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Then the heat of formation of CO is
2Na(s)+2HCl(g)→2NaCl(s)+ (g);
∆H= -152 kcal then,
and are diatomic molecules. If the bond enthalpies of and are in the ratio , then the bond enthalpy of is ____ (Nearest integer)
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NEET • CHEMISTRY • THERMODYNAMICS
Thermodynamics practice · Mixed difficulty
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31
Easy
31 of 560
Application
+4 | -1
If 1 mole of gas expands from 1 litre to 5 litre against constant atmospheric
pressure than calculate the work done.
pressure than calculate the work done.
View answer and explanation
Correct option: A
Option A (-4) is correct because the work done against constant pressure is calculated as , where is the change in volume.
32
Medium
32 of 560
Understanding
+4 | -1
Hess’s law of constant heat summation is an application of :
View answer and explanation
Correct option: B
First law of thermodynamics
The correct answer is B (First law of thermodynamics) because Hess's law is a direct consequence of the fact that enthalpy is a state function, which is a restatement of the first law.
33
Medium
33 of 560
Understanding
+4 | -1
In which of the following cases entropy decreases?
View answer and explanation
Correct option: D
Polymerisation
Option D, Polymerisation, is correct because it involves combining many small molecules into a large one, decreasing the number of possible arrangements and thus decreasing entropy.
34
Hard
34 of 560
Analyzing
+4 | -1
An ideal gas was subjected to a reversible adiabatic expansion then its initial volume was restored by reversible, isothermal compression. If ' ' denotes the heat added to the system and the work done on the system for the overall process then
View answer and explanation
Correct option: A
w , q
Option A (w < 0, q < 0) is correct because in a reversible adiabatic expansion, no heat is exchanged (q=0), and the gas does work, so w < 0. During isothermal compression, work is done on the system, and heat is released, so w < 0 and q < 0 overall.
35
Easy
35 of 560
Analyzing
+4 | -1
Select the incorrect statement
View answer and explanation
Correct option: B
For a closed system with P-V work only, an isobaric process that has q=+ve must have ∆T=+ve.
Option B is the incorrect statement. For an isobaric process with positive heat (q > 0), the temperature change (∆T) must be positive according to the first law of thermodynamics.
36
Medium
36 of 560
Application
+4 | -1
One mono atomic gas is expanded adiabatically from to at external pressure, find (in )
View answer and explanation
Correct option: A
Option A (-8) is correct because the change in internal energy is equal to the work done in an adiabatic process against constant external pressure, which is -PΔV.
37
Hard
37 of 560
Application
+4 | -1
A piece of iron (specific heat capacity ) at is immersed in water of .
What will be the temperature of this system at thermal equilibrium?
What will be the temperature of this system at thermal equilibrium?
View answer and explanation
Correct option: A
31
Option A (31) is correct because it is calculated using the principle of calorimetry, where heat lost by iron equals heat gained by water until thermal equilibrium is reached.
38
Medium
38 of 560
Analyzing
+4 | -1
Then the heat of formation of CO is
View answer and explanation
Correct option: A
X-Y
Option A (X-Y) is correct because the heat of formation of CO can be found by subtracting the heat of the second reaction from the heat of the first reaction based on Hess's Law.
39
Medium
39 of 560
Remembering
+4 | -1
Which one among the following is the correct option for right relationship between for one mole of ideal gas?
View answer and explanation
Correct option: D
Option D, , is the correct relationship between and for one mole of an ideal gas.
40
Easy
40 of 560
Remembering
+4 | -1
Which is the correct Gibb's Helmholtz
equation?
equation?
View answer and explanation
Correct option: C
Option C, , is the correct Gibb's Helmholtz equation relating change in Gibbs free energy, enthalpy, and entropy at a given temperature.
41
Medium
41 of 560
Application
+4 | -1
The dissociation energy of and are respectively 360 and 620 kcal/mol. The bond energy of C-C bond is:
View answer and explanation
Correct option: D
80 kcal/mol
Option D (80 kcal/mol) is correct. It is calculated by using the given dissociation energies and assuming that the dissociation energy is the sum of the bond energies.
42
Easy
42 of 560
Understanding
+4 | -1
Which of the following is not a thermodynamic property of a system?
View answer and explanation
Correct option: D
Option D () is correct because work () is a path function, not a thermodynamic property of the system itself, while H, P, and E are state functions.
43
Hard
43 of 560
Application
+4 | -1
The enthalpy of vaporization of a liquid is 30 kJ and entropy of vaporization is 75 J . The boiling point of liquid at 1 atm is :
View answer and explanation
Correct option: B
400 K
Option B (400 K) is correct. It is calculated using the Clausius-Clapeyron equation, where boiling point is determined by the enthalpy and entropy of vaporization.
44
Easy
44 of 560
Remembering
+4 | -1
For one mole of an ideal gas
View answer and explanation
Correct option: A
Option A () is correct because it represents the relationship between molar heat capacities at constant pressure and volume for an ideal gas.
45
Easy
45 of 560
Application
+4 | -1
The enthalpy and entropy change for a chemical reaction are respectively.Predict that nature of reaction at 298 K is:
View answer and explanation
Correct option: A
Spontaneous
Option A (Spontaneous) is correct because the Gibbs free energy change is negative, indicating a spontaneous reaction at 298 K.
