SPEED TEST 6
Type: Custom Test Questions: 135 Total Marks: 540 Difficulty: Easy, Medium
Topics covered
Chemistry
Thermodynamics (Chemistry) - Importance and Limitations of Thermodynamics, Basic Terminology of Thermodynamics,
Work and Heat , Internal Energy and Its Characteristics, First Law of Thermodynamics (c), Enthalpy, Calorimeter, Joule –
Thomson Effect, Thermochemistry, Heat of Reaction, Laws of Thermochemistry, Types of Heat of Reactions, Statements of
Second Law of Thermodynamics , Spontaneity, Entropy (c), Gibbs Free Energy and Standard Free Energy Change, Gibbs
Energy Change and Equilibrium, Zeroth Law of Thermodynamics, Absolute Entropy and Third Law of Thermodynamics,
Coupled reactions, Miscellaneous questions on Thermodynamics
Biology
Cell : The Unit of Life - Cell Theory - Introduction, Overview of Cell, Prokaryotic Cell, Eukaryotic Cell, Cell Membrane, Cell
Wall, Endo Membrane System, Mitochondria, Plastids, Ribosomes, Cytoskeleton, Cilia and Flagella, Centriole, Nucleus,
Microbodies
Biomolecules - Introduction, Primary and Secondary Metabolites, Biomacromolecules, Proteins, Polysaccharides, Nucleic
Acid, Structure of Proteins, Nature of Bond Linking Monomers in a Polymer, Dynamic State of Body Constituents Concept of
Metabolism, Metabolic Basis for Living , The Living State, Enzymes
1. Calculate the change of entropy for the process, water 3. Assertion: For isothermal expansion of an ideal gas,
(liquid) to water (vapour) involving its enthalpy decreases.
ΔH vap
= 40850 J mol at 373 K .
−1
(1) ΔS vap = 98.5 J K
−1
mol
−1
Reason: Enthalpy depends on temperature.
(2) ΔS vap = 109.52 J K
−1
mol
−1
(1) Both Assertion and Reason are correct and Reason
(3) ΔS vap = 89 J K
−1
mol
−1
is the correct explanation of the Assertion.
(4) ΔS vap = 72 J K
−1
mol
−1 (2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
2. Read the Assertion and Reason carefully to mark (3) Assertion is correct but Reason is incorrect.
correct option given below: (4) Assertion is incorrect but Reason is correct.
Assertion: The laws of thermodynamics are applicable 4. Read the Assertion and Reason carefully to mark
only when a system is in a non-equilibrium state. correct option given below:
Reason: Thermodynamics is concerned with the state Assertion: The standard enthalpy of combustion is
functions which describe equilibrium states. always more negative than the standard enthalpy of
(1) Both Assertion and Reason are correct and Reason formation.
is the correct explanation of the Assertion.
Reason: Combustion is a specific type of reaction that
(2) Both Assertion and Reason are correct but Reason
involves the reaction of a substance with O2 to form
is not the correct explanation of the Assertion.
combustion products.
(3) Assertion is correct but Reason is incorrect.
(1) Both Assertion and Reason are correct and Reason
(4) Assertion is incorrect but reason is correct.
is the correct explanation of the Assertion.
(2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
(3) Assertion is correct but Reason is incorrect.
(4) Assertion is incorrect but reason is correct.
,5. Given 9. Heat of neutralisation of a weak dibasic acid by NaOH
∘ −1
C + 2 S → CS 2 , Δ f H = +117. 0 kJ mol
is −26kcalmol . Hence, its dissociation energy is
−1
∘ −1
C+O 2 → CO 2 , Δ f H = −393. 0 kJ mol
(1) 0.7 k cal mol
−1
∘ −1
S+O 2 → SO 2 , Δ f H = −297. 0 kJ mol
(2) 1.4 k cal mol
−1
The heat of combustion of
CS 2 + 3O 2 → CO 2 + 2SO 2 is (3) 13.0 k cal mol
−1
(1) −807 kJ mol
−1 (4) None of these
(2) −1104 kJ mol
−1
10. ΔH and ΔE for the reaction,
(3) +1104 kJ mol
−1
Fe 2 O 3 ( s) + 3H 2 ( g) → 2Fe(s) + H 2 O(l) at
constant temperature are related as
(4) +807 kJ mol
−1
(1) ΔH = ΔE
6. Read the Statement - A and Statement - B carefully to
(2) ΔH = ΔE + RT
mark the correct option given below:
(3) ΔH = ΔE + 3RT
For a spontaneous For a spontaneous
(4) ΔH = ΔE − 3RT
process, the entropy of process, the entropy of
the system always the universe always
11. The following graph indicates the system containing 1
increases. increases.
mole of gas involving various steps. When it moves
(1) Both Statement A and Statement B are true. from Z to X, the type of undergoing process is
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(4) Both Statement A and Statement B are false.
