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JEE • CHEMISTRY • THERMODYNAMICS
Thermodynamics practice · Medium
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Practice format and chapter quiz
These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.
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Showing 241-270 of 329 questions
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241
Medium
241 of 329
Understanding
+4 | -1
On increasing temperature, the equilibrium constant of exothermic and endothermic reactions repectively
View answer and explanation
Correct option: B
Decreases and increases
Option B (Decreases and increases) is correct because increasing temperature favors the endothermic reaction, shifting the equilibrium and increasing its constant, while it disfavors the exothermic reaction, decreasing its constant.
242
Medium
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Application
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1 Mole of ideal gas expanded reversibly and isothermally from 1 lit to 10 lit.
then find out the entropy change in cal/Kelvin.
then find out the entropy change in cal/Kelvin.
View answer and explanation
Correct option: D
4.606
Option D (4.606) is correct because the entropy change for an isothermal reversible expansion is calculated using the formula .
243
Medium
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Application
+4 | -1
For a certain reaction the change in enthalpy and change in entropy are and respectively. What is the value of at
View answer and explanation
Correct option: A
Option A () is correct because it is the calculated value of Δ G using the Gibbs free energy equation.
244
Medium
244 of 329
Analyzing
+4 | -1
Which statement(s) is/are correct?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given statements regarding thermodynamic properties are correct and follow standard thermodynamic relationships.
245
Medium
245 of 329
Remembering
+4 | -1
Standard molar enthalpy of formation at is zero for
View answer and explanation
Correct option: A
Option A (Br₂(l)) is correct because the standard molar enthalpy of formation is zero for an element in its standard state at a specified temperature (298 K). Bromine exists as a liquid at 298K.
246
Medium
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Analyzing
+4 | -1
Given,
Select the correct statement :
Select the correct statement :
View answer and explanation
Correct option: B
can be reduced by spontaneously
Option B ( can be reduced by spontaneously) is correct because the overall Gibbs free energy change is negative, indicating spontaneity.
247
Medium
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Application
+4 | -1
ice absorbs of heat to melt at . The entropy change will be
View answer and explanation
Correct option: C
Option C () is correct because it is the entropy change calculated using the formula for the melting of ice.
248
Medium
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Analyzing
+4 | -1
Statement I: Heat evolved in the neutralization of either 1 mole of or 1 mole of with is same.
Statement II: Both and are strong acids.
Statement II: Both and are strong acids.
View answer and explanation
Correct option: D
If Statement I is INCORRECT, but Statement II is CORRECT
Option D is correct because the heat evolved in the neutralization of 1 mole of with is different from that of 1 mole of because is a dibasic acid, while is a monobasic acid.
249
Medium
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Understanding
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Among the following the set of parameters that represents path functions, is
(A) (B) (C) (D)
(A) (B) (C) (D)
View answer and explanation
Correct option: D
(B) and (C)
Option D ((B) and (C)) is correct because heat () and work () are path functions, meaning their values depend on the path taken during a process.
250
Medium
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Remembering
+4 | -1
Identify the correct statement regarding entropy
View answer and explanation
Correct option: C
At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero
Option C (At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero) is correct because the Third Law of Thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero (0 K) is zero.
251
Medium
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Application
+4 | -1
The heat of combustion of and are and respectively. The heat of formation of methane is
View answer and explanation
Correct option: C
Option C (-80 kcal) is correct because it correctly applies Hess's Law to calculate the heat of formation.
252
Medium
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Application
+4 | -1
The entropy change for vaporisation of liquid water to steam at 100℃ is…Given that heat of vaporisation is
View answer and explanation
Correct option: A
109.38
Option A (109.38) is correct because the entropy change is calculated by dividing the heat of vaporization by the temperature in Kelvin.
253
Medium
253 of 329
Remembering
+4 | -1
Specific heat is defined as:
View answer and explanation
Correct option: A
Heat capacity/g
Option A (Heat capacity/g) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
254
Medium
254 of 329
Remembering
+4 | -1
The enthalpy of a monoatomic gas at Kelvin is:
View answer and explanation
Correct option: D
Option D () is correct because the enthalpy of a monoatomic gas is given by H = E + PV = .
255
Medium
255 of 329
Application
+4 | -1
At 1 mole of butane upon heating forms and . The work done is
View answer and explanation
Correct option: B
atm
Option B () is correct because it involves calculating the work done using the change in the number of moles of gas and the ideal gas constant.
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256
Medium
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Application
+4 | -1
The enthalpy of neutralisation of and is and enthalpy of neutralisation of with strong base is . The enthalpy of ionisation of will be
View answer and explanation
Correct option: C
Option C () is correct because it is calculated by considering the enthalpy of neutralization of a weak acid with a weak base and a weak acid with a strong base.
