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

Explanation

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:

1. \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), \quad K_1 \)
2. \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g), \quad K_2 \)
3. \( H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g), \quad K_3 \)

### Step 1: Reverse the first equation
Since we need \( 2NH_3 \) on the reactant side, we reverse the first equation:

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

The equilibrium constant for the reversed reaction is:

\[ K_4 = \frac{1}{K_1} \]

### Step 2: Use the second equation as is
The second equation is already in the required form:

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

The equilibrium constant remains:

\[ K_2 \]

### Step 3: Multiply the third equation by 3
We need \( 3H_2O \) on the product side, so we multiply the third equation by 3:

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

The equilibrium constant for this modified reaction is:

\[ K_5 = K_3^3 \]

### Step 4: Combine the equations
Now we combine the modified equations:

1. From the reversed first equation:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \quad (K_4 = \frac{1}{K_1}) \]

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

3. From the modified third equation:
\[ 3H_2(g) + \frac{3}{2}O_2(g) \rightleftharpoons 3H_2O(g) \quad (K_5 = K_3^3) \]

### Step 5: Add the equations
When we add the equations, 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 combined reaction is given by:

\[ K = K_4 \cdot K_2 \cdot K_5 = \left(\frac{1}{K_1}\right) \cdot K_2 \cdot (K_3^3) \]

Thus, we have:

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

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

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

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