Chemical Thermodynamics Quiz-7

CHEMICAL THERMODYNAMICS QUIZ-7

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Q1. Given, C+O₂→CO₂+94.2 kcal ….(i)
H₂+1/2 O₂→H₂O+68.3 kcal ….(ii)
CH₄+2O₂→CO₂+2H₂O+210.8 kcal ….(iii)
The heat of formation in kcal will be :

•  45.9
•  47.8
•  - 20.0
•  47.3

Solution
(c) C+2H₂→CH₄; ∆H= ?
Find ∆H by eqs. (i) + 2 × (ii) – (iii)

Q2.Which of the following expressions represents the first law of thermodynamics?

•  ∆E=-q + W
•  ∆E=q - W
•  ∆E=q + W
•  ∆E=-q - W

Solution
c)

Q3.  The first law of thermodynamic is expressed as

•  q - W=∆E
•  ∆E=q - W
•  q=∆E - W
•  W=q + ∆E

Solution
(c) The first law of thermodynamics can be expressed as :
∆E=q+W
q=∆E-W

Q4. A heat engine absorbs heat Q₁ at temperature T₁ and heat Q₂ at temperature T₂ work
done by the engine is (Q₁+Q₂) this data

•  Violates 1st law of thermodynamics
•  Violates 1st law of thermodynamics if a₁ is –ve
•  Violates 1st law of thermodynamics if a₂ is -ve
•  Does not violates 1st law of thermodynamics

Solution
(d) It does not violate the first law of thermodynamics but violates the II law of thermodynamics

Q5.Mark the correct statement

•  For a chemical reaction to be feasible, ∆G should be zero
•  Entropy is a measure of order in a system
•  For a chemical reaction to be feasible, ∆G should be positive
•  The total energy of an isolated system is constant

Solution
  (d)

Q6. The correct thermochemical equation is :

•  C+O₂⟶CO₂; ∆H= -94 kcal
•  C+O₂ ⟶ CO₂; ∆H= +94 kcal
•  C(s)+O₂(g) ⟶ CO₂(g); ∆H= -94 kcal
•  C(s)+O₂ (g)⟶CO₂(g); ∆H= +94 kcal

Solution
(c) It provides information about physical states of reactants and products as well as about thermal changes. (d) is wrong because combustion is exothermic.

Q7.Entropy change of fusion at constant pressure is given by:

•  ∆S_(f)=∆H_f/T
•  ∆S_(f)=∆G_f/T
•  ∆S_(f)=(∆H_f)/∆T
•  None of these

Solution
(a) This is derived formula.

Q8.Gibbs energy G, enthalpy H and entropy S are related by:

•  G=H + TS
•  G=H - TS
•  G - TS=H
•  S=H - G

Solution
(b) Gibbs energy change ∆G is given by:
∆G=∆H-T∆G
Also, G=H-TS

Q9. If, combustion of 4 g of CH₄ liberates 2.5 kcal of heat, the heat of combustion of CH₄ is :

•  -2 kcal mol^-
•  -10 kcal mol^-
•  2.5 kcal mol^-1
•  -5 kcal mol^-1

Solution
(b) ∆H=(2.5 ×16)/4= -10 kcl mol^-1

Q10. The ∆H_f^o of O₃,CO₂,NH₃ and HI are 142.2 -393.3,-46.2 and +25.9 kJ per mol respectively.
The order of their increasing stabilities will be

•  O₃,CO₂,NH₃,HI
•  CO₂,NH₃,HI,O₃
•  O₃,HI,NH₃,CO₂
•  NH₃,HI,CO₂,O₃

Solution
(c) Energy absorbed ∝1/(stability of compound)
Energy released ∝ stability of compound
Thus, the order of stability is
142.2>25.9>-46.2>-393.2
ie,O₃>HI>NH₃>CO₂