Calculate the expression for `K_(C) "and" K_(P)` if initially a moles of `N_(2) "and" B "moles of" H_(2)` is taken for the following reaction. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) (Deltanlt0)` (P,T,V given)

Explanation

To solve the problem, we need to calculate the equilibrium constants \( K_c \) and \( K_p \) for the reaction:

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

Given that initially there are \( a \) moles of \( N_2 \) and \( b \) moles of \( H_2 \).

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:

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

### Step 2: Determine the change in moles
From the balanced equation, we can see that:

- 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \) to produce 2 moles of \( NH_3 \).
- If we start with \( a \) moles of \( N_2 \) and \( b \) moles of \( H_2 \), the reaction will proceed until one of the reactants is completely consumed.

### Step 3: Set up the expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}} \]

For our reaction, this becomes:

\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 4: Calculate the concentrations at equilibrium
Let \( x \) be the change in moles of \( N_2 \) that react at equilibrium. Therefore:

- Moles of \( N_2 \) at equilibrium = \( a - x \)
- Moles of \( H_2 \) at equilibrium = \( b - 3x \)
- Moles of \( NH_3 \) at equilibrium = \( 2x \)

The equilibrium concentrations can be expressed as:

\[ [N_2] = \frac{a - x}{V}, \quad [H_2] = \frac{b - 3x}{V}, \quad [NH_3] = \frac{2x}{V} \]

### Step 5: Substitute into the \( K_c \) expression
Substituting the equilibrium concentrations into the \( K_c \) expression:

\[ K_c = \frac{\left(\frac{2x}{V}\right)^2}{\left(\frac{a - x}{V}\right)\left(\frac{b - 3x}{V}\right)^3} \]

This simplifies to:

\[ K_c = \frac{4x^2}{(a - x)(b - 3x)^3} \cdot V^2 \]

### Step 6: Calculate \( K_p \)
The relationship between \( K_p \) and \( K_c \) is given by:

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

Where \( \Delta n \) is the change in moles of gas, calculated as:

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - (1 + 3) = 2 - 4 = -2 \]

Thus, \( K_p \) can be expressed as:

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

### Final Expressions
1. \( K_c = \frac{4x^2 \cdot V^2}{(a - x)(b - 3x)^3} \)
2. \( K_p = K_c \cdot \frac{1}{(RT)^2} \)