_p \):

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

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

**Step 5: Calculate \( \Delta n \)**

For the reaction:

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

- Moles of products = 2 (from \( 2NH_3 \))
- Moles of reactants = 1 + 3 = 4 (from \( N_2 + 3H_2 \))

Thus,

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

**Step 6: Convert temperature to Kelvin**

Convert \( 527^\circ C \) to Kelvin:

\[ T = 527 + 273.15 = 800.15 \approx 800 \text{ K} \]

**Step 7: Substitute values into the equation for \( K_p \)**

Now we can substitute the values into the equation:

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

Substituting the values:

\[ K_p = \frac{1}{16} \cdot (0.0821 \cdot 800)^{-2} \]

Calculating \( RT \):

\[ RT = 0.0821 \cdot 800 = 65.68 \]

Now substituting back:

\[ K_p = \frac{1}{16} \cdot (65.68)^{-2} \]

Calculating \( (65.68)^{-2} \):

\[ (65.68)^{-2} \approx 0.000230 \]

Finally,

\[ K_p \approx \frac{1}{16} \cdot 0.000230 \approx 0.000014375 \]

### Final Result

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

\[ K_p \approx 0.000014375 \]