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CHEMISTRY CHEMICAL THERMODYNAMICS QUIZ-1

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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.
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Q1. The free energy change for a reversible reaction at equilibrium is
  •  Large, positive
  •  Small, negative
  •  Small, positive
  •  0
Solution
d)

Q2.In a flask, colourless N2O4(g) is in equilibrium with brown coloured NO2 (g). At equilibrium when the flask is heated to
100℃, the brown colour deepens and on cooling it becomes coloured.Which statement is incorrect about this observation?
  •  The ∆H for the reaction N2O4 (g)⇌2NO2(g) is +ve
  •  Paramagnetism increases on cooling
  •  The ∆H-∆U at 100℃ is equal to 200 cal
  •  Dimerisation is reduced on heating
Solution
(c) ∆H-∆U=∆nRT=1×2×373=746 cal.
Q3.  The second law of thermodynamics introduced the concept of:
  •  Third law of thermodynamics
  •  Work
  •  Entropy
  •  Internal energy
Solution
c) The second law of thermodynamics has been defined as – the entropy of universe is always increasing in the course of every spontaneous process.

Q4. ∆G°for the reaction X+Y⇌Z is -4.606 kcal. The value of equilibrium constant of the reaction at 227℃ is (R=2.0 cal mol-1 K-1)
  •  100
  •  10
  •  2
  •  0.01
Solution
a) ∆G=-2.303 RT log⁡K -4.606=-2.303×0.002×500 log⁡K log⁡K=2,K=100

Q5.The standard change is Gibbs energy for the reaction,
H2O⇌H++OH- at 25℃ is:
  •  100 kJ
  •  -90 kJ
  •  90 kJ
  •  -100 kJ
Solution
 b) ∆G°=-2.303 RT log⁡K (K for H2 O=10-14/55.6) =-2.303×8.314×298×log⁡10-14/55.6 =-89.84 kJ

Q6.In an adiabatic process
  •  p.∆V=0
  •  q=+W
  •  ∆E=q
  •  q=0
Solution
d) In the adiabatic process no heat enters or leaves the system i.e., q=0

Q7.What is ∆E for system that does 500 cal of work on surrounding and 300 cal of heat is absorbed by the system?
  •  -200 cal
  •  -300 cal
  •  +200 cal
  •  +300 cal
Solution
(a) From first law of thermodynamic. ∆E=q+W Given,q=+300 cal (∵Heat is absorbed) W=-500 cal (∵Work is done on surroundings) ∴ ∆E=q+W=300+(-500) =-200 cal

Q8.An ideal gas undergoing expansion in vacuum shows:
  •  ∆U=0
  •  W=0
  •   q=0
  •  All of these
Solution
d) Ideal gas does not show intermolecular forces of attractions

Q9.The bond dissociation energy of B-F in BF-3 is 646 kJ mol-1, whereas that of C-F in CF-4 is 515 kJ mol-1.
The correct reason for higher B-F bond dissociation energy as compared to that of C-F is:
  •  Stronger σ bond between B and F in BF3 as compared to that between C and F in CF4
  •  Significant pπ-pπ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4
  •  Lower degree of pπ-pπ interaction between B and F in BF3 than that between C and F in CF4
  •  Smaller size of B-atom as compared to that of C-atom
  • Solution
    b) In BF3 pπ—pπ interaction leads to back bonding due to vacant p–orbitals of boron and completely filled p-orbitals of F.

    Q10. Boiling point of a liquid is 50 K at 1 atm and ∆Hvap=460.6 cal mol-1. What will be its b.p. at 10 atm?
    •  150 K
    •  75 K
    •  100 K
    •  200 K
    Solution
    (c)

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