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CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-10

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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.
.

Q1. A strong acid is titrated with weak base. At equivalence point, pH will be
  •  <7
  •   >7
  •  =0
  •  Cannot predict
Solution
a) When a strong acid is titrated with weak base, solution will be acidic in nature. Since, the pH of acidic solution is always less than 7, therefore, pH of the given solution will also be less than 7.

Q2.When a strong acid-strong base or their salt are dissolved in water, they are completely ionised. If a strong acid is added to
a strong base, H+ ions from the former combines with OH^- ions of the latter forming water. The formation of each water
molecule liberates a certain quantity of energy and the reaction is exothermic. The heat liberated when one mole of water
is formed by combining hydrochloric acid and sodium hydroxide is 13.7 kcal. The heat liberated when one mole of water is
formed by combining sulphuric acid and sodium hydroxide is:
  •  34 kcal
  •  13.7 kcal
  •  8.5 kcal
  •  25.5 kcal
Solution
(b) H++OH-⟶H2O ∆H=-13.7 kcal.
Q3.  The degree of dissociation of a weak acid is 1.34% at 0.1 M concentration. Its dissociation constant is:
  •  8×10-6
  •  1.79×10-5
  •  0.182×10-5
  •  8×10-5
Solution
(b) Ka=Cα2=0.1×(1.34×10-2)2=1.79×10-5.

Q4. The indicator used in titration of oxalic acid with caustic soda solution is
  •  Methyl orange
  •  Methyl red
  •  Fluorescein
  •  Phenolphthalein
Solution
d) In the titrationof weak acid with strong base, phenolphthalein is used

Q5.0.5 M ammonium benzoate is hydrolysed to 0.25 percent. Hence, its hydrolysis constant is
  •  2.5×10-5
  •  1.25×10-5
  •  3.125×10-6
  •  6.25×10-6
Solution
 c) Kh=Ch2 =0.5×(0.25/100)2 =3.125×10-6

Q6.According to Le-Chatelier’s principle, the addition of temperature to the following reaction
CO2(g)+2H2O(g)→CH4(g)+2O2(g)
will cause it to the right. This reaction is, therefore
  •  Exothermic
  •  Unimolecular
  •  Endothermic
  •  Spontaneous
Solution
c) CO2 (g)+2H2O(g)→CH4(g)+2O2(g) According to Le-Chatelier’s principle, addition of temperature shifts a endothermic reaction towards right. The addition of temperature to the above reaction will cause it to right, hence it is an endothermic reaction. (∆H=+ve).

Q7.If 0.1 mole of I2 is introduced into 1.0 litre flask at 1000 K, at equilibrium (Kc=10-6), which one is correct?
  •  [I2(g)]>[I(g)]
  •  [I2(g)]<[I(g)]
  •  [I2(g)]=[I(g)]
  •  [I2(g)]=1/2[I(g)]
Solution
(a)

Q8.1.6 moles of PCl5(g) is placed in 4 dm^3 closed vessel. When the temperature is raised to 500 K, it decomposes and at
equilibrium 1.2 moles of PCl5(g) remains.
What is the Kc value for the decomposition of PCl5 (g) to PCl3 (g) and Cl2 (g) at 500 K?
  •   0.013
  •  0.050
  •  0.033
  •  0.067
Solution
c)

Q9.The addition of which salt will decrease the H+ concentration of HCN solution?
  •  NH4Cl
  •  Al2(SO4 )3
  •  AgNO3
  •  NaCN
Solution
d) The dissociation of HCN will decrease in presence of NaCN due to common ion effect

Q10. 10-6 M NaOH is diluted 100 times. The pH of the diluted base is
  •  Between 7 and 8
  •  Between 5 and 6
  •  Between 6 and 7
  •  Between 10 and 11
Solution
(b) [OH-] in the diluted base=10-6/102 =10-8 Total [OH-]=10-8+[OH-] of water =(10-8+10-7)M =11×10-8 M pOH=-log⁡11×10-8 =-log⁡11+8 log⁡10 =6.9586 pH=14-6.9586 =7.0414

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