{"id":17591,"date":"2022-05-24T08:30:26","date_gmt":"2022-05-24T03:00:26","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17591"},"modified":"2022-05-16T15:24:07","modified_gmt":"2022-05-16T09:54:07","slug":"mcq-questions-for-class-11-chemistry-chapter-7","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-chemistry-chapter-7\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 7 Equilibrium with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 7 Equilibrium with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers<\/a> are prepared as per the Latest Exam Pattern. Students can solve these Equilibrium Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

Equilibrium Class 11 MCQs Questions with Answers<\/h2>\n

Solving the Equilibrium Multiple Choice Questions of Class 11 Chemistry Chapter 7 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Equilibrium Class 11 with answers pave for a quick revision of the Chapter thereby helping you to enhance subject knowledge. Have a glance at the MCQ of Chapter 7 Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Question 1.
\nWhich of the following fluoro-compounds is most likely to behave as a Lewis base?
\n(a) BF3<\/sub>
\n(b) PF3<\/sub>
\n(c) CF4<\/sub>
\n(d) SiF4<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) PF3<\/sub>
\nExplanation:
\nBF3<\/sub> \u2192 Lewis acid (incomplete octet)
\nPF3<\/sub> \u2192 Lewis base (presence of lone pair on p atom)
\nCF4<\/sub> \u2192 Complete octet
\nSiF4<\/sub> \u2192 Lewis acid (empty d-orbital in Si-atom)<\/p>\n<\/details>\n

\n

Question 2.
\nCalculate the pOH of a solution at 25\u00b0C that contains 1 \u00d7 10-10<\/sup> M of hydronium ions, i.e. H3<\/sub>O+<\/sup>.
\n(a) 4.000
\n(b) 9.000
\n(c) 1.000
\n(d) 7.000<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 4.000
\nExplanation:
\nGiven H3<\/sub>O+<\/sup> ion concentration = 1 \u00d7 10-10<\/sup>
\npH = \u2212log[H+<\/sup>], pH = \u2212log[1 \u00d7 10-10<\/sup>],
\npH= + 10log10, pH = 10
\nWe know that, pH + pOH = 14 …….. (i)
\nPut the value of pH in eq. (i)
\n10 + pOH = 14
\npOH = 4.<\/p>\n<\/details>\n

\n

Question 3.
\nWhen two reactants, A and B are mixed to give products C and D, the reaction quotient, Q, at the initial stages of the reaction
\n(a) is zero
\n(b) Decreases With Time
\n(c) Is Independent Of Time
\n(d) Increases With Time<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Increases With Time
\nExplanation:
\nReaction quotient, Qc is equal to ratio of concentration of products to concentration of reactants at any instant of time. In the beginning of reaction, the concentration of products is negligible as compared to reactants. Therefore, the value of reaction quotient is very less.<\/p>\n<\/details>\n

\n

Question 4.
\n1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is
\n(a) Not a buffer solution with pH < 7
\n(b) Not a buffer solution with pH > 7
\n(c) A buffer solution with pH < 7
\n(d) A buffer solution with pH > 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Not a buffer solution with pH < 7
\nExplanation:
\nBuffer can accept and donate protons at the same time and HCl is an acid. So, it has pH < 7
\nSo, this is not a buffer and the solution will be acidic.<\/p>\n<\/details>\n

\n

Question 5.
\nIf, in the reaction N2<\/sub>O4<\/sub> 2NO2<\/sub>, x is that part of N2<\/sub>O4<\/sub> which dissociates, then the number of molecules at equilibrium will be
\n(a) 1
\n(b) 3
\n(c) (1 + x)
\n(d) (1 + xy)\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1
\nExplanation:
\nN2<\/sub>O4<\/sub> \u2194 2NO2<\/sub>
\n1 0 (1 – x) 2x
\nSo total number of moles at equilibrium (if initially 1 mole of N2<\/sub>O4<\/sub> was taken) = (1 – x) + 2x
\n= (1 + x)<\/p>\n<\/details>\n

\n

Question 6.
\nThe solubility product of a salt having general formula MX2<\/sub>. In water is : 4 \u00d7 10-12<\/sup>. The concentration of M2+<\/sup>ions in the aqueous solution of the salt is
\n(a) 4.0 \u00d7 10-10<\/sup> M
\n(b) 1.6 \u00d7 10-4<\/sup> M
\n(c) 1.0 \u00d7 10-4<\/sup> M
\n(d) 2.0 \u00d7 10-6<\/sup> M<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1.0 \u00d7 10-4<\/sup> M
\nExplanation:
\n(MX2) \u21d4 M2+<\/sup>s<\/sub> + 2X–<\/sup>2s<\/sub>
\nKSP<\/sub>MS2<\/sub> = 4S\u00b3
\n= 4 \u00d7 10-12<\/sup>
\nTherefore, S = 10-4<\/sup>
\nTherefore [M2+<\/sup>] = 1.0 \u00d7 10-4<\/sup>M<\/p>\n<\/details>\n

