1. K₂K₃³/K₁ 2. K₂K₃/K₁ 3. K₂³K₃/K₁ 4. K₁K₃³/K₂

Answer: 1. K₂K₃³/K₁ Key idea: combine reactions; reversing gives 1/K , multiplying by n gives Kⁿ , adding gives the product of K values. Target: 2NH₃ + 5/2 O₂ → 2NO + 3H₂O . Reverse reaction 1: 2NH₃ → N₂ + 3H₂ , constant 1/K₁ . Reaction 2 as given: N₂+O₂→ 2NO , constant K₂ . Reaction 3 times 3: 3H₂+3/2O₂→ 3H₂O , constant K₃³ . Adding them gives the target, so K = K₂K₃³/K₁ .

Topic: Law of chemical equilibrium, Kp and Kc· Equilibrium

The equilibrium constants of the following are :N_2 + 3H_2 ⇌ 2NH_3 K_1N_2 + O_2 ⇌ 2NO K_2H_2 + 1/2O_2 → H_2O K_3The equilibrium constant (K) of the reaction : 2NH_3 + 5/2O_2 K⇌ 2NO + 3H_2O, will be :

1. \(K_2K_3^3/K_1\)

2. \(K_2K_3/K_1\)

3. \(K_2^3K_3/K_1\)

4. \(K_1K_3^3/K_2\)

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. \(K_2K_3^3/K_1\)

Key idea: combine reactions; reversing gives \(1/K\), multiplying by n gives \(K^n\), adding gives the product of K values.

Target: \(2NH_3 + \tfrac52 O_2 \rightarrow 2NO + 3H_2O\).

Reverse reaction 1: \(2NH_3 \rightarrow N_2 + 3H_2\), constant \(1/K_1\).
Reaction 2 as given: \(N_2+O_2\rightarrow 2NO\), constant \(K_2\).
Reaction 3 times 3: \(3H_2+\tfrac32O_2\rightarrow 3H_2O\), constant \(K_3^3\).

Adding them gives the target, so \(K = \dfrac{K_2K_3^3}{K_1}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

The equilibrium constants of the following are :N_2 + 3H_2 ⇌ 2NH_3 K_1N_2 + O_2 ⇌ 2NO K_2H_2 + 1/2O_2 → H_2O K_3The equilibrium constant (K) of the reaction : 2NH_3 + 5/2O_2 K⇌ 2NO + 3H_2O, will be :

1 answer

Your answer