257
Medium
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Application
+4 | -1
Given that ;
, the standard molar enthalpy of formation in of is
, the standard molar enthalpy of formation in of is
View answer and explanation
Correct option: D
-46
Option D (-46) is correct because the standard molar enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed.
258
Medium
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Analyzing
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Butane exists in various conformations in nature. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed and fully eclipsed conformation is and respectively. If the molar enthalpy of combustion of natural butane is at then calculate the enthalpy of combustion of butane if all the butane molecules are in gauche conformation. The potential energy of gauche form is higher than anti form but it is less than eclipsed form and less than fully eclipsed form.
View answer and explanation
Correct option: C
Option C (-691 kcal/mol) is correct because it calculates the enthalpy of combustion for the gauche conformation based on the given probabilities and energy differences.
259
Medium
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Understanding
+4 | -1
According to the first law of thermodynamics which of the following quantities represents change in a state function?
View answer and explanation
Correct option: D
Option D () is correct because, according to the first law of thermodynamics, the change in internal energy, a state function, is equal to the sum of heat and work.
260
Medium
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Application
+4 | -1
1 mol of an ideal gas expanded reversibly from 10 L to 100 L at Work done involved in the process is
View answer and explanation
Correct option: B
-5.74 kJ
Option B (-5.74 kJ) is correct because the work done in a reversible isothermal expansion is calculated using the formula W = -2.303 nRT log(V2/V1).
261
Medium
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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 (An adiabatic process) is correct because an adiabatic process is defined as a thermodynamic process in which there is no heat exchange between the system and its surroundings (ΔQ = 0).
262
Medium
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Application
+4 | -1
The enthalpy changes for the following processes are listed below :
Given that the standard states for iodine and chlorine are and , the standard enthalpy of formation for is :
Given that the standard states for iodine and chlorine are and , the standard enthalpy of formation for is :
View answer and explanation
Correct option: A
Option A (+16.8 kJ mol⁻¹) is correct because it is calculated using Hess's Law and the given enthalpy changes.
263
Medium
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Application
+4 | -1
The heat of formations of and are and respectively. The heat of combustion of carbon monoxide will be
View answer and explanation
Correct option: D
Option D (-67.6 kcal) is correct because it is the result of subtracting the heat of formation of CO from that of CO2.
264
Medium
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Application
+4 | -1
Consider a reversible isontropic expansion of 1.0 mole of an ideal monoatomic gas from to . If the initial pressure was , final pressure is
View answer and explanation
Correct option: A
1.474 bar
Option A (1.474 bar) is correct because it is calculated using the isentropic process formula relating pressure and temperature.
265
Medium
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Application
+4 | -1
If the bond energies of and are 433,192 and mol respectively the for the reaction is:
View answer and explanation
Correct option: C
Option C (-103 kJ) is correct because it calculates the enthalpy change of the reaction using bond energies of reactants and products.
266
Medium
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Understanding
+4 | -1
Which of the following on combustion gives
maximum energy?
maximum energy?
View answer and explanation
Correct option: D
Butane
Option D (Butane) is correct because, for alkanes, the heat of combustion increases with the number of carbon atoms.
267
Medium
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Understanding
+4 | -1
A process which is unfavourable with respect to enthalpy,
but favourable with respect to entropy
but favourable with respect to entropy
View answer and explanation
Correct option: B
could occur at high temperatures, but not at low
temperatures
temperatures
Option B is correct because a process unfavorable in enthalpy but favorable in entropy can occur at high temperatures where the entropy term dominates the Gibbs free energy equation.
268
Medium
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Understanding
+4 | -1
Heat absorbed at constant volume is measured in a calorimeter in which steel vessel is immersed to ensure that
View answer and explanation
Correct option: C
No heat is lost to the surrounding
Option C is correct because a calorimeter is designed to prevent heat exchange with the surroundings, ensuring accurate measurement of heat absorbed at constant volume.
269
Medium
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Analyzing
+4 | -1
Assertion (A): The dissolution of in water is endothermic but it dissolves in water.
Reason (R): On dissolution of , entropy increases.
Reason (R): On dissolution of , entropy increases.
View answer and explanation
Correct option: B
Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion
Option B (Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion) is correct because while NH4Cl dissolution is endothermic and dissolves in water, the driving force is entropy increase, not a direct consequence of endothermicity.
270
Medium
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Remembering
+4 | -1
The molar heat capacity, of an ideal gas whose energy is that of translational motion only is
View answer and explanation
Correct option: B
Option B () is correct because for an ideal gas with only translational motion, Cv = (3/2)R, which equals 12.47 J/mol·K.
Revision notes and key takeaways
Review THERMODYNAMICS in your textbook or class notes, then use this set to check your understanding.
Showing 241-270 of 329 questions
