CHEMICAL THERMODYNAMICS QUIZ-1

CHEMISTRY CHEMICAL THERMODYNAMICS QUIZ-1

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 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 ∆H_vap=460.6 cal mol^-1. What will be its b.p. at 10 atm?

•  150 K
•  75 K
•  100 K
•  200 K

Solution
(c)