(1) $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$, $$K_1$$
(2) $$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$$, $$K_2$$
(3) $$H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g)$$, $$K_3$$
The equation for the equilibrium constant of the reaction $$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$$, $$(K_4)$$ in terms of $$K_1$$, $$K_2$$ and $$K_3$$ is :

We are given three reactions with their equilibrium constants:

(1) $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$, with equilibrium constant $$K_1$$

(2) $$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$$, with equilibrium constant $$K_2$$

(3) $$H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g)$$, with equilibrium constant $$K_3$$

We need to find the equilibrium constant $$K_4$$ for the reaction:

$$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$$

in terms of $$K_1$$, $$K_2$$, and $$K_3$$.

Recall that for a general reaction $$aA + bB \rightleftharpoons cC + dD$$, the equilibrium constant is $$K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$. Also, when manipulating reactions:

• Reversing a reaction gives $$K_{\text{new}} = \frac{1}{K_{\text{original}}}$$.
• Multiplying a reaction by a factor $$n$$ gives $$K_{\text{new}} = (K_{\text{original}})^n$$.
• Adding reactions gives $$K_{\text{new}} = K_1 \times K_2 \times \cdots$$ for the combined reaction.

First, write the expressions for the given equilibrium constants:

For reaction (1): $$K_1 = \frac{[NH_3]^2}{[N_2] [H_2]^3}$$

For reaction (2): $$K_2 = \frac{[NO]^2}{[N_2] [O_2]}$$

For reaction (3): $$K_3 = \frac{[H_2O]}{[H_2] [O_2]^{1/2}}$$

For the target reaction: $$K_4 = \frac{[NO]^2 [H_2O]^3}{[NH_3]^2 [O_2]^{5/2}}$$

To obtain the target reaction, we manipulate the given reactions:

Step 1: Reverse reaction (1) to get $$NH_3$$ as a reactant:

$$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$$, with equilibrium constant $$\frac{1}{K_1}$$.

Step 2: Use reaction (2) as is, since it produces $$NO$$:

$$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$$, with equilibrium constant $$K_2$$.

Step 3: Multiply reaction (3) by 3 to get three $$H_2O$$ molecules:

$$3H_2(g) + \frac{3}{2}O_2(g) \rightleftharpoons 3H_2O(g)$$, with equilibrium constant $$(K_3)^3 = K_3^3$$.

Now, add these three reactions together:

Reversed (1): $$2NH_3 \rightleftharpoons N_2 + 3H_2$$

Reaction (2): $$N_2 + O_2 \rightleftharpoons 2NO$$

Multiplied (3): $$3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O$$

When adding, the intermediates $$N_2$$ and $$3H_2$$ cancel out:

• $$N_2$$ is produced in reversed (1) and consumed in reaction (2).
• $$3H_2$$ is produced in reversed (1) and consumed in multiplied (3).

The net reaction is:

Left side: $$2NH_3 + O_2 + \frac{3}{2}O_2 = 2NH_3 + \frac{5}{2}O_2$$

Right side: $$2NO + 3H_2O$$

So, the net reaction is:

$$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$$

which is exactly the target reaction.

Since we added the reactions, the equilibrium constant for the net reaction is the product of the constants for the steps:

$$K_4 = \left( \frac{1}{K_1} \right) \times K_2 \times K_3^3 = \frac{K_2 \cdot K_3^3}{K_1}$$

Comparing with the options:

A. $$\frac{K_1 \cdot K_2}{K_3}$$

B. $$\frac{K_1 \cdot K_2^2}{K_2}$$ which simplifies to $$K_1 K_2$$

C. $$K_1 K_2 K_3$$

D. $$\frac{K_2 \cdot K_3^3}{K_1}$$

Option D matches our derived expression.

Hence, the correct answer is Option D.