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CHEMICAL THERMODYNAMICS QUIZ-8

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. The enthalpy of vaporisation of a substance is 840 J/mol and its boiling point is -173℃.
Its entropy of vaporisation is
  •  4.8 J/mol/K
  •  12 J/mol/K
  •  210 J/mol/K
  •  49 J/mol/K
Solution
a) ∆Svap=(∆Hvap)/Tb =840/173=4.8 J/mol/K
Q2.When ammonium chloride is dissolved in water, the solution becomes cold. The change is:
  •  Endothermic
  •  Exothermic
  •  Super cooling
  •  None of these
Solution
a) The heat is provided by solvent and thus, feels cooler.
Q3.  Heat of combustion ∆H for C(s),H2(g) and CH4(g) are -94,-68 and -213 kcal/mole
then ∆H for C(s)+2H2(g)→CH4(g)is∶
  •  -17 kcal
  •  -111 kcal
  •  -170 kcal
  •  -85 kcal
Solution
a) Solve using Hess’s law

Q4. For the gas phase reaction,
PCl5(g)⇌PCl3(g)+Cl2(g)
Which of the following conditions are correct?
  •  ∆H>0 and ∆S>0
  •  ∆H<0 and ∆S<0
  •  ∆H>0 and ∆S<0
  •  ∆H=0 and ∆S<0
Solution
a)
∆H=+ve and ∆S=+ve; the disorder increases with increase in moles.
PCl5⇌PCl3+Cl2;∆H=+ve (dissociation)
Q5.Given that standard heat enthalpy of CH4,C2H4 and C3H8 are -17.9,12.5,-24.8 kcal/mol.
The ∆H for CH4+C2H4→C3H8 is:
  •  - 55.2 kcal mol-1
  •  - 30.2 kcal mol-1
  •  55.2 kcal mol-1
  •  - 19.4 kcal mol-1
Solution
  (d) ∆H=HC3H8-HCH4-HC2H4
= -24.8-(-17.9)-12.5
= 19.4 kcal/mol
Q6. Bond energies of (H─H), (O〓 O) and (O─H) are 105, 120 and 220 kcal/mol respectively,
then ∆H in the reaction , 2H2(g)+O2(g)→2H2O(l):
  •  - 115
  •  - 130
  •   - 118
  •  - 550
Solution
(d)
∆H= -2×[2× eO─H]+2 × eH─H+eO─O
= -4 × 220+2 × 105+120= -550 kJ
Q7.The temperature of the system increases during an:
  •  Isothermal expansion
  •  Adiabatic compression
  •  Adiabatic expansion
  •  Isothermal compression
Solution
(b) q=∆U-W, if q=0 for adiabatic process, than -∆U=-W or ∆U=W,i.e., work done on the system or work of compression brings in an increase in temperature
Q8.If CH3COOH+OH-=CH3COO-+H2O+q1 and H+ +OH-=H2O+q2,then the enthalpy change
for the reaction, CH3COOH=CH3COO- + H+ is equal to:
  •  q1 + q2
  •  q1 - q2
  •  q2 - q1
  •  -q1 - q2
Solution
(b)
Q9. Work done by the system on surroundings is:
  •  Positive
  •  Negative
  •  Zero
  •  None of these
Solution
b) Work done by the system or work of expansion is negative. Work done on the system or work of compression positive.
Q10. The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume
of 5 dm3 isothermally until the volume become 25dm3 is:
  •   2.01×103 J
  •  +2.81×103 J
  •  2.01×10-3 J
  •  +2.01×10-6 J
Solution
(a)
-W=+2.303 nRT log⁡V2/V1
-W=2.303×16/32×300×8.314 log⁡ 25/5
-W=2.01×103 J

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