\n

Question 7.
\nEquimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH?
\n(a) CaCl2<\/sub>
\n(b) SrCl2<\/sub>
\n(c) BaCl2<\/sub>
\n(d) MgCl2<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) BaCl2<\/sub>
\nExplanation:
\nEquimolar solutions of the given chlorides when prepared in water forms their respective hydroxides.
\nThe pH of salt BaCl2<\/sub> = 7, whereas SrCl2<\/sub> and CaCl2<\/sub> = 7 and MgCl2<\/sub> < 7.<\/p>\n<\/details>\n

\n

Question 8.
\nOxidation number of Iodine varies from
\n(a) -1 to +1
\n(b) -1 to +7
\n(c) +3 to +5
\n(d) -1 to +5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) -1 to +7
\nExplanation:
\nVarious oxidation numbers of Iodine are -1, 0, +1, +3, +5, +7.
\nSo, Iodine shows -1 to +7 oxidation state.<\/p>\n<\/details>\n

\n

Question 9.
\nWhich of the following molecualr species has unpaired electrons?
\n(a) N2<\/sub>
\n(b) F2<\/sub>
\n(c) O–<\/sup>2<\/sub>
\n(d) O2<\/sub>-2<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) O–<\/sup>2<\/sub>
\nExplanation:
\nIn O2<\/sub>–<\/sup> total electrons are 17 moiecular orbital configuration.
\nO–<\/sup>2<\/sub> has one unpaired electron.<\/p>\n<\/details>\n

\n

Question 10.
\nA certain buffer solution contains equal concentration of X–<\/sup> and HX. The ka for HX is 10-8<\/sup>. The pH of the buffer is
\n(a) 3
\n(b) 8
\n(c) 11
\n(d) 14<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 8
\nExplanation:
\nka<\/sub> kb<\/sub> = kw<\/sub>
\nkb<\/sub> = 10-8<\/sup>
\nka<\/sub> \u00d7 10-8<\/sup> = 10-14<\/sup>
\nka<\/sub> = 10-6<\/sup> = [H+<\/sup>]
\npH = \u2212log [H+<\/sup>]
\npH = \u2212 log 10-6<\/sup>
\n= 6pH + pOH
\n= 14 pOH
\n= 14 \u2212 6
\n= 8<\/p>\n<\/details>\n

\n

Question 11.
\nAmong the following the weakest Bronsted base is
\n(a) F–<\/sup>
\n(b) Cl–<\/sup>
\n(c) Br–<\/sup>
\n(d) I–<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) I–<\/sup>
\nExplanation:
\nAccording to this theory, an acid is a proton donor and a base is a proton acceptor. Every strong Bronsted acid has a weak conjugate base and every strong base has a weak conjugate acid. The acidity increases in halogen group atoms,
\nHF < HCl < HBr < HI.
\nSo, HI is highly acidic and their conjugate bases decrease in order F–<\/sup> > Cl–<\/sup> > Br–<\/sup> > I–<\/sup>.<\/p>\n<\/details>\n

\n

Question 12.
\nWhich of the following statements is correct about the equilibrium constant?
\n(a) Its value increases by increase in temperature
\n(b) Its value decreases by decrease in temperature
\n(c) Its value may increase or decrease with increase in temperature
\n(d) Its value is constant at all temperatures<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Its value may increase or decrease with increase in temperature
\nExplanation:
\nIncrease in the temperature decreases the value of equilibrium constant because forward reaction is exothermic. When the forward reaction is endothermic increase the temperature increases the value of equilibrium constant. These occurs when chemical equilibrium shifts toward the products or reactants.<\/p>\n<\/details>\n

\n

Question 13.
\npH value of which one of the following is NOT equal to one?
\n(a) 0.1 M CH3<\/sub>COOH
\n(b) 0.1 M HNO3<\/sub>
\n(c) 0.05 M H2<\/sub>SO4<\/sub>
\n(d) 50 cm\u00b3 0.4 M HCl + 50 cm\u00b3 0.2 M NaOH<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0.1 M CH3<\/sub>COOH
\nExplanation:
\nSince CH3<\/sub>COOH does not dissociate completely, its 10-1<\/sup> M solution does not have pH = 1<\/p>\n<\/details>\n

\n

Question 14.
\n[ OH–<\/sup>] in a solution is 1 mol L–<\/sup>. The pH the solution is
\n(a) 1
\n(b) 0
\n(c) 14
\n(d) 10-14<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 14
\nExplanation:
\n[H3<\/sub>O+<\/sup>] = (Kw<\/sub>)\/ [OH–<\/sup>]
\n= (10-14<\/sup>)\/ (1)
\npH = 14<\/p>\n<\/details>\n

