Question

The equilibrium constants of the following are.
\(N_2+3H_2⇋NH_3\)     \(K_1\)
\(N_2+O_2⇋2No\)    \(K_2\)
\(H_2+\frac 12O_2⇋H_2O\)    \(K_3\)
The equilibrium constant \((K)\) of the reaction:
\(2NH_3+\frac 52O_2⇋2NO+3H_2O\)
will be

• A. \(\frac {K_1K_3^3}{K_2}\)
• B. \(\frac {K_2K_3^3}{K_1}\)
• C. \(\frac {K_2K_3}{K_1}\)
• D. \(\frac {K_2^3K_3}{K_1}\)

Answer 1

The Correct Option is B

To determine the equilibrium constant \( K \) for the reaction \( 2NH_3 + \frac{5}{2}O_2 ⇋ 2NO + 3H_2O \) in terms of the given equilibrium constants \( K_1, K_2, \) and \( K_3 \), we need to manipulate the reactions provided and their given equilibrium constants.

1. Write the equilibrium expressions for the given reactions:
   • The reaction \( N_2 + 3H_2 ⇋ 2NH_3 \) has an equilibrium constant \( K_1 \).
   • The reaction \( N_2 + O_2 ⇋ 2NO \) has an equilibrium constant \( K_2 \).
   • The reaction \( H_2 + \frac{1}{2}O_2 ⇋ H_2O \) has an equilibrium constant \( K_3 \).
2. To obtain the desired reaction \( 2NH_3 + \frac{5}{2}O_2 ⇋ 2NO + 3H_2O \), manipulate and combine the given reactions:
   • Reverse the first equation: \( 2NH_3 ⇋ N_2 + 3H_2 \). The equilibrium constant for the reverse reaction is \( \frac{1}{K_1} \).
   • The second reaction \( N_2 + O_2 ⇋ 2NO \) is used as is with \( K_2 \).
   • Multiply the third reaction by 3: \( 3H_2 + \frac{3}{2}O_2 ⇋ 3H_2O \). The equilibrium constant for this reaction is \( K_3^3 \) (due to cubing of \( K_3 \) since the reaction is multiplied by 3).
3. Combine the manipulated reactions:
   • From the first reaction (reversed): \( 2NH_3 ⇋ N_2 + 3H_2 \) with \(\frac{1}{K_1}\).
   • From the second reaction: \( N_2 + O_2 ⇋ 2NO \) with \( K_2 \).
   • From the third reaction multiplied: \( 3H_2 + \frac{3}{2}O_2 ⇋ 3H_2O \) with \( K_3^3 \).
4. The combined equation is:
   • Add up: \( 2NH_3 + \frac{5}{2}O_2 ⇋ 2NO + 3H_2O \).
   • The equilibrium constant \( K \) for this reaction is obtained by multiplying the equilibrium constants of each step: \[ K = \left( \frac{1}{K_1} \right) \times K_2 \times K_3^3 = \frac{K_2 K_3^3}{K_1} \]

Thus, the equilibrium constant \( K \) for the reaction \( 2NH_3 + \frac{5}{2}O_2 ⇋ 2NO + 3H_2O \) is \(\frac{K_2 K_3^3}{K_1}\).