Please refer to CBSE Class 11 Chemistry HOTs Equilibrium. Download HOTS questions and answers for Class 11 Chemistry. Read CBSE Class 11 Chemistry HOTs for Chapter 07 Equilibrium below and download in pdf. High Order Thinking Skills questions come in exams for Chemistry in Class 11 and if prepared properly can help you to score more marks. You can refer to more chapter wise Class 11 Chemistry HOTS Questions with solutions and also get latest topic wise important study material as per NCERT book for Class 11 Chemistry and all other subjects for free on Studiestoday designed as per latest CBSE, NCERT and KVS syllabus and pattern for Class 11
Chapter 07 Equilibrium Class 11 Chemistry HOTS
Class 11 Chemistry students should refer to the following high order thinking skills questions with answers for Chapter 07 Equilibrium in Class 11. These HOTS questions with answers for Class 11 Chemistry will come in exams and help you to score good marks
HOTS Questions Chapter 07 Equilibrium Class 11 Chemistry with Answers
Question. For the synthesis of ammonia by the reaction N2 + 3H2 ⇌ 2NH3 in the Haber process, the attainment of equilibrium is correctly predicted by the curve
Answer : A
Question.PCl5 is dissociating 50% at 250°C at a total pressure of P atm. If equilibrium constant is Kp, then which of the following relation is numerically correct –
(a) Kp = 3P
(b) P = 3Kp
(c) P = 2KP/3
(d) Kp = 2P/3
Answer : B
Question.The value of Kp for the equilibrium reaction N2O4 (g) ⇌ 2NO2 (g) is 2.
The percentage dissociation of N2O4(g) at a pressure of 0.5 atm is
(a) 25
(b) 88
(c) 50
(d) 71
Answer : D
Question.Two moles of PCl5 were heated in a closed vessel of 2L. At equilibrium 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is
(a) 0.53
(b) 0.267
(c) 2.63
(d) 5.3
Answer : B
Question.K1 and K2 are equilibrium constant for reactions (i) and (ii)
N2(g) + O2 (g) ⇌ 2 NO (g) ...(i)
NO(g) ⇌ (1/2) N2 (g) + (1/2) O2 (g) ...(ii)
Then,
(a) K1 = (1/K2)2
(b) K1 = K22
(c) K1 = 1/K2
(d) K1 = (K2)0
Answer : A
Question.For the decomposition of the compound, represented as NH2COONH4(s) ⇌ 2NH3(g) + CO2(g) the Kp = 2.9 × 10–5 atm3. If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :
(a) 1.94 × 10–2 atm
(b) 5.82 × 10–2 atm
(c) 7.66 × 10–2 atm
(d) 38.8 × 10–2 atm
Answer : B
Question.1.0 mole of AB5(g) is placed in a closed container under one atmosphere and at 300K. It is heated to 600K, when 20% by mass of it dissociates as AB5 (g) → AB(g) + 2 B2 (g). The resultant pressure is
(a) 1.2 atm
(b) 2.4 atm
(c) 1.4 atm
(d) 2.8 atm
Answer : D
Question.For the reaction C(s) + CO2(g) → 2CO(g), Kp = 63 atm at 1000 K. If at equilibrium : PCO = 10
PCO2, then the total pressure of the gases at equilibrium is
(a) 6.3 atm
(b) 6.93 atm
(c) 0.63 atm
(d) 0.693 atm
Answer : B
Question.The equilibrium constant for a reaction, N2(g) + O2(g) ⇆ 2NO(g) is 4 × 10–4 at 2000 K. In the presence of catalyst, the equilibrium is attained 10 times faster. The equilibrium constant in presence of catalyst at 2000 K is :
(a) 10 × 10–4
(b) 4 × 10–2
(c) 4 × 10–4
(d) 40 × 10–4
Answer : C
Question.The figure shows the change in concentration of species A and B as a function of time.
The equilibrium constant Kc for the reaction
A(g) ⇌ 2B(g) is :
(a) Kc > 1
(b) K < 1
(c) K = 1
(d) data insufficient
Answer : A
Question. Air containing 79% of nitrogen and 21% of oxygen by volume is heated at 2200 K and 1 atm until equilibrium is established according to the reaction N2(g) + O2(g) ⇆ 2NO(g) If the KP of the reaction is 1.1 × 10–3, calculate the amount of nitric oxide produced in terms of volume percent.
(a) 1.67 %
(b) 1.23 %
(c) 1.33 %
(d) 1.54%
Answer : C
Question.At a certain temperature and 2 atm pressure equilibrium constant (KP) is 25 for the reaction SO2 (g) + NO2 (g) ⇆ SO3 (g) + NO(g) Initially if we take 2 moles of each of the four gases and 2 moles of inert gas, what would be the equilibrium partial pressure of NO2?
(a) 1.33 atm
(b) 0.1665 atm
(c) 0.133 atm
(d) None of these
Answer : C
Question.The degree of dissociation of acetic acid in a 0.1 M solution is 1.32 × 10–2. Find out the dissociation constant of the acid.
(a) 1.50 × 10–4
(b) 1.80 × 10–16
(c) 1.76 × 10–5
(d) 1.2 × 10–3
Answer : C
Question.Among the following, the correct statement is:
(a) pH decreases when solid ammonium chloride is added to a dilute aqueous solution of NH3
(b) pH decreases when solid sodium acetate is added to a dilute aqueous solution of acetic acid
(c) pH decreases when solid NaCl is added to a dilute aqueous solution of NaOH
(d) pH decreases when solid sodium oxalate is added to a dilute aqueous solution of oxalic acid
Answer : A
Question. For the manufacture of ammonia by the reaction N2 + 3H2 ⇆ 2NH3 + 2 kcal the favourable conditions are
(a) low temperature, low pressure and catalyst
(b) low temperature, high pressure and catalyst
(c) high temperature, low pressure and catalyst
(d) high temperature, high pressure and catalyst
Answer : B
Question.For the reaction XCO3 (s) ⇆ XO(s) + CO2 (g), Kp = 1.642 atm at 727°C. If 4 moles of XCO3(s) was put into a 50 litre container and heated to 727°C. What mole percent of the XCO3 remains unreacted at equilibrium?
