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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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151 Easy
151 of 560
Analyzing +4 | -1
2MnO4−+16H++10Cl−→2Mn2++5Cl2(g)+8H2O2 {MnO}_4^{-}+16 {H}^{+}+10 {Cl}^{-} \rightarrow 2 {Mn}^{2+}+5 {Cl}_2(\mathrm{g})+8 {H}_2 {O}

The above reaction is endothermic and hence, the actual temperature of the reaction vessel (isolated from the surrounding) may be different from that expected. Given that the initial temperature of the reaction vessel was used in the calculations, how would this affect the predicted value of moles of Cl2(n){Cl}_2(n) according to the equation n=PV/RT?n=P V/ R T?
View answer and explanation
Correct option: B
It would be less than the actual value
It would be less than the actual value. The reaction is endothermic, so it absorbs heat from the surroundings (the reaction vessel). The temperature decreases, and since n = PV/RT, a lower temperature leads to a lower calculated value of n.
152 Medium
152 of 560
Remembering +4 | -1
Which represents the largest amount of energy?
View answer and explanation
Correct option: A
Calorie
Option A, Calorie, represents the largest amount of energy among the given options.
153 Easy
153 of 560
Understanding +4 | -1
During an isothermal expansion of an ideal gas its:
View answer and explanation
Correct option: D
Internal energy remains constant
Option D is correct because, for an ideal gas undergoing isothermal expansion, the internal energy remains constant as it only depends on temperature, which remains constant in an isothermal process.
154 Medium
154 of 560
Understanding +4 | -1
Internal energy and pressure of a gas of unit volume are related as:
View answer and explanation
Correct option: A
P=23UP = \frac{2}{3}U
Option A is correct because the internal energy (U) of a gas is related to its pressure (P) and volume as P=23UP = \frac{2}{3}U for a monoatomic ideal gas.
155 Easy
155 of 560
Understanding +4 | -1
Which among the following is an extensive property?
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Correct option: A
Volume
Option A (Volume) is correct because volume is an extensive property that depends on the amount of substance.
156 Easy
156 of 560
Understanding +4 | -1
Thermochemistry is the study of relationship between heat energy and :
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Correct option: A
Chemical energy
Option A (Chemical energy) is correct because thermochemistry deals with the relationship between heat and chemical changes.
157 Medium
157 of 560
Remembering +4 | -1
The equilibrium constant (K)(K) of a reaction may be written as,
View answer and explanation
Correct option: D
K=e−ΔG∘/RTK=e^{-\Delta G^{\circ} / R T}
Option D (K=e−ΔG∘/RTK=e^{-\Delta G^{\circ} / R T}) is correct because it accurately represents the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (ΔG∘\Delta G^{\circ}).
158 Easy
158 of 560
Understanding +4 | -1
Which of the following are path functions ?
a) Heat capacity b) work c) Entropy
View answer and explanation
Correct option: C
a and b only
Option C (a and b only) is correct because heat capacity and work are path-dependent, while entropy is a state function.
159 Medium
159 of 560
Analyzing +4 | -1
Consider the following reactions :

C(s)+O2( g)⟶CO2( g)+x kJ\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+x \mathrm{~kJ}

CO(g)+12O2(g)⟶CO2(g)+y kJ\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+y \mathrm{~kJ}

