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Kelvi

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The equilibrium constants of the following are $\\ \begin{array}{c} N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1 \\ N_2 + O_2 \rightleftharpoons 2NO \quad K_2 \\ H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3 \end{array} \\$ The equilibrium constant $(K)$ of the reaction $\\ 2NH_3 + \frac{5}{2} O_2 \xrightleftharpoons{K} 2NO + 3H_2O$ will be

hard

Equilibrium

2017

chemistry

$\frac{K_2 K_3^3}{K_1}$

Explanation

To find the equilibrium constant $K$ for the reaction $2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O,$

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$\frac{K_2 K_3}{K_1}$

$\frac{K_2^2 K_3}{K_1}$

$\frac{K_1 K_3^3}{K_2}$

we need to relate it to the given reactions:

\\

1.

N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1 \\

2.

N_2 + O_2 \rightleftharpoons 2NO \quad K_2 \\

3.

H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3 \\

We need to manipulate these reactions to obtain the target reaction.

\\

Step 1: Reverse the first reaction to get

2NH_3 \rightleftharpoons N_2 + 3H_2. \\

The equilibrium constant for the reverse reaction is

\frac{1}{K_1}. \\

Step 2: Use the second reaction as it is:

N_2 + O_2 \rightleftharpoons 2NO. \\

The equilibrium constant remains

K_2. \\

Step 3: Multiply the third reaction by 3 to get

3H_2 + \frac{3}{2} O_2 \rightleftharpoons 3H_2O. \\

The equilibrium constant for this modified reaction is

K_3^3. \\

Now, add these reactions:

\\

•

2NH_3 \rightleftharpoons N_2 + 3H_2 \quad \left(\frac{1}{K_1}\right) \\

•

N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \\

•

3H_2 + \frac{3}{2} O_2 \rightleftharpoons 3H_2O \quad (K_3^3) \\

Adding these gives:

\\ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \\

The equilibrium constant for the overall reaction is the product of the individual constants:

\\ K = \frac{1}{K_1} \cdot K_2 \cdot K_3^3 \\

Simplifying, we get:

\\ K = \frac{K_2 K_3^3}{K_1} \\

Therefore, the correct option is Option 1:

\frac{K_2 K_3^3}{K_1}.