(a) 20
(b) 25
(c) 50
(d) None of these
Answer : D
Question.Solubility products of CuI and Ag2CrO4 have almost the same value (~ 4 ´ 10-12). The ratio of solubilities of the two salts (CuI:Ag2CrO4) is closest to
(a) 0.01
(b) 0.02
(c) 0.03
(d) 0.10
Answer : B
Question.At 25° C, the solubility product of Hg2Cl2 in water is 3.2 × 10–17 mol3 dm–9. What is the solubility of Hg2Cl2 in water at 25° C?
(a) 1.2 × 10–12 M
(b) 3.0 × 10–6 M
(c) 2 × 10–6 M
(d) 1.2 × 10–16 M
Answer : C
Question.Solid AgNO3 is slowly added to a solution containing each of 0.01 M NaCl and 0.001 M NaBr.
What will be the concentration of Cl– ions in solution when AgBr will just start to precipitate?
Ksp (AgBr) = 3.6 × 10–13, Ksp (AgCl) = 1.8 × 10–10.
(a) 1.8 × 10–7
(b) 3.6 × 10–10
(c) 0.01
(d) 2 × 10–4
Answer : C
Question.A buffered solution is prepared by mixing equal volumes of
(a) 0.2 M NH4OH and 0.1 M HCl
(b) 0.2 M NH4OH and 0.2 M HCl
(c) 0.2 M NaOH and 0.1 M CH3COOH
(d) 0.1 M NH4OH and 0.2 M HCl
Answer : A
Numeric Value Answer
Question. In the reaction PCl5 ⇆ PCl3 + Cl2, the amount of each PCl5, PCl3 and Cl2 is 2 mole at equilibrium and total pressure is 3 atm. What will be the value of Kp?
Answer : 1
Question.For the reaction : SnO2(s)+2H2(g) ⇆ 2H2O(g) + Sn(s)
The value of 2 x KP at 900 K where the equilibrium steam hydrogen mixture was 45% H2 by volume is
Answer : 3
Question. A soft drink was bottled with a partial prssure of CO2 of 3 bar over the liquid at room temperature. The partial pressure of CO2 over the solution approaches a value of 30 bar when 44g of CO2 is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ × 10–1.
(First dissociation constant of H2CO3 = 4.0 × 10–7; log 2 = 0.3; density of the soft drink = 1g mL–1)
Answer : 3
Question.Calculate the pH at the equivalence point when a solution of 0.01 M CH3COOH is titrated with a solution of 0.01 M NaOH. pKa of CH3COOH is 4.74.
Answer : 8.22
Question.A buffer solution is prepared by mixing 0.1 M ammonia and 1.0 M ammonium chloride. At 298 K, the pKb of NH4OH is 5.0. The pH of the buffer is
Answer : 8
Question.If the solubility product of AB2 is 3.20 × 10–11M3, then the solubility of AB2 in pure water is ______ × 10–4 mol L–1.
[Assuming that neither kind of ion reacts with water]
Answer : 2.0
Question.Calculate pH of a resultant solution of 25 mL of 0.1 M HCl, 50 mL of 0.02 M HNO3 and 25 mL of 0.1 M NaOH.
Answer : 2
Question.Calculate the pOH of a solution at 25°C that contains 1 x 10– 10 M of hydronium ions, i.e. H3O+.
Answer : 4
Question.For a reaction X + Y ⇆ 2Z, 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L–1. The equilibrium constant of the reaction is . x/15 The value of x is _________.
Answer : 16
Question.0.1 M NaOH is titrated with 0.1 M HA till the end point; Ka for HA is 5.6 × 10–6 and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point
Answer : 8.98
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HOTS for Chapter 07 Equilibrium Chemistry Class 11
Expert teachers of studiestoday have referred to NCERT book for Class 11 Chemistry to develop the Chemistry Class 11 HOTS. If you download HOTS with answers for the above chapter you will get higher and better marks in Class 11 test and exams in the current year as you will be able to have stronger understanding of all concepts. High Order Thinking Skills questions practice of Chemistry and its study material will help students to have stronger understanding of all concepts and also make them expert on all critical topics. You can easily download and save all HOTS for Class 11 Chemistry also from www.studiestoday.com without paying anything in Pdf format. After solving the questions given in the HOTS which have been developed as per latest course books also refer to the NCERT solutions for Class 11 Chemistry designed by our teachers. We have also provided lot of MCQ questions for Class 11 Chemistry in the HOTS so that you can solve questions relating to all topics given in each chapter. After solving these you should also refer to Class 11 Chemistry MCQ Test for the same chapter
You can download the CBSE HOTS for Class 11 Chemistry Chapter 07 Equilibrium for latest session from StudiesToday.com
Yes, the HOTS issued by CBSE for Class 11 Chemistry Chapter 07 Equilibrium have been made available here for latest academic session
HOTS stands for "Higher Order Thinking Skills" in Chapter 07 Equilibrium Class 11 Chemistry. It refers to questions that require critical thinking, analysis, and application of knowledge
Regular revision of HOTS given on studiestoday for Class 11 subject Chemistry Chapter 07 Equilibrium can help you to score better marks in exams
Yes, HOTS questions are important for Chapter 07 Equilibrium Class 11 Chemistry exams as it helps to assess your ability to think critically, apply concepts, and display understanding of the subject.