Given
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K_(1)`
`N_(2)(g)+O_(2)(g)hArr2NO(g),K_(2)`
`H_(2)(g)+(1)/(2)O_(2)hArrH_(2)O(g),K_(3)`
The equilibrium constant for
`2NH_(3)(g)+(5)/(2)O_(2)(g)hArr2NO(g)+3H_(2)O(g)`
will be

Options

A. `(K_(1)K_(3)^(3))/(K_(2))`
B. `(K_(2)K_(3)^(3))/(K_(1))`
C. `(K_(2)K_(3))/(K_(1))`
D. `(K_(2)^(3)K_(3))/(K_(1))`

Answer

To find the equilibrium constant for the reaction:

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

we will use the given reactions and their equilibrium constants. Let's break it down step by step.

### Step 1: Write the given reactions and 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 \( K_3 \)

### Step 2: Reverse the first reaction

To obtain \( 2NH_3 \) on the reactant side, we need to reverse the first reaction:

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

The equilibrium constant for the reversed reaction is:

\[ K_1' = \frac{1}{K_1} \]

### Step 3: Multiply the third reaction by 3

We need \( 3H_2O \) in the products, so we multiply the third reaction by 3:

\[ 3H_2(g) + \frac{3}{2}O_2(g) \rightleftharpoons 3H_2O(g) \]

The equilibrium constant for this modified reaction is:

\[ K_3' = K_3^3 \]

### Step 4: Use the second reaction as it is

The second reaction is already in the desired form for \( 2NO \):

\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

The equilibrium constant remains:

\[ K_2' = K_2 \]

### Step 5: Combine the reactions

Now, we will add the modified reactions together:

1. \( 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \) (with \( K_1' = \frac{1}{K_1} \))
2. \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \) (with \( K_2' = K_2 \))
3. \( 3H_2(g) + \frac{3}{2}O_2(g) \rightleftharpoons 3H_2O(g) \) (with \( K_3' = K_3^3 \))

When we add these reactions, we get:

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

### Step 6: Calculate the overall equilibrium constant

The overall equilibrium constant \( K \) for the final reaction is the product of the equilibrium constants of the individual reactions:

\[ K = K_1' \cdot K_2' \cdot K_3' = \left(\frac{1}{K_1}\right) \cdot K_2 \cdot K_3^3 \]

Thus, the equilibrium constant for the reaction:

\[ K = \frac{K_2 \cdot K_3^3}{K_1} \]

### Final Answer

The equilibrium constant for the reaction \( 2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g) \) is:

\[ K = \frac{K_2 \cdot K_3^3}{K_1} \]