7. The equation,
1 1
H2 + Cl 2 → HCl; ΔH 298
2 2
= −22.060 kcal means
(1) The heat absorbed when one gram molecule of HCl
is formed from its elements at 25 C is 22.060 kcal.
∘
(2) The heat given out when one gram molecule of HCl
is formed from its elements at 298 K is 22.060 kcal.
(3) The heat absorbed when one atom of hydrogen
reacts with one atom of chlorine to form one (1) cyclic
molecule of HCl at 25 C and one atom of chlorine
∘
(2) isothermal
to form one molecule of HCl at ∘
25 C and one (3) isochoric
atmosphere pressure is 22.060 kcal.
(4) isobaric
(4) The heat absorbed when one gram equivalent of
HCl is formed from its elements at 298 K is 22.060 12. Read the Statement - A and Statement - B carefully to
kcal . mark the correct option given below:
8. The work done when 2 moles of an ideal gas A reaction with a positive The spontaneity of a
expands reversibly and isothermally from a ΔS and a negative ΔH reaction can be reversed
volume of 1 L to 10 L at 300 K is will always be by changing the pressure
( R = 0.0083 kJKmol ) −1
spontaneous. conditions.
(1) 0.115 kJ
(1) Both Statement A and Statement B are true.
(2) 58.5 kJ
(2) Statement A is true, but Statement B is false.
(3) 11.5 kJ (3) Statement A is false, but Statement B is true.
(4) 5.8 kJ
(4) Both Statement A and Statement B are false.
, 13. At a particular temperature 18. The volume of gas is reduced to half from its original
H
+
+ OH
−
→ H 2 O (l) ; ΔH = −57.1 kJ volume. The specific heat will be
(aq) (aq)
The approximate heat evolved when 400 mL of (1) reduce to half
0.2 M H 2 SO 4 is mixed with 600 mL of 0.1 M KOH (2) be doubled
solution will be (3) remain constant
(1) 3.426 kJ (4) increase four times
(2) 13.7 kJ 19. The enthalpy changes for the following processes are
(3) 5.2 kJ
listed below
−1
Cl 2 ( g) → 2Cl(g), 242.3 kJ mol
(4) 55 kJ
−1
I 2 ( g) → 2I(g), 151.0 kJ mol
14. Molar heat capacity of water in equilibrium with ice at ICl(g) → I(g) + Cl(g), 211.3 kJ mol
−1
constant pressure is I 2 ( s) → I 2 ( g), 62.76 kJ mol
−1
(1) 0 Given that the standard states for iodine and chlorine
(2) ∞ are I ( s) and Cl ( g), the standard enthalpy of
2 2
(3) 1 formation for ICl(g) is
(4) 0.5 (1) −14.6 kJ mol
−1
15. A chemical process is carried out in a thermostat (2) −16.8 kJ mol
−1
maintained at 25 ∘
, it is known as
C (3) +16.8 kJ mol
−1
(1) isothermal (4) +244.8 kJ mol
−1
(2) isobaric
20. This graph expresses the various steps of the system
(3) adiabatic
containing 1 mole of gas. Which type of process does
(4) isotropic
the system have when it moves from C to A ?
16. Entropy changes for the process, H 2 O(l) → H 2 O(s)
at normal pressure and 274 K are given below
ΔS system = −22.13, ΔS surr = +22.05 , the process
is non-spontaneous because
(1) ΔS system is-ve
(2) ΔS surr is + ve
(3) ΔS u is -ve
(4) ΔS system ≠ ΔS surr
('surr' stands for surroundings and 'u ' stands for
universe) (1) Isochoric
(2) Isobaric
17. Assertion: Enthalpy of graphite is lower than that of
(3) Isothermal
diamond.
(4) Cyclic
Reason: Entropy of graphite is greater than that of
diamond.
(1) Both Assertion and Reason are correct and Reason
is the correct explanation of the Assertion.
(2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
(3) Assertion is correct but Reason is incorrect.
(4) Assertion is incorrect but Reason is correct.
