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

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 manipulate the given reactions and their equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \).

### Step 1: Reverse the first reaction

The first reaction is:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad (K_1) \]

To use \( NH_3 \) on the left side, we reverse this equation:

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

The equilibrium constant for the reversed reaction is:

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

### Step 2: Use the second reaction as is

The second reaction is:

\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad (K_2) \]

We will keep this reaction as it is.

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

The third reaction is:

\[ H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g) \quad (K_3) \]

Since we need \( 3H_2O \), we multiply this entire reaction by 3:

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

The equilibrium constant for this modified reaction becomes:

\[ K'' = K_3^3 \]

### Step 4: Combine the reactions

Now we will add the modified reactions together:

1. From Step 1:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \quad (K' = \frac{1}{K_1}) \]

2. From Step 2:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad (K_2) \]

3. From Step 3:
\[ 3H_2(g) + \frac{3}{2}O_2(g) \rightleftharpoons 3H_2O(g) \quad (K'' = K_3^3) \]

Now, we add these reactions:

- The \( N_2(g) \) cancels out.
- The \( 3H_2(g) \) from the third reaction cancels with \( 3H_2(g) \) from the first reaction.
- We combine the \( O_2 \) terms: \( O_2(g) + \frac{3}{2}O_2(g) = \frac{5}{2}O_2(g) \).

The resulting reaction is:

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

### Step 5: Write the equilibrium constant for the final reaction

The equilibrium constant for the final reaction is given by the product of the constants of the individual reactions:

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

Thus, we can write:

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

### Final Answer

The equilibrium constant for the reaction

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

is

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