The heat of formation of CO(g)\mathrm{CO}(g) is :
View answer and explanation
Correct option: C
(y−x)kJ/mol(y-x) \mathrm{kJ} / \mathrm{mol}
Option C is correct. The heat of formation of CO(g) can be calculated by subtracting the second reaction from the first, resulting in (y-x) kJ/mol.
160 Easy
160 of 560
Understanding +4 | -1
Entropy of an adiabatic reversible process is:-
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Correct option: D
Constant
Option D (Constant) is correct because entropy remains constant in a reversible adiabatic process, as there is no heat exchange with the surroundings.
161 Hard
161 of 560
Application +4 | -1
Two moles of helium gas expanded isothermally and irreversible at 27℃ from volume 1 dm3d{m^3} to 1 m3{m^3} at constant pressure of 100 k Pa. Calculate the work done.
View answer and explanation
Correct option: B
99900 J
Option B (99900 J) is correct because it is calculated using the formula for work done in an irreversible isothermal expansion against constant pressure.
162 Easy
162 of 560
Application +4 | -1
For the reaction aA(s)+bB(g)→ dD(s)+cC(g),aA\left( s \right) + bB\left( g \right)\xrightarrow{\,}dD\left( s \right) + cC\left( g \right), then
View answer and explanation
Correct option: B
ΔH−ΔE=(c−b)RT\Delta H - \Delta E = \left( {c - b} \right)RT
Option B (ΔH−ΔE=(c−b)RT\Delta H - \Delta E = (c - b)RT) is correct because it correctly applies the relationship between enthalpy change, internal energy change, and the change in the number of moles of gas.
163 Medium
163 of 560
Understanding +4 | -1
When hydrogen and oxygen burn to form water in an oxyhydrogen torch, the entropy change is:
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Correct option: A
Negative
Option A (Negative) is correct because the formation of water from hydrogen and oxygen involves a decrease in the number of gas molecules, leading to a decrease in entropy.
164 Medium
164 of 560
Remembering +4 | -1
Which of the following relation is correct
regarding reversible adiabatic process?
View answer and explanation
Correct option: C
TiViγ−1=TfVfγ−1\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^{\gamma-1}=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^{\gamma-1}
Option C (TiViγ−1=TfVfγ−1\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^{\gamma-1}=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^{\gamma-1}) is correct because it represents the correct relationship between temperature and volume in a reversible adiabatic process.
165 Medium
165 of 560
Remembering +4 | -1
Which of the following represents the first law
of thermodynamics
View answer and explanation
Correct option: A
ΔU=q+W\Delta U=q+W
The correct answer is A (ΔU=q+W\Delta U=q+W) because it accurately represents the first law of thermodynamics, stating that the change in internal energy equals heat added plus work done on the system.
166 Easy
166 of 560
Remembering +4 | -1
Entropy of a perfectly crystalline solid at 0 K is
View answer and explanation
Correct option: C
zero
Option C (zero) is correct because the third law of thermodynamics states that the entropy of a perfectly crystalline solid at 0 K is zero.
167 Medium
167 of 560
Application +4 | -1
For a reaction, A(l)→4B(g)A\left( l \right) \to 4B\left( g \right) if ΔU=4.5 kJ\Delta U = 4.5\,kJ and ΔS=60 JK−1\Delta S = 60\,J{K^{ - 1}} at 270C{27^0}C then ΔG\Delta G will be
View answer and explanation
Correct option: C
–3.5 kJ
Option C (–3.5 kJ) is correct because it is calculated using the formula ΔG=ΔU+Δ(PV)−TΔS=ΔU+ΔnRT−TΔS\Delta G = \Delta U + \Delta(PV) - T\Delta S = \Delta U + \Delta nRT - T\Delta S, where Δn\Delta n is the change in the number of moles of gas.
168 Easy
168 of 560
Understanding +4 | -1
Predict the change in internal energy for an isolated system at constant volume
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because in an isolated system at constant volume, no heat or work is exchanged with the surroundings, thus the internal energy remains constant.
169 Easy
169 of 560
Understanding +4 | -1
A well stoppered thermos flask containing some ice cubes is an example of:
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Correct option: C
Isolated system
Option C is correct because a thermos flask prevents both heat and mass exchange with the surroundings, making it an isolated system.
170 Easy
170 of 560
Understanding +4 | -1
For the process, CO2(s)→CO2(g):\mathrm{C}{\mathrm{O}_2}\left( \mathrm{s} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right):
View answer and explanation
Correct option: A
Both ∆H and ∆S are +ve
The correct answer is A (Both ∆H and ∆S are +ve) because sublimation requires energy input (endothermic, +∆H) and increases disorder (+∆S).
171 Easy
171 of 560
Remembering +4 | -1
For a process, entropy change of a system is
expressed a
View answer and explanation
Correct option: B
qrev T\frac{q_{\text {rev }}}{T}
Option B (qrev T\frac{{q_{\text{{rev }}}}}{{T}}) is correct because it represents the definition of entropy change for a reversible process.
172 Easy
172 of 560
Remembering +4 | -1
In isolated system:
View answer and explanation
Correct option: D
Neither energy nor mass transfer take place
Option D (Neither energy nor mass transfer take place) is correct because an isolated system, by definition, does not exchange energy or mass with its surroundings.
173 Easy
173 of 560
Application +4 | -1
An ideal gas absorb 50 cal50\, \mathrm{cal} heat and expand in the vaccum from 5 L5 \mathrm{~L} to 15 L15 \mathrm{~L}. Change in its internal energy is :-
View answer and explanation
Correct option: B
50 cal50\, \mathrm{cal}
Option B (50 cal) is correct because in a vacuum expansion, no external pressure exists, so no work is done, and the heat absorbed equals the change in internal energy.
174 Hard
174 of 560
Analyzing +4 | -1
The bond dissociation energies of N≡N,N=N,O=O & N=ON \equiv N, N=N, O=O \,\&\, N=O are 946,418,498946,418,498 and 607 KJ/mole607 \mathrm{~KJ/ mole} respectively. If heat of formation of N2O{N}_2 {O} is 82 KJ/mole82 \mathrm{~KJ/ mole} then the magnitude of resonance energy of N2O{N}_2 {O} in KJ/mole\mathrm{KJ} / \mathrm{mole} is ___
View answer and explanation
Correct option: C
88 KJ/mole88 \mathrm{~KJ} / \mathrm{mole}
Option C (88 kJ/mole) is correct because the resonance energy is calculated by subtracting the experimental heat of formation from the calculated heat of formation based on bond energies.
175 Medium
175 of 560
Understanding +4 | -1
Consider an endothermic reaction X→YX \to Y with the activation energies, respectively in general
View answer and explanation
Correct option: A
Eb<Ef{E_b} < {E_f}
Option A (Eb<Ef{E_b} < {E_f}) is correct because in an endothermic reaction, the energy of the products is higher than the energy of the reactants, thus the forward activation energy is greater than the backward activation energy.
176 Medium
176 of 560
Analyzing +4 | -1
The heats of neutralization of CH3COOH,   HCOOH,  HCNC{H_3}COOH,\,\,\,HCOOH,\,\,HCN and H2S{H_2}S are −13.2, − 13.4, −2.9- 13.2,\, - \,13.4,\, - 2.9 and - 3.8 kcal per equivalent respectively. Arrange the increasing order of strength.
View answer and explanation
Correct option: A
HCCOH>CH3COOH>H2S>HCNHCCOH > C{H_3}COOH > {H_2}S > HCN
Option A (HCCOH>CH3COOH>H2S>HCNHCCOH > C{H_3}COOH > {H_2}S > HCN) is correct because the strength of an acid is inversely proportional to its heat of neutralization; lower the heat of neutralization, stronger the acid.
177 Medium
177 of 560
Analyzing +4 | -1
At 25∘C25^{\circ} \mathrm{C}, the enthalpy of the following processes are given

