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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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421 Easy
421 of 560
Understanding +4 | -1
The heats of neutralization of four acids A,B,C and D are – 13.7, –9.4, –11.2 and –12.4 kcal respectively when they are neutralized by a common base. The acidic character obeys the order :
View answer and explanation
Correct option: B
A>D>C>B
Option A (A>B>C>D) is correct because the acid with the most negative heat of neutralization is the strongest acid.
422 Medium
422 of 560
Analyzing +4 | -1
In which case will the efficiency of a Carnot cycle be higher, when the hot body temperature is increased by ΔT\Delta T or when the cold body temperature is decreased by the same magnitude?
View answer and explanation
Correct option: B
On decreasing the temperature of cold body
Option B is correct because decreasing the cold body temperature has a greater impact on increasing the Carnot cycle efficiency.
423 Hard
423 of 560
Application +4 | -1
A sample of argon gas at 1 atm1 \mathrm{~atm} pressure and 27∘C27^{\circ} \mathrm{C} expands reversibly and adiabatically from 1.25 dm31.25 \mathrm{~dm}^3 to 2.50 dm3.2.50 \mathrm{~dm}^3. Calculate the enthalpy change in this process. Cv,m\mathrm{C}_{\mathrm{v}, \mathrm{m}} for argon is 12.48 JK−1 mol−112.48 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}
View answer and explanation
Correct option: A
−116 J-116 \text{ J}
Option A (-116 J) is correct because it is the enthalpy change calculated using the adiabatic process equation and the given heat capacity.
424 Medium
424 of 560
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 this is the statement of the Third Law of Thermodynamics.
425 Medium
425 of 560
Analyzing +4 | -1
STATEMENT-1 : The magnitude of the work involved in an isothermal expansion is greater than that involved in an adiabatic expansion.

STATEMENT2 : P−VP-V curve ( PP on yy-axis and VV on xx-axis) decrease more rapidly for reversible adiabatic exparision compared to reversible isothermal expansion starting from same initial state.
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. The magnitude of work in isothermal expansion is greater than in adiabatic expansion because the P-V curve decreases more rapidly in adiabatic expansion, meaning less area under the curve and thus less work done. Statement 2 correctly explains Statement 1.
426 Medium
426 of 560
Understanding +4 | -1
Which one of the following is an exothermic reaction?
View answer and explanation
Correct option: B
N2(g)+3H2(g)−92kJ→2NH3(g){\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) - 92\mathrm{kJ} \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)
Option B is the exothermic reaction because the energy term is on the product side of the equation, indicating heat is released.
427 Medium
427 of 560
Understanding +4 | -1
Which are correct to express work terms?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given expressions correctly represent work in terms of capacity and intensity factors.
428 Easy
428 of 560
Application +4 | -1
For which one of the following system ΔH>ΔE\Delta H > \Delta E
View answer and explanation
Correct option: C
2NH3(g)→N2(g)+3H2(g)2N{H_{3\left( g \right)}} \to {N_{2\left( g \right)}} + 3{H_{2\left( g \right)}}
Option C (2NH3(g)→N2(g)+3H2(g)2N{H_{3(g)}} \to {N_{2(g)}} + 3{H_{2(g)}}) is correct because the number of moles of gaseous products is greater than the number of moles of gaseous reactants, leading to a positive ΔH−ΔE\Delta H - \Delta E.
429 Easy
429 of 560
Application +4 | -1
What is entropy change in 2 mol N22 {~mol} {~N}_2, when its temperature is taken from 400 K400 {~K} to 800 K800 {~K}, adiabatically.
View answer and explanation
Correct option: C
40JK40 \frac{{J}}{{K}}
The correct answer is 40 J/K. Since the process is adiabatic, q = 0. The entropy change is calculated using the formula ΔS=nCvln⁡(T2/T1)\Delta S = nC_v \ln(T_2/T_1) for an ideal gas.
430 Easy
430 of 560
Remembering +4 | -1
Which among the following is not a state function?
View answer and explanation
Correct option: D
Work
Option D (Work) is correct because work is a path function, meaning its value depends on the path taken, while Gibbs energy, pressure, and temperature are state functions.
431 Easy
431 of 560
Understanding +4 | -1
A mixture of hydrogen and chlorine on exposure to ultra violet sunlight reacts with explosion. The step involved in the initiation of the reaction is :
View answer and explanation
Correct option: D
Cl2→Cl∙+Cl∙\mathrm{C}{\mathrm{l}_2} \to \mathrm{C}{\mathrm{l}^ \bullet } + \mathrm{C}{\mathrm{l}^ \bullet }
The correct answer is D (Cl2→Cl∙+Cl∙\mathrm{C}{\mathrm{l}_2} \to \mathrm{C}{\mathrm{l}^ \bullet } + \mathrm{C}{\mathrm{l}^ \bullet }) because the initiation step in the reaction of hydrogen and chlorine involves the homolytic cleavage of chlorine molecules into chlorine radicals.
