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 find the \( K_p \) for the reaction

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

given that \( K_c = 4 \) for the reaction

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

at \( 527^\circ C \), we can follow these steps:

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

The 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 \( K_c \) for the desired reaction, we first reverse the given reaction. The reversed reaction is:

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

### Step 3: Calculate \( K_c \) for the reversed reaction

The equilibrium constant for the reversed reaction is the reciprocal of the original \( K_c \):

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

### Step 4: Multiply the reversed reaction by 2

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

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

### Step 5: Calculate \( K_c \) for the new reaction

When we multiply a reaction by a factor, the equilibrium constant is raised to the power of that factor. Therefore, for our new reaction:

\[ K_c'' = (K_c')^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 6: Use the relation between \( K_p \) and \( K_c \)

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

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

where \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

### Step 7: 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 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4

Thus,

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

### Step 8: Convert temperature to Kelvin

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

\[ T(K) = 527 + 273 = 800 \, K \]

### Step 9: Substitute values into the \( K_p \) equation

Now we can substitute \( K_c \), \( R \), and \( T \) into the equation for \( K_p \):

\[ K_p = K_c''(RT)^{\Delta n} = \frac{1}{16} \cdot (R \cdot 800)^{-2} \]

### Step 10: Final expression for \( K_p \)

Thus, we have:

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

This is the expression for \( K_p \) for the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \).

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