## CHEMISTRY THERMODYNAMICS QUIZ-12

IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything.
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Q1.Heat change occurring during conversion of 1 kg of CaSO4∙2H2O(s) (molar mass 172 g mol-1) of CaSO4∙1/2H2O(s) is equal to:
•  484 kJ mol-1
•  400 kJ
• -484.0 kJ mol-1
• -1000 kJ
Solution
∆H = ∆HP - ∆HR; (for 1 mol)
= [-1575.0 kJ mol-1 - 3/2×241.8]
- [-2021.0 kJ mol-1]
= +83.3 kJ mol-1
For 1 kg CaSO4⋅2H2O
Number of moles = 1000/172 = 5.81
∴ Heat change for 5.81 mol of CaSO4⋅2H2O
= 5.81×83.3 kJ mol-1
= 484 kJ mol-1

Q2.The work of expansion in adiabatic process (wadi) is related to work of expansion in isothermal process (wiso) as
Solution
Work in reversible isothermal expansion is greater than work done in adiabatic expansion
Q3.For the spontaneity of a reaction, which statement is true?
•  ∆G =+ve ; ∆H = +ve
•  ∆H =+ve ; ∆S=-ve
•  ∆G = -ve ; ∆S = -ve
•  ∆H = -ve ; ∆S = +ve
Solution
∆H = -ve and ∆S = +ve , both favour the process

Q4.The heat energy released during neutralization of 1 eq. of NaOH and with 1 eq. of CH3COOH and 1 eq. of NaOH are respectively :
•  - 16.4 kcal, -12.0 kcal
•  - 12.0 kcal, -10 kcal
•  - 13.7 kcal in both
•  - 12.0 kcal, -16.4 kcal
Solution
Heat released during neutralization of weak acid < -13.7 in
HF, hydration of F- ion is responsible for higher value.

Q5.Select the correct order of the entropy change
•  sysS > 0, ∆surrS = 0
•  sysS < 0, ∆surrS > 0
•  sysS > 0, ∆surrS < 0
•  sysS > 0, ∆surrS = 0
Solution
In expansion of CO2 gas molecule, entropy of system ∆sysS
increases or ∆sysS > 0 and entropy of surrounding decreases ∆surrS < 0

Q6.Arrange N-H, O-H and F-H bonds in the decreasing order of bond energy
•  F-H > O-H > N-H
•  N-H > O-H > F-H
• O-H > N-H > F-H
•  F-H > N-H > O-H
Solution
Fluorine is more electron-negative than oxygen and oxygen is
more electro-negative than nitrogen and hence bond energy
between F-H in greater than O-H which is greater than N-H

Q7.The free energy for a reaction having ∆H = 31400 cal, ∆S = 32 cal K-1mol-1 at 1000℃ is
•  -9336 cal
•  -7006 cal
•  -2936 cal
•  +9006 cal
Solution
∆G = ∆H - T∆S

Q8.The entropy change in an adiabatic process is
•  Zero
•  Positive
•  Negative
•  Remains true
Solution
In adiabatic process, q = 0
and ∆S = q/T
∴ ∆S = 0

Q9.If w1,w2,w3 and w4 are work done in isothermal, adiabatic, isobaric, and isochoric reversible processes, respectively then the correct sequence (for expansion) would be
•  w1 < w2 < w3 < w4
•  w1 = w2 = w3 = w4
•  w3 < w2 < w4 < w1
•  w3 > w1 > w2 > w4
Solution
w = Work done = Area under curve
Q10. In which of the following cases ∆H and ∆U are not equal to each other?
•  The reaction involves no gaseous reactant and product
•  The number of moles of gaseous reactants and gaseous product is not equal to each other
•  The number of moles of gaseous reactants and gaseous products is equal to each other
• The process is carried out in closed vessel
Solution
If ∆ng = 0, then only ∆H = ∆U.

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