432 Easy
432 of 560
Understanding +4 | -1
An adiabatic process occurs in
View answer and explanation
Correct option: C
Isolated system
The correct answer is C (Isolated system) because an adiabatic process requires no heat exchange with the surroundings, which is only possible in an isolated system.
433 Medium
433 of 560
Analyzing +4 | -1
Following reaction occurs in an automobile 2C8H18(g)+25 O2(g)→16 CO2(g)+18H2O(g)2{C_8}{H_{18}}(g) + 25\,{O_2}(g) \to 16\,C{O_2}(g) + 18{H_2}O(g) The sign of ΔH, ΔS and  ΔG\Delta H,\,\Delta S\mathrm{ }\,\mathrm{and}\,\,\mathrm{ }\Delta G would be
View answer and explanation
Correct option: B
−,+,−- , + , -
Option B ( - , + , - ) is correct because combustion reactions are exothermic (negative ΔH), increase disorder (positive ΔS), and are spontaneous (negative ΔG).
434 Medium
434 of 560
Analyzing +4 | -1
Statement I : When enthalpy factor is absent then randomness factor decides spontaneity of a process
Statement II : When randomness factor is absent then enthalpy factor decides spontaneity of aprocess
View answer and explanation
Correct option: C
Statement I is correct but statement II is incorrect
Option C is correct. Statement I is correct because if enthalpy is absent (ΔH=0), a positive ΔS leads to a spontaneous process. Statement II is incorrect because exothermic reactions can be non-spontaneous if randomness decreases significantly.
435 Hard
435 of 560
Application +4 | -1
Given the bond energies of N ≡ N, H─H and N─H bonds as 945, 436 and 391 kJ  mol−1kJ\,\,mo{l^{ - 1}}respectively, the enthalpy of the reaction, 2N2(g)+3H2(g)→2NH3(g)2{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) is :
View answer and explanation
Correct option: A
-93 kJ
Option A (-93 kJ) is correct because it correctly calculates the enthalpy of the reaction using the given bond energies and the formula ΔH=∑BEreactants−∑BEproducts\Delta H = \sum{BE_{reactants}} - \sum{BE_{products}}.
436 Medium
436 of 560
Application +4 | -1
Combustion of glucose takes place according to the equation C6H12O6+CO2→6CO2+6H2O;ΔH=−72  kcal{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + \mathrm{C}{\mathrm{O}_2} \to 6\mathrm{C}{\mathrm{O}_2} + 6{\mathrm{H}_2}\mathrm{O};\Delta H = - 72\;kcal How much energy will be required for the production of 1.6 g of glucose (Molecular mass of glucose = 180)?
View answer and explanation
Correct option: B
0.64 kcal
Option B (0.64 kcal) is correct because it is calculated based on the stoichiometry of the reaction and the given enthalpy change.
437 Medium
437 of 560
Analyzing +4 | -1
Consider following statements:
(a) The absolute value of internal energy cannot be
evaluated.
(b) Entropy is a state function and extensive
property
(c) Temperature is an extensive property
(d) The standard enthalpy of formation of an element (ΔHf)\left(\Delta \mathrm{H}_{\mathrm{f}}\right) is always considered to be zero

(e) For an adiabatic process ΔS>0\Delta S>0
wrong statements are :
View answer and explanation
Correct option: C
(c) and (e)
Option C ((c) and (e)) is correct. Temperature is an intensive property, not extensive, and for an adiabatic process, ∆S can be zero (reversible) or > 0 (irreversible).
438 Easy
438 of 560
Understanding +4 | -1
When the change of entropy is greater, then the ability for work is:
View answer and explanation
Correct option: A
Maximum
Option A (Maximum) is correct because a greater change in entropy implies a greater amount of energy is unavailable to do work, thus the ability for work is maximum.
439 Hard
439 of 560
Application +4 | -1
1 mole of an ideal gas is allowed to expand isothermally and reversibly at 27∘C27^{\circ} \mathrm{C} until its volume is tripled. Calculate ΔSsys\Delta \mathrm{S}_{\mathrm{sys}} (Give answer approximately).
View answer and explanation
Correct option: A
9
Option A (9) is correct because the entropy change for an isothermal reversible expansion is calculated using the formula ΔS = nRln(V2/V1).
440 Hard
440 of 560
Application +4 | -1
Consider the following cases of standard enthalpy of reaction (ΔHr∘\left(\Delta \mathrm{H}_{\mathrm{r}}^{\circ}\right. in kJ mol−1)\left.\mathrm{kJ} \mathrm{~mol}^{-1}\right)

C2H6(g)+72O2(g)→2CO2(g)+3H2O(ℓ);ΔH1∘=−1550C(graphite)+O2(g)→CO2(g);ΔH2∘=−393.5H2(g)+12O2(g)→H2O(ℓ);ΔH3∘=−286\begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{g})+\frac{7}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\ell) ;\Delta \mathrm{H}_1^{\circ}=-1550 \\ & \mathrm{C}(\text {graphite})+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) ;\Delta \mathrm{H}_2^{\circ}=-393.5 \\ & \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\ell) ;\Delta \mathrm{H}_3^{\circ}=-286 \end{aligned}

The magnitude of ΔHf C2H6(g)∘\Delta \mathrm{H}_{\text{f }\mathrm{C}_2 \mathrm{H}_6(\mathrm{g})}^{\circ} is ____ kJ mol−1\mathrm{kJ} \mathrm{~mol}^{-1} (Nearest integer).
