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CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-6

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.
.

Q1. When rain is accompanied by a thunderstorm the collected rain water will have a pH value:
  •  Uninfluenced by occurrence of thunderstorm
  •  Depending on the amount of dust in air
  •  Slightly lower than that of rain water without thunderstorm
  •  Slightly higher than that when the thunderstorm is not there
Solution
c) Thunderstorm produces acidic oxides of N,S which produce acidic rain on dissolution in water.

Q2.What is the pH of a 1M CH3COONa solution? Ka of acetic acid =1.8×10-5,Kw=10-14 mol2 litre-2
  •  2.4
  •  3.6
  •   4.8
  •  9.4
Solution
d) CH3 COO-+H2 O⇌CH3 COOH+OH- ∴[OH-]=c∙h=c√([Kh/c] ) =√([Kw/Ka∙c] ) =√((10-14×1)/(1.8×10-5 ))=2.35×10-5 ∴ pOH=4.6289 ∴ pH=9.3710
Q3.  Ammonia under a pressure of 15 atm at 27℃ is heated to 347℃ in a closed vessel in the presence of catalyst. Under the
conditions, NH3 is partially decomposed according to the equation, 2NH3⇌N2+3H2.
The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the
percentage of NH3 actually decomposed
  •  61.3%
  •  63.5%
  •  65.3%
  •  66.6%
Solution
a) 2NH3⇌N2+3H2 Initial moles a 0 0 At equilibrium(a-2x) x 3x Initial pressure of NH3 of ‘a’mole = 15 atm at 27℃. The pressure of ’a’ mole of NH3=p atm at 347℃ ∴15/300=p/620 ∴p=31 atm At constant volume and at 347℃, mole ∝ pressure a∝31 (before equilibrium) ∴(a-2x)∝50 (after equilibrium) ∴ ((a-2x))/a=50/31 ∴x=19/62 a ∴% of NH3 decomposed =2x/a×100 =(2×19a)/(62×a)×100 =61.33%

Q4. In the equilibrium,2SO2 (g)+O2 (g) ⇌2SO3 (g), the partial pressure of SO2,O2 and SO3 are 0.662, 0.101 and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO2 and SO3 are equal?
  •  0.4 atm
  •  1.0 atm
  •  0.8 atm
  •  0.25 atm
Solution
a) Kp=(pSO3 )2/((p(SO2 ) )2 (p(O2 ) ) )=(0.331)2/((0.662)2 (0.101))=2.5 Now, Kp=(p(SO3 ) )2/((pSO2 )2 PO2 );
If pSO3= p(SO2 )
Then, p(O2 )=1/Kp =1/2.5=0.4 atm


Q5.The exothermic formation of ClF3 is represented by the equation
Cl2 (g)+3F2 (g)⇌2ClF3 (g); ∆H=-329 kJ
Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2,F2 and ClF3?
  •  Adding F2
  •  Increasing the volume of the container
  •  Removing Cl2
  •  Increasing the temperature
Solution
 a) Reaction is exothermic. By Le-Chatelier’s principle, a reaction is spontaneous in forward side (in the direction of formation of more ClF_3) if F_2 is added, temperature is lowered and ClF_3 is removed.

Q6.In a lime kiln, to get higher yield of CO2, the measure that can be taken is
  •  To remove CaO
  •  To add more CaCO3
  •  To maintain high temperature
  •  To pump out CO2
Solution
d) CaCO(s)⇌CaO(s)+CO2 (g) The equilibrium constant for this reaction is given by K=[CO2 ](as CaCO3 and CaO are solid). Hence, to get more CO2, we need to pump out continuously the CO2 gas.

Q7.The dissociation equilibrium of a gas AB2 can be represented as:
2AB2 (g)⇌2AB(g)+B2 (g)
The degree of dissociation is 'x' and is small compared to 1. The expression
relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is:
  •  (2Kp/ P )1/3
  •   (2Kp/ P )1/2
  •  Kp/P
  •  2Kp/P
Solution
a)

Q8.The solubility of AgCl in water at 10℃ is 6.2×10-6 mol/litre. The Ksp of AgCl is:
  •   [6.2×10-6]1/2
  •  6.2×(10-6)2
  •  (6.2)2×10-6
  •  [6.2×10-6]2
Solution
d) Ksp for AgCl=s2.

Q9.The correct relationship between Kc and Kp in gaseous equilibrium is :
  •  Kc= Kp(RT)∆n
  •   Kp=Kc(RT)∆n
  •  Kc/RT = (Kp)∆n
  •  d) KP/RT=(Kc)∆n
Solution
b) Kp=Kc x (RT)∆n. Where ∆n= mole of products – mole of reactants

Q10. In a chemical equilibrium, the rate constant of the backward reaction is 7.5×10-4 and the equilibrium constant is 1.5. So, the rate constant of the forward reaction is
  •   1.125×10-3
  •   2.225×10-3
  •   3.335×10-5
  •  1.125×10-1
Solution
a)Kc=kf/kb ∴kf=Kc×kb=1.5×7.5×10-4=1.125×10-3

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