back to directory
NEET CHEMISTRYEquilibriumMedium

Question

The following equilibrium constants are given: N2+3H22NH3N_2 + 3H_2 \rightleftharpoons 2NH_3; K1K_1 N2+O22NON_2 + O_2 \rightleftharpoons 2NO; K2K_2 H2+12O2H2OH_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K3K_3 The equilibrium constant for the oxidation of NH3NH_3 by oxygen to give NO is:

A

K2K33K1\frac{K_2K_3^3}{K_1}

B

K2K32K1\frac{K_2K_3^2}{K_1}

C

K22K3K1\frac{K_2^2K_3}{K_1}

D

K1K2K3\frac{K_1K_2}{K_3}

Step-by-Step Solution

The target reaction is the oxidation of NH3NH_3 to NO: 2NH3+52O22NO+3H2O2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O This reaction can be obtained by adding the following reactions:

  1. Reverse of reaction 1: 2NH3N2+3H22NH_3 \rightleftharpoons N_2 + 3H_2; K1=1K1K'_1 = \frac{1}{K_1}
  2. Reaction 2: N2+O22NON_2 + O_2 \rightleftharpoons 2NO; K2K_2
  3. Reaction 3 multiplied by 3: 3H2+32O23H2O3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O; K3=K33K'_3 = K_3^3 Adding the three reactions, we get: 2NH3+52O22NO+3H2O2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O The equilibrium constant for this net reaction is the product of the individual equilibrium constants : K=K1×K2×K3=1K1×K2×K33=K2K33K1K = K'_1 \times K_2 \times K'_3 = \frac{1}{K_1} \times K_2 \times K_3^3 = \frac{K_2K_3^3}{K_1}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumfollowingequilibriumconstantsrightleftharpoonsrightleftharpoons

More Equilibrium Questions

View all

Consider the following reaction: $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$. At equilibrium, the concentrations of $[\text{A}_2] = 3.0 \times 10^{–3} \text{ M}$; $[\text{B}_2] = 4.2 \times 10^{–3} \text{ M}$ and $[\text{AB}] = 2.8 \times 10^{–3} \text{ M}$. The value of $K_c$ for the above-given reaction in a sealed container at $527^\circ\text{C}$ is:

A.3.9
B.0.6
C.4.5
D.2
MediumSolve

Amongst the given options, which of the following molecules/ions acts as a Lewis acid?

A.$\text{OH}^-$
B.$\text{NH}_3$
C.$\text{H}_2\text{O}$
D.$\text{BF}_3$
EasySolve

Boric acid is an acid because its molecule

A.contains replaceable H⁺ ion
B.gives up a proton
C.accepts OH⁻ from water releasing proton
D.combines with proton from water molecule
MediumSolve

The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. pH of which one of them will be equal to 1?

A.$60 \text{ mL } \frac{M}{10} \text{ HCl } + 40 \text{ mL } \frac{M}{10} \text{ NaOH}$
B.$55 \text{ mL } \frac{M}{10} \text{ HCl } + 45 \text{ mL } \frac{M}{10} \text{ NaOH}$
C.$75 \text{ mL } \frac{M}{5} \text{ HCl } + 25 \text{ mL } \frac{M}{5} \text{ NaOH}$
D.$100 \text{ mL } \frac{M}{10} \text{ HCl } + 100 \text{ mL } \frac{M}{10} \text{ NaOH}$
MediumSolve

The tendency of $BF_3$, $BCl_3$ and $BBr_3$ to behave as Lewis acid decreases in the sequence:

A.$BCl_3 > BF_3 > BBr_3$
B.$BBr_3 > BCl_3 > BF_3$
C.$BBr_3 > BF_3 > BCl_3$
D.$BF_3 > BCl_3 > BBr_3$
MediumSolve

A compound $\text{BA}_2$ has $K_{sp} = 4 \times 10^{-12}$. Solubility of this compound will be:

A.$10^{-3} \text{ mol/L}$
B.$10^{-4} \text{ mol/L}$
C.$10^{-5} \text{ mol/L}$
D.$10^{-6} \text{ mol/L}$
EasySolve

What is the molarity of the saturated solution if the solubility product for a salt of type AB is $4 \times 10^{-8}$?

A.$2 \times 10^{-4} \text{ mol/L}$
B.$16 \times 10^{-16} \text{ mol/L}$
C.$2 \times 10^{-16} \text{ mol/L}$
D.$4 \times 10^{-4} \text{ mol/L}$
EasySolve

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$. Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$. What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ($K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$, $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$)

A.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}$
B.$[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}$
C.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}$
D.$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$
MediumSolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →