Chemical Equilibrium Quiz-3

CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-3

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.  Which of the following will supress the ionisation of acetic acid is aqueous solution?

•  NaCl
•  HCl
•  KCl
•  Unpredictable

Solution
b) HCl provides common H⁺ ions to CH₃COOH CH₃COOH⇌CH₃COO^-+H⁺

Q2.Which among the following is the strongest acid?

•  H(ClO)O₂
•  H(ClO)O₃
•  H(ClO)O
•  H(ClO)

Solution
(b) For oxoacids of the same element, the acidic strength increases with increase in the oxidation number of the element

Q3.  For the gaseous phase reaction, 2NO ⇌ N2 + O2, ∆H^o = - 43.5 kcal mol-1 , which statement is correct for,
N2(g) + O2(g) ⇌ 2NO(g)?

•   K is independent of temperature
•  K increases as temperature decreases
•  K decreases as temperature decreases
•  K varies with addition of NO

Solution
(c) 2.303 log K₂/K₁ =∆H/R ([T₂-T₁])/(T₁T₂) , ∆H= +ve for the reaction, N₂+O₂⇌2NO.

Q4. The solubility of a saturated solution of calcium fluoride is 2×10^-4 moles per litre. Its solubility product is:

•  32×10^-10
•  32×10^-8
•  32×10^-14
•  32×10^-12

Solution
d) K_sp of CaF₂=4s³=4×(2×10^-4)³=32×10^-12

Q5.Four moles of PCl₅ are heated in a closed 4 dm³ container to reach equilibrium at 400 K. at equilibrium 50% of PCl₅ is dissociated. What is the value of K_c for the dissociation of PCl₅ into PCl₃ and Cl₂ at 400 K?

•  0.50
•  1.00
•  1.15
•  0.05

Solution
 a) PCl₅⇌PCl₃+Cl₂ Initial conc. 4/4 0 0 Equili. conc. 2/4 2/4 2/4 K=[PCl₃][Cl₂]/[PCl₅ ] =(2×2×4)/(4×4×2)=1/2=0.5

Q6.The conjugate base of H₂SO₄ in the following reaction is: H₂SO₄+H₂O⇌H₃O⁺+HSO₄^-

•  H₂O
•  HSO₄^-
•  H₃O⁺
•  SO₄^(2-)

Solution
b) H₂SO₄(acid) →^{(-H+ )}HSO_4^- (Conjugate base).

Q7. The weakest base among the following is∶

•  H^-
•  CH₃^-
•  CH₃O^-
•  Cl^-

Solution
d) HCl a mineral acid is strong acid and thus, Cl^- is weak base

Q8.Solubility product of a salt AB is 1×10^(-8) M² in a solution in which the concentration of A⁺ ions is 10^(-3) M. The salt will
precipitate when the concentration of B^- ions is kept

•  Between 10^-8 to 10^-7 M
•  Between 10^-7 M to 10^-8 M
•  >10^-5 M
•  <10^-8 M

Solution
(c) A salt is precipitated only when the product of ionic concentration is more than its solubility product. K_sp=1×10^-8 [A⁺]=10^-3 M [B^-]=(1×10^-8)/10^-3 =10^-5 M So, AB will be more precipitated only when the concentration of [B^-] is more than 10^-5 M.

Q9.For which reaction is K_p=K_c ?

•  2NOCl(g) ⇌2NO(g)+Cl₂ (g)
•  N₂(g)+3H₂(g)⇌2NH₃(g)
•  H₂(g)+Cl₂(g) ⇌2HCl(g)
•  2SO₂(g)+O₂ (g)⇌2SO₃ (g)

Solution
(c) ∆n=0 and thus, K_p=K_c.

Q10. A saturated solution of Ag₂SO₄ is 2.5×10^-2 M. The value of its solubility product is

•  62.5×10^-6
•   6.25×10^-4
•   15.625×10^-6
•   3.125×10^-6

Solution
a) Ag₂SO₄→2Ag⁺+SO₄^2- Ag₂SO₄ is a 2:1 type electrolyte i.e., it gives two Ag⁺ ions and one SO₄^2- ion on ionisation. Given, s=2.5×10^-2 M Hence, its K_sp=4s³ (where,s= molar solubility) K_sp=4(2.5×10^-2)³ =4×15.63×10^-6 =62.5×10^-6