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46
Easy
46 of 560
Understanding
+4 | -1
A reaction occurs spontaneously if:
View answer and explanation
Correct option: B
T∆S>∆H and both ∆H and ∆S are positive
Option B (T∆S>∆H and both ∆H and ∆S are positive) is correct because for a reaction to be spontaneous, the change in Gibbs free energy (∆G) must be negative, which occurs when T∆S > ∆H, given that both ∆H and ∆S are positive.
47
Medium
47 of 560
Application
+4 | -1
2Na(s)+2HCl(g)→2NaCl(s)+ (g);
∆H= -152 kcal then,
View answer and explanation
Correct option: C
Option C (-98 kcal) is correct. By manipulating the given equations using Hess's Law, the enthalpy change for the desired reaction is calculated.
48
Medium
48 of 560
Understanding
+4 | -1
Which statements are correct?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the statements regarding the Clausius-Clapeyron equation, Trouton's rule, and the definition of entropy are accurate.
49
Medium
49 of 560
Remembering
+4 | -1
A process in which no heat change takes place is called.
View answer and explanation
Correct option: B
An adiabatic process
Option B is correct because an adiabatic process is defined as one in which no heat is exchanged between the system and its surroundings.
50
Easy
50 of 560
Remembering
+4 | -1
Which is true for internal energy of system is
View answer and explanation
Correct option: A
State function
Option A (State function) is correct because internal energy is a property that depends only on the current state of the system, not on the path taken to reach that state.
51
Hard
51 of 560
Analyzing
+4 | -1
Statement I: A reaction which is spontaneous and
accompanied by decrease of randomness must be
exothermic.
Statement II: All exothermic reactions are
accompanied by decrease of randomness.
accompanied by decrease of randomness must be
exothermic.
Statement II: All exothermic reactions are
accompanied by decrease of randomness.
View answer and explanation
Correct option: C
If Statement I is CORRECT, but Statement II is INCORRECT
Option C is correct. Statement I is correct because for a spontaneous reaction with decreasing randomness (negative ΔS), the reaction must be exothermic (negative ΔH) to ensure ΔG is negative. Statement II is incorrect because not all exothermic reactions decrease randomness.
52
Easy
52 of 560
Understanding
+4 | -1
A boiled egg show a/an …in entropy.
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because boiling an egg involves denaturing proteins, which increases the disorder and thus the entropy of the system.
53
Medium
53 of 560
Analyzing
+4 | -1
Statement I: The final temperature of an ideal
gas in adiabatic compression is less in reversible
compression than in irreversible compression
against a constant external pressure.
Statement II: The magnitude of work done on
an ideal gas in adiabatic compression is more in
irreversible process than that in reversible process.
gas in adiabatic compression is less in reversible
compression than in irreversible compression
against a constant external pressure.
Statement II: The magnitude of work done on
an ideal gas in adiabatic compression is more in
irreversible process than that in reversible process.
View answer and explanation
Correct option: A
If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I
Option A is correct. Irreversible adiabatic compression results in a higher final temperature and more work done compared to reversible adiabatic compression because energy is dissipated as heat in the irreversible process.
54
Medium
54 of 560
Analyzing
+4 | -1
For one mole of an ideal gas, which of these statements must be true?
A) each depends only on temperature
B) Compressibility factor Z is not equal to 1.
C)
D) for any process.
A) each depends only on temperature
B) Compressibility factor Z is not equal to 1.
C)
D) for any process.
View answer and explanation
Correct option: C
A, C and D
Option C (A, C and D) is correct because for one mole of an ideal gas, U and H depend only on temperature, , and for any process.
55
Easy
55 of 560
Application
+4 | -1
and are diatomic molecules. If the bond enthalpies of and are in the ratio , then the bond enthalpy of is ____ (Nearest integer)
View answer and explanation
Correct option: A
400
Option A (400) is correct because the bond enthalpy of A2 is calculated by using the given ratio of bond enthalpies and the enthalpy of formation.
56
Easy
56 of 560
Application
+4 | -1
A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ∆H is the enthalpy change and ∆E is the change in internal energy, then
View answer and explanation
Correct option: B
∆H<∆E
∆H < ∆E because the reaction results in a decrease in the number of moles of gas, leading to a negative work term.
57
Medium
57 of 560
Analyzing
+4 | -1
The heat of neutralization of by is – 55.9 KJ/mol. If the heat of neutralization of by is – 12.1 KJ/mol. The heat of dissociation of is
View answer and explanation
Correct option: B
43.8 KJ
Option B (43.8 KJ) is correct because the heat of dissociation of HCN is the difference between the heat of neutralization of a strong acid and the heat of neutralization of HCN.
58
Medium
58 of 560
Application
+4 | -1
The heat of combustion of methane is . If 3.2 g of methane is burnt:
View answer and explanation
Correct option: A
- 176 kJ of heat is evolved
Option A (- 176 kJ of heat is evolved) is correct because the heat evolved is directly proportional to the moles of methane burnt.
59
Medium
59 of 560
Remembering
+4 | -1
The second law of thermodynamics introduced the concept of:
View answer and explanation
Correct option: C
Entropy
Option C (Entropy) is correct because the second law of thermodynamics introduces the concept of entropy, which describes the degree of disorder in a system.
60
Easy
60 of 560
Analyzing
+4 | -1
STATEMENT-1 : There cannot be chemical equilibrium in an open system.
STATEMENT-2 : There is no fixed mass in an open system.
STATEMENT-2 : There is no fixed mass in an open system.
View answer and explanation
Correct option: A
If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
Option A is correct because chemical equilibrium requires a closed system with fixed mass, which is absent in an open system.
Revision notes and key takeaways
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