Question

At \(298\ K\)
\(N_2( g)+3H_2( g)⇌2NH_3( g),K_1=4×10^5\)
\(N_2( g)+O_2( g)⇌2NO(g),K_2=1.6×10^{12}\)
\(H_2( g)+1 2O_2( g)⇌H_2O(g),K_3=1.0×10^{−13}\)
Based on above equilibria, the equilibrium constant of the reaction, 
\(2NH_3( g)\)+\(\frac5 {2}\) \(O_2( g)⇌2NO(g)+3H_2O(g)\) is _______ \(×10^{−33 }\). (Nearest integer)

Answer 1

Correct Answer: 4

 To find the equilibrium constant for the reaction \(2NH_3(g) + \frac{5}{2}O_2(g) ⇌ 2NO(g) + 3H_2O(g)\), we need to manipulate the given equilibria.
1. Utilize the given reactions and constants:
\(N_2 + 3H_2 ⇌ 2NH_3,\ K_1 = 4 \times 10^5\)
\(N_2 + O_2 ⇌ 2NO,\ K_2 = 1.6 \times 10^{12}\)
\(H_2 + \frac{1}{2}O_2 ⇌ H_2O,\ K_3 = 1.0 \times 10^{-13}\)
2. Reverse and adjust the stoichiometry:
- Reverse the first equilibrium to match \(2NH_3(g)\) on the reactant side:\)
- Keep the second as is:
\(N_2 + O_2 ⇌ 2NO,\ K_2 = 1.6 \times 10^{12}\)
- Triple the third reaction:
\(3H_2 + \frac{3}{2}O_2 ⇌ 3H_2O,\ K_{3t} = (1.0 \times 10^{-13})^3\ =\ 1.0 \times 10^{-39}\)
3. Combine the modified reactions to match the target equation:
\(2NH_3 + \frac{5}{2}O_2 ⇌ 2NO + 3H_2O\)
4. Calculate the new equilibrium constant \(K\):
\(K = K_{1r} \times K_2 \times K_{3t} = \frac{1}{4 \times 10^5} \times 1.6 \times 10^{12} \times 1.0 \times 10^{-39}\)
\(K = \frac{1.6 \times 10^{12}}{4 \times 10^5} \times 1.0 \times 10^{-39}\)
\(K = 4.0 \times 10^{-33}\)
Conclusion:
The equilibrium constant for the given reaction is \(4 \times 10^{-33}\), which fits the range of 4,4 (within the constraint that the coefficient should be \(4\ \times 10^{-33}\)).