CHEMICAL EQUILBRIUM QUIZ-4

CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-4

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.

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Q1.  The pH of 10^-10M NaOH solution is nearest to

•  -4
•  -10
•  4
•  7

Solution
(d) Given, concentration of NaOH=10^-10 M NaOH →Na⁺ + OH^- 10^-10 10^-10 10^-10 ∴[OH^-] from NaOH=10^-10 ∴ Total [OH^-]=10^-7+10^-10 =10^-7(1+0.001) =10^-7(1001/1000) =10^-10×1001 pOH=-log⁡[OH^-] =-log⁡(1001×10^-10) =-3.004+10 =6.9996 ∵ pH+pOH=14 ∴ pH=14-6.9996 =7.0004=7

Q2.HSO₄^-+OH^-→SO₄^2-+H₂O
Which is correct about conjugate acid base pair?

•  HSO₄^- is conjugate acid of base SO₄^2-
•  HSO₄^- is conjugate base of base SO₄^2-
•  SO₄^2- is conjugate acid of acid HSO₄^-
•  None of the above

Solution
a) The residual part of the acid after removal of a proton from the molecule of an acid, is called its conjugate base. Thus, conjugate base=Acid-H^+ and a base after accepting a proton gets converted into its conjugate acid. Thus, HSO₄^- is conjugate acid of base SO₄^2-.

Q3.  Which of the following is a Lewis acid?

•  AlCl₃
•  Cl^-
•  CO
•  C₂H₂

Solution
(a) According to Lewis acid is any species (molecule, radial or ion) that can accept an electron pair to form a coordinate covalent bond. Thus, acid is an electron deficient species e.g., BF₃,AlCl₃,SO₃ and all cations etc

Q4. The equilibrium constant for the reaction N₂(g)+O₂(g)⇌2NO(g) at temperature Tis 4×10^-4. The value of K_c for the reaction NO(g)⇌1/2 N₂(g)+1/2 O₂(g) at the same temperature is

•  25
•  50
•  75
•  100

Solution
b) N₂ (g)+O₂ (g)⇌2NO(g) K_c=[NO]²/[N₂ ][O₂] =4×10^-4 NO(g)⇌1/2 N_{2 (g)+1/2 O2 (g) Kc^'=([N2 ]^{(1/2) [O2 ](1/2))/([NO]) =1/√(Kc )=1/√(4×10-4 ) =1/(2×10-2 )=100/2=50}}

Q5.The equilibrium constant for the reaction, 2X(g)+Y(g)⇌2Z(g) is 2.25 litre mol^-1. What would be the concentration of Y at equilibrium with 2.0 mole of X and 3.0 mole of Z in one litre vessel?

•  1.0 M
•  2.25 M
•  2.0 M
•  4.0 M

Solution
 a) K_c=[Z]²/([X]² [Y] )=(3)²/((2)² (Y))=2.25,∴[Y]=1M

Q6.For a hypothetical equilibrium: 4A+5B⇌4x+6y; the equilibrium constant K_c has the unit:

•  mol²litre^-2
•  litre mol^-1
•  litre²mol^-2
•  mol litre^-1

Solution
d) Unit of K_c=[ ]^∆n.∆n=+1.

Q7.The correct statement about buffer solution is:

•  It contains a weak acid and its conjugate base
•  It contains a weak base and its conjugate acid
•  It shows little change in pH on adding small amount of an acid or base
•  All of the above

Solution
d)Buffer is (CH₃ COOH)⁡ + (CHCOO^-) and has pH fixed.

Q8.In the reaction, 3A+2B→2C, the equilibrium constant K_c is given by

•  ([3A]×[2B])/([C])
•  ([A]³×[B])/([C])
•  [C]²/([A]³×[B]²)
•  ([C])/([3A][2B])

Solution
c) 3A+2B→2C K_c=(concentration of products)/(concnetration of reactants) =[C]²/([A]³×[B]² )

Q9.NaOH(aq),HCl(aq)andNaCl(aq) concentration of each is 10^-3 M. Their pH will be respectively _p=K_c ?

•  10, 6, 2
•  11, 3, 7
•  10, 2, 6
•  3, 4, 7

Solution
b) NaOH=[OH^- ]=10^(-3) [H⁺ ][OH^- ]=10^-14 [H⁺ ]=10^-11 pH=-log⁡[H⁺] =-log⁡[10^-11 ]=11 HCl(aq)=[H⁺ ]=10^-3 pH=-log⁡[10^-3 ]=3 NaCl(aq)=Neutral; [H⁺ ]=[OH^- ]=10^-7 ie,pH=7

Q10. In the following reversible reaction, 2SO₂+O₂⇌2SO₃+Q cal
Most suitable condition for the higher production of SO₃ is

•   Low temperature and high pressure
•   Low temperature and low pressure
•   High temperature and high pressure
•  High temperature and low pressure

Solution
a) Reaction is exothermic and volume is decreasing from left to right, so for higher production of SO_3, there should be low temperature and high pressure