Type: Custom Test Questions: 135 Total Marks: 540 Difficulty: Easy, Medium
Topics covered
Chemistry
Thermodynamics (Chemistry) - Importance and Limitations of Thermodynamics, Basic Terminology of Thermodynamics,
Work and Heat , Internal Energy and Its Characteristics, First Law of Thermodynamics (c), Enthalpy, Calorimeter, Joule –
Thomson Effect, Thermochemistry, Heat of Reaction, Laws of Thermochemistry, Types of Heat of Reactions, Statements of
Second Law of Thermodynamics , Spontaneity, Entropy (c), Gibbs Free Energy and Standard Free Energy Change, Gibbs
Energy Change and Equilibrium, Zeroth Law of Thermodynamics, Absolute Entropy and Third Law of Thermodynamics,
Coupled reactions, Miscellaneous questions on Thermodynamics
Biology
Cell : The Unit of Life - Cell Theory - Introduction, Overview of Cell, Prokaryotic Cell, Eukaryotic Cell, Cell Membrane, Cell
Wall, Endo Membrane System, Mitochondria, Plastids, Ribosomes, Cytoskeleton, Cilia and Flagella, Centriole, Nucleus,
Microbodies
Biomolecules - Introduction, Primary and Secondary Metabolites, Biomacromolecules, Proteins, Polysaccharides, Nucleic
Acid, Structure of Proteins, Nature of Bond Linking Monomers in a Polymer, Dynamic State of Body Constituents Concept of
Metabolism, Metabolic Basis for Living , The Living State, Enzymes
1. Calculate the change of entropy for the process, water 3. Assertion: For isothermal expansion of an ideal gas,
(liquid) to water (vapour) involving its enthalpy decreases.
ΔH vap
= 40850 J mol at 373 K .
−1
(1) ΔS vap = 98.5 J K
−1
mol
−1
Reason: Enthalpy depends on temperature.
(2) ΔS vap = 109.52 J K
−1
mol
−1
(1) Both Assertion and Reason are correct and Reason
(3) ΔS vap = 89 J K
−1
mol
−1
is the correct explanation of the Assertion.
(4) ΔS vap = 72 J K
−1
mol
−1 (2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
2. Read the Assertion and Reason carefully to mark (3) Assertion is correct but Reason is incorrect.
correct option given below: (4) Assertion is incorrect but Reason is correct.
Assertion: The laws of thermodynamics are applicable 4. Read the Assertion and Reason carefully to mark
only when a system is in a non-equilibrium state. correct option given below:
Reason: Thermodynamics is concerned with the state Assertion: The standard enthalpy of combustion is
functions which describe equilibrium states. always more negative than the standard enthalpy of
(1) Both Assertion and Reason are correct and Reason formation.
is the correct explanation of the Assertion.
Reason: Combustion is a specific type of reaction that
(2) Both Assertion and Reason are correct but Reason
involves the reaction of a substance with O2 to form
is not the correct explanation of the Assertion.
combustion products.
(3) Assertion is correct but Reason is incorrect.
(1) Both Assertion and Reason are correct and Reason
(4) Assertion is incorrect but reason is correct.
is the correct explanation of the Assertion.
(2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
(3) Assertion is correct but Reason is incorrect.
(4) Assertion is incorrect but reason is correct.
,5. Given 9. Heat of neutralisation of a weak dibasic acid by NaOH
∘ −1
C + 2 S → CS 2 , Δ f H = +117. 0 kJ mol
is −26kcalmol . Hence, its dissociation energy is
−1
∘ −1
C+O 2 → CO 2 , Δ f H = −393. 0 kJ mol
(1) 0.7 k cal mol
−1
∘ −1
S+O 2 → SO 2 , Δ f H = −297. 0 kJ mol
(2) 1.4 k cal mol
−1
The heat of combustion of
CS 2 + 3O 2 → CO 2 + 2SO 2 is (3) 13.0 k cal mol
−1
(1) −807 kJ mol
−1 (4) None of these
(2) −1104 kJ mol
−1
10. ΔH and ΔE for the reaction,
(3) +1104 kJ mol
−1
Fe 2 O 3 ( s) + 3H 2 ( g) → 2Fe(s) + H 2 O(l) at
constant temperature are related as
(4) +807 kJ mol
−1
(1) ΔH = ΔE
6. Read the Statement - A and Statement - B carefully to
(2) ΔH = ΔE + RT
mark the correct option given below:
(3) ΔH = ΔE + 3RT
For a spontaneous For a spontaneous
(4) ΔH = ΔE − 3RT
process, the entropy of process, the entropy of
the system always the universe always
11. The following graph indicates the system containing 1
increases. increases.
mole of gas involving various steps. When it moves
(1) Both Statement A and Statement B are true. from Z to X, the type of undergoing process is
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(4) Both Statement A and Statement B are false.
