THERMODYNAMICS Quiz-11

CHEMISTRY THERMODYNAMICS QUIZ-11

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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..

Q1.The standard heat of formation of U₃O₈ is -853.5 kcal mol^-1 and standard heat of the reaction, 3UO₂ + O₂ → U₃O₈ is - 76.01 kcal. The standard, heat of formation of UO₂ is/are ::

•  -1083 kJ mol^-1
•  -1102 kJ mol^-1
•  -259 kcal mol^-1
•  -302 kcal mol^-1

Solution
-259 kcal mol^-1

Q2.

•  14.24 kJ
•  16.56 kJ
•  -7.2 kJ
•  4.88 kJ

Solution
(d)
Use Hess’ law

Q3.All natural processes proceed spontaneously in a direction which:

•  Increases entropy
•  Increases free energy
•  Decreases entropy
•  Decreases free energy

Solution
Decreases free energy

Q4.Hess’ law is applicable for determination of enthalpy of

•  Reaction
•  Formation
•  Transition
•  All of these

Solution
All of these

Q5.Which is an irreversible process?

•  Mixing of two gases by diffusion
•  Evaporation of water at 373 K and 1 atm pressure
•  Dissolution of NaCl in water
•  All of the above

Solution
 All of the above

Q6.Which of the following statements is/are false?

•  Work is a state function
•  Temperature is a state function
• Change in the state is completely defined when the initial and final states are specified
•  Work appears at the boundary of the system

Solution
Work is not a state function, but it is a path function

Q7.Work done on the gas in single stage compression is

•  144 bar-L
•  98 bar-L
•  54 bar-L
•  121 bar-L

Solution
pV = nRT
2×8 = 2×0.080×T
W_irr = -p_ext(V2-V1)
= -20(nRT/p₂ - nRT/p₁) = 144 bar-L

Q8.The specific heat of a gas at constant volume is 0.075 cal/g. Predict the atomicity of the gas. Molar mass of gas is 40 g mol^-1

•  3
•  2
•  1
•  None of these

Solution
C_V = 0.075*40 = 3 cal
C_p-C_V = 4
C_p - 3 = 2
C_p = 5 cal
γ = C_p/C_V = 5/3 = 1.66
Thus, gas is monatomic

Q9.When CaCO₃ is heated to a high temperature, it undergoes decomposition into CaO and CO₂ whereas it is quite stable at room temperature. The most likely explanation of it is

•  The enthalpy of reaction (∆H) overweigths the term T∆S at high temperature
•  The term T∆S overweigths the enthalpy of reaction at high temperature
•  At high temperature, both enthalpy of reaction and entropy change become negative
•  None of these

Solution
∆G = ∆H - T∆S, at high temperature T∆S factor dominates of ∆H and hence ∆G becomes negative and reaction occurs spontaneously

Q10. Given that:
A(s) ⟶ A(l) ; ∆H = x
A(l) ⟶ A(g) ; ∆H = y
The heat of sublimation of A will be

•  x - y
•  x + y
•  x or y
• -(x+ y)

Solution
A(s) ⟶ A(l) ; ∆H = x
A(l) ⟶ A(g) ; ∆H = y
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On addition A(s) ⟶ A(g) ; ∆H = x + y
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