View answer and explanation
Correct option: A
95
The correct answer is 95 because it involves using Hess's Law to calculate the enthalpy of formation of C2H6 from the given combustion and formation enthalpies.
441 Hard
441 of 560
Application +4 | -1
The standard enthalpies of formation of CO2(g),  H2O(l)C{O_2}\left( g \right),\,\,{H_2}O\left( l \right) and glucose (s) at 25o  C{25^\mathrm{o}}\,\,C are −400  KJ⋅mol−1,  −300  KJ⋅mol−1- 400\,\,KJ \cdot mo{l^{ - 1}},\,\, - 300\,\,KJ \cdot mo{l^{ - 1}} and −1300  KJ/mol- 1300\,\,KJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 25o  C{25^\mathrm{o}}\,\,C is
View answer and explanation
Correct option: C
- 16.11 KJ
Option C (-16.11 KJ) is correct because it is calculated by finding the enthalpy of combustion and dividing by the molar mass of glucose.
442 Easy
442 of 560
Understanding +4 | -1
When a gas undergoes adiabatic expansion, it gets cooled due to
View answer and explanation
Correct option: A
Loss of kinetic energy
Option A (Loss of kinetic energy) is correct because adiabatic expansion involves work done by the gas, which comes at the expense of its internal energy and thus kinetic energy.
443 Medium
443 of 560
Remembering +4 | -1
The enthalpies of all the elements in their standard states are
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because, by convention, the enthalpy of elements in their standard states is defined as zero.
444 Easy
444 of 560
Understanding +4 | -1
The change in the enthalpy during the reaction,  NaOH+HCl→  NaCl+H2O,\;\mathrm{NaOH} + \mathrm{HCl} \to \;\mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O}, is called :
View answer and explanation
Correct option: B
Heat of neutralization
Option B (Heat of neutralization) is correct because the reaction involves the neutralization of an acid (HCl) by a base (NaOH).
445 Easy
445 of 560
Understanding +4 | -1
∆S is positive for the change:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because mixing gases, boiling liquids, and melting solids all increase disorder, resulting in a positive change in entropy.
446 Easy
446 of 560
Remembering +4 | -1
It is impossible to attain the lowest temperature known as zero degree absolute. This is a simple statement of:
View answer and explanation
Correct option: C
Third law of thermodynamics
Option C (Third law of thermodynamics) is correct because the third law states that the entropy of a system approaches a minimum value as the temperature approaches absolute zero, making it impossible to reach absolute zero in a finite number of steps.
447 Medium
447 of 560
Application +4 | -1
Given that,
3C(s)+2Fe2O3(s)→4Fe(s)+3CO2(g)ΔHo=−93657  cal3\mathrm{C}\left( s \right) + 2\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}\left( s \right) \to 4\mathrm{Fe}\left( s \right) + 3\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)\Delta {H^o} = - 93657\,\,cal
C(s)+O2(g)→CO2(g);ΔHo=−94050\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta {H^o} = - 94050 cal
If both the values of ∆H are at 25 ∘C25\,^\circ C then calculate ?HFe2O3o?H_{\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}}^\mathrm{o}
View answer and explanation
Correct option: C
−94.25  kcal- 94.25\,\,\mathrm{kcal}
Option C (-94.25 kcal) is the correct answer. It is calculated using Hess's Law by manipulating the given thermochemical equations to solve for the enthalpy of formation of Fe2O3.
448 Hard
448 of 560
Understanding +4 | -1
Heat of reaction at constant pp or constant VV varies with temperature as given by Kirchoff's equation is/are
View answer and explanation
Correct option: A
) ΔH2=ΔH1+ΔCp(T2−T1)\Delta H_2=\Delta H_1+\Delta C_p\left(T_2-T_1\right)
Option A (ΔH2=ΔH1+ΔCp(T2−T1)\Delta H_2=\Delta H_1+\Delta C_p(T_2-T_1)) is correct because it represents Kirchoff's equation for the change in enthalpy with temperature.
449 Easy
449 of 560
Remembering +4 | -1
We believe in the laws of thermodynamics because they are
View answer and explanation
Correct option: D
Mere statements
Option C (Empirical and nobody disproved) is correct because the laws of thermodynamics are based on experimental observations and have consistently held true.
450 Medium
450 of 560
Remembering +4 | -1
The state of a gas can be described by quoting the relationship between
View answer and explanation
Correct option: D
Pressure, volume, temperature and amount
Option D (Pressure, volume, temperature and amount) is correct because the state of a gas is defined by the relationship between pressure, volume, temperature, and the amount of gas (number of moles).
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