7. The equation,
1 1
H2 + Cl 2 → HCl; ΔH 298
2 2
= −22.060 kcal means
(1) The heat absorbed when one gram molecule of HCl
is formed from its elements at 25 C is 22.060 kcal.
∘
(2) The heat given out when one gram molecule of HCl
is formed from its elements at 298 K is 22.060 kcal.
(3) The heat absorbed when one atom of hydrogen
reacts with one atom of chlorine to form one (1) cyclic
molecule of HCl at 25 C and one atom of chlorine
∘
(2) isothermal
to form one molecule of HCl at ∘
25 C and one (3) isochoric
atmosphere pressure is 22.060 kcal.
(4) isobaric
(4) The heat absorbed when one gram equivalent of
HCl is formed from its elements at 298 K is 22.060 12. Read the Statement - A and Statement - B carefully to
kcal . mark the correct option given below:
8. The work done when 2 moles of an ideal gas A reaction with a positive The spontaneity of a
expands reversibly and isothermally from a ΔS and a negative ΔH reaction can be reversed
volume of 1 L to 10 L at 300 K is will always be by changing the pressure
( R = 0.0083 kJKmol ) −1
spontaneous. conditions.
(1) 0.115 kJ
(1) Both Statement A and Statement B are true.
(2) 58.5 kJ
(2) Statement A is true, but Statement B is false.
(3) 11.5 kJ (3) Statement A is false, but Statement B is true.
(4) 5.8 kJ
(4) Both Statement A and Statement B are false.
, 13. At a particular temperature 18. The volume of gas is reduced to half from its original
H
+
+ OH
−
→ H 2 O (l) ; ΔH = −57.1 kJ volume. The specific heat will be
(aq) (aq)
The approximate heat evolved when 400 mL of (1) reduce to half
0.2 M H 2 SO 4 is mixed with 600 mL of 0.1 M KOH (2) be doubled
solution will be (3) remain constant
(1) 3.426 kJ (4) increase four times
(2) 13.7 kJ 19. The enthalpy changes for the following processes are
(3) 5.2 kJ
listed below
−1
Cl 2 ( g) → 2Cl(g), 242.3 kJ mol
(4) 55 kJ
−1
I 2 ( g) → 2I(g), 151.0 kJ mol
14. Molar heat capacity of water in equilibrium with ice at ICl(g) → I(g) + Cl(g), 211.3 kJ mol
−1
constant pressure is I 2 ( s) → I 2 ( g), 62.76 kJ mol
−1
(1) 0 Given that the standard states for iodine and chlorine
(2) ∞ are I ( s) and Cl ( g), the standard enthalpy of
2 2
(3) 1 formation for ICl(g) is
(4) 0.5 (1) −14.6 kJ mol
−1
15. A chemical process is carried out in a thermostat (2) −16.8 kJ mol
−1
maintained at 25 ∘
, it is known as
C (3) +16.8 kJ mol
−1
(1) isothermal (4) +244.8 kJ mol
−1
(2) isobaric
20. This graph expresses the various steps of the system
(3) adiabatic
containing 1 mole of gas. Which type of process does
(4) isotropic
the system have when it moves from C to A ?
16. Entropy changes for the process, H 2 O(l) → H 2 O(s)
at normal pressure and 274 K are given below
ΔS system = −22.13, ΔS surr = +22.05 , the process
is non-spontaneous because
(1) ΔS system is-ve
(2) ΔS surr is + ve
(3) ΔS u is -ve
(4) ΔS system ≠ ΔS surr
('surr' stands for surroundings and 'u ' stands for
universe) (1) Isochoric
(2) Isobaric
17. Assertion: Enthalpy of graphite is lower than that of
(3) Isothermal
diamond.
(4) Cyclic
Reason: Entropy of graphite is greater than that of
diamond.
(1) Both Assertion and Reason are correct and Reason
is the correct explanation of the Assertion.
(2) Both Assertion and Reason are correct but Reason
is not the correct explanation of the Assertion.
(3) Assertion is correct but Reason is incorrect.
(4) Assertion is incorrect but Reason is correct.