H2(g)+O2(g)→2OH(g)ΔH∘=78 kJ mol−1\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =78 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

H2(g)+12O2(g)→H2O(g)ΔH∘=−242 kJ mol−1\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =-242 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

H2(g)→2H(g)ΔH∘=436 kJ mol−1\mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=436 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

12O2(g)→O(g)ΔH∘=249 kJ mol−1\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=249 \,\mathrm{kJ} \,\mathrm{mol}^{-1}
What would be the value of XX for the following reaction? _____ (Nearest integer)
H2O(g)→H(g)+OH(g)ΔH∘=X kJ mol−1\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=\mathrm{X} \,\mathrm{kJ} \,\mathrm{mol}^{-1}
View answer and explanation
Correct option: A
499
The correct answer is 499. By manipulating the given equations and applying Hess's Law, we can calculate the enthalpy change for the desired reaction.
178 Medium
178 of 560
Understanding +4 | -1
A closed flask contains water in all its three states, solids, liquid and vapour at 0℃. In this situation the average KE of the water molecule will be:
View answer and explanation
Correct option: D
Same in all the three states
Option D (Same in all the three states) is correct because the average KE of water molecules depends only on temperature, which is constant at 0°C.
179 Easy
179 of 560
Analyzing +4 | -1
For the reactions: C+O2→CO2;ΔH=−393kJ\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H = - 393\mathrm{kJ} 2Zn+O2→2ZnO;ΔH=−412kJ2\mathrm{Zn} + {\mathrm{O}_2} \to 2\mathrm{ZnO};\Delta H = - 412\mathrm{kJ} which one is correct?
View answer and explanation
Correct option: A
Carbon can reduce ZnO to Zn
Option D (Zn liberates more heat than carbon during oxidation) is correct because the enthalpy change for the oxidation of Zn is more negative than that of carbon, indicating more heat is released.
180 Hard
180 of 560
Application +4 | -1
An ideal gas undergoes isothermal expansion from (10 atm,1 L)(10 \mathrm{~atm}, 1 \mathrm{~L}) to (1 atm,10 L)(1 \mathrm{~atm}, 10 \mathrm{~L}) either by path-I (infinite stage expansion) or by path-II (first against 5 atm5 \mathrm{~atm} and then against 1 atm).1 \mathrm{~atm}). The value of (qpath-Iqpath-II)\left(\frac{q_{\text {path-I}}}{q_{\text {path-II}}}\right) is
View answer and explanation
Correct option: A
2.3031.3\frac{2.303}{1.3}
Option A (2.3031.3\frac{{2.303}}{{1.3}}) is correct because it represents the ratio of heat absorbed in an infinite stage expansion to that in a two-stage expansion, calculated using the isothermal work formula.
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