\n

Question 15.
\nWhat is the pH of a 0.10 M solution of barium hydroxide, Ba (OH)2<\/sub>?
\n(a) 11.31
\n(b) 11.7
\n(c) 13.30
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 13.30
\nExplanation:
\npH = 13.30
\nBarium hydroxide is a strong base for both stages of dissociation:
\nBa (OH)2<\/sub> (s) \u2192 Ba2+<\/sup> + 2OH–<\/sup>
\nSo the solution will have 0.20 M hydroxide ions. Now use the auto dissociation product for water:
\n[H+<\/sup>][OH–<\/sup>] = 1.0 \u00d7 10-14<\/sup>M
\n[OH–<\/sup>] = 2.0 \u00d7 10-1<\/sup>M
\n[H+<\/sup>] = 5.0 \u00d7 10-14<\/sup>M
\nAnd then pH = \u2212log10<\/sub> ([H+<\/sup>] = 5.0 \u00d7 10-14<\/sup>)
\n= 13.30.<\/p>\n<\/details>\n

\n

Question 16.
\nThe Ksp<\/sub> for Cr (OH)3<\/sub> is 1.6 \u00d7 10-30<\/sup>. The molar solubility of this compound in water is:
\n(a) \u221a (1.6 \u00d7 10-30<\/sup>)
\n(b) (\u221a(1.6 \u00d7 10-30<\/sup>))(1\/4)<\/sup>
\n(c) (\u221a(1.6 \u00d7 10-30<\/sup>)\/(27)))(1\/4)<\/sup>
\n(d) (1.6 \u00d7 10-30<\/sup>)\/(27)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) (\u221a(1.6 \u00d7 10-30<\/sup>)\/(27)))(1\/4)<\/sup><\/p>\n<\/details>\n

\n

Question 17.
\nThe solubility product of CuS, Ag2<\/sub>S and HgS are 10-31<\/sup>, 10-44<\/sup> and 10-54<\/sup> respectively. The solubilities of these sulphides are in the order
\n(a) HgS > Ag2<\/sub>S > CuS
\n(b) CuS > Ag2<\/sub>S > HgS
\n(c) Ag2<\/sub>S > CuS > HgS
\n(d) AgS > HgS > CuS<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Ag2<\/sub>S > CuS > HgS
\nExplanation:
\nFor CuS & HgS Ksp<\/sub> = S\u00b2 (where s= solubility) & for Ag2<\/sub>S ; Ksp<\/sub> = 4S3<\/sup> (s= solubility) now put the values of Ksp<\/sub> and check in which case the value of S is highest & lowest.
\nSolubility of CuS : (10-31<\/sup>)(1\/2)<\/sup> = ( 10(-31\/2)<\/sup>);
\nAg2<\/sub>S: (10-44<\/sup>)(1\/3)<\/sup> =( 10(-44\/3)<\/sup>);
\nHgS;
\n(10-54<\/sup>)(1\/2)<\/sup> = (10-54\/2<\/sup>) = (10-27<\/sup>);
\nThe order of solubility will be as per above values:
\nHence , the order of solubility is: Ag2<\/sub>S > CuS > HgS<\/p>\n<\/details>\n

\n

Question 18.
\nBuffer solutions have constant acidity and alkalinity because
\n(a) They have large excess of H+<\/sup> or OH–<\/sup> ions
\n(b) They have fixed value of pH
\n(c) These give unionised acid or base on reaction with added acid or alkali
\n(d) Acids and alkalies in these solutions are shielded from attack by other ions<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) These give unionised acid or base on reaction with added acid or alkali
\nExplanation:
\nConsider a buffer of CH3<\/sub>COOH + CH3<\/sub>COONa
\nAddition of acid: H+<\/sup> + CH3<\/sub>COO – \u2192 CH3<\/sub>COOH
\nWeak acid
\nAddition of alkali:
\nOH–<\/sup> + CH3<\/sub>COOH \u2192 H2<\/sub>O + CH3<\/sub>COO–<\/sup>
\nWeak electrolyte
\nThus, the addition of acid or alkali does not cause any change in pH.<\/p>\n<\/details>\n

\n

Question 19.
\nThe position of some metals in the electrochemical series in decreasing electropositive character is Mg > Al > Zn > Cu > Ag. In a chemical factory, a worker by accident used a copper rod to stir a solution of aluminium nitrate; he was scared that now there would be some reaction in the solution, so he hurriedly removed the rod from the solution and observed that
\n(a) The rod was coated with Al
\n(b) An alloy of Cu and Al was being formed.
\n(c) The solution turned blue in colour
\n(d) There was no reaction.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) There was no reaction.
\nExplanation:
\nMg > Al > Zn > Cu > Ag Copper rod is used to stir solution of Al(NO3<\/sub>)3<\/sub> so no reaction as one above in series will displace metal lower in series.<\/p>\n<\/details>\n

\n

Question 20.
\nAn amount of solid NH4<\/sub>HS is placed in a flask already contaniing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3<\/sub> and H2<\/sub>S gases in the flask. When the decomposition reaction reaches equilibrium , the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4<\/sub>HS decomposition at this temperature is
\n(a) 0.11
\n(b) 0.17
\n(c) 0.18
\n(d) 0.30<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0.11<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 7 Equilibrium with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers are prepared as per the Latest Exam Pattern. Students can solve these Equilibrium Class 11 MCQs Questions with Answers and assess their preparation level. …<\/p>\n

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