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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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301 Easy
301 of 560
Understanding +4 | -1
Which is not a spontaneous process?
View answer and explanation
Correct option: C
Heat flowing from colder body to a hotter body
Option C (Heat flowing from colder body to a hotter body) is correct because heat naturally flows from a hotter body to a colder body, not the other way around. This violates the second law of thermodynamics.
302 Easy
302 of 560
Understanding +4 | -1
For the precipitation of AgCl by Ag+\mathrm{A}{\mathrm{g}^ + } ions and HCl:
View answer and explanation
Correct option: C
∆G=-ve
Option C (∆G=-ve) is correct because the precipitation of AgCl is a spontaneous process under normal conditions, and spontaneous processes have a negative change in Gibbs free energy.
303 Hard
303 of 560
Application +4 | -1
When 110 g110 \mathrm{~g} of manganese (At. mass =55=55 ) dissolves in dilute HNO3\mathrm{HNO}_3 at 27∘C27^{\circ} \mathrm{C} under atmospheric pressure, the work done in the process is :
View answer and explanation
Correct option: C
−4988.4 J-4988.4 \mathrm{~J}
Option C (−4988.4 J-4988.4 \mathrm{~J}) is correct because the work done is negative, indicating expansion against atmospheric pressure, and is calculated using W=−PΔV=−ΔnRTW = -P\Delta V = -\Delta nRT.
304 Easy
304 of 560
Remembering +4 | -1
Calorie is equivalent to:
View answer and explanation
Correct option: B
4.184 Joule
Option B (4.184 Joule) is correct because 1 calorie is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius, which is equivalent to 4.184 Joules.
305 Easy
305 of 560
Understanding +4 | -1
The value of ΔE\Delta E for an open and isolated system respectively is
View answer and explanation
Correct option: C
Non-zero; zero
Option C (Non-zero; zero) is correct because an open system can exchange both energy and matter with the surroundings, so its internal energy can change, while an isolated system cannot exchange either, so its internal energy remains constant.
306 Hard
306 of 560
Understanding +4 | -1
Consider the following reactions:
(i)   H+(aq)+OH−(aq)→H2O(l),  ΔH=−X1  kJ  mol−1\mathrm{(i)}\,\,\,{H^ + }(aq) + O{H^ - }(aq) \to {H_2}O(l),\,\,\Delta H = - {X_1}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(ii)   H2(g)+12O2(g)→H2O(l),  ΔH=−X2  kJ  mol−1\mathrm{(ii)}\,\,\,{H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l),\,\,\Delta H = - {X_2}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(iii)   CO2(g)+H2(g)→CO(g)+H2O,  ΔH=−X3  kJ  mol−1\mathrm{(iii)}\,\,\,C{O_2}(g) + {H_2}(g) \to CO(g) + {H_2}O,\,\,\Delta H = - {X_3}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(iv)   C2H2(g)+52O2(g)→2CO2(g)+H2O(l),  ΔH=+4X4  kJ  mol−1\mathrm{(iv)}\,\,\,{C_2}{H_2}(g) + \frac{5}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(l),\,\,\Delta H = +4 {X_4}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Enthalpy of formation of H2O(l){H_2}{O_{(l)}} is
View answer and explanation
Correct option: D
−X2  kJ  mol−1- {X_2}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Option D (−X2 kJ mol−1- {X_2}\,\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because reaction (ii) directly represents the formation of one mole of H2O(l) from its elements in their standard states.
307 Easy
307 of 560
Application +4 | -1
If H++OH−→H2O+13.7{\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } \to {\mathrm{H}_2}\mathrm{O} + 13.7 kcal then the heat of neutralization for complete neutralization of one mole of H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} by a base will be :
View answer and explanation
Correct option: B
−27.4kcal- 27.4\mathrm{kcal}
Option B (-27.4 kcal) is correct because the heat of neutralization for a strong acid and strong base is additive, and sulfuric acid has two acidic protons.
308 Medium
308 of 560
Understanding +4 | -1
Heat of neutralisation of which acid-base reaction is - 57.32 kJ for?
View answer and explanation
Correct option: D
HNO3+LiOH\mathrm{HN}{\mathrm{O}_3} + \mathrm{LiOH}
Option D (HNO3+LiOH\mathrm{HNO_3} + \mathrm{LiOH}) is correct because it represents the neutralization of a strong acid (HNO3\mathrm{HNO_3}) with a strong base (LiOH\mathrm{LiOH}), which results in a heat of neutralization of -57.32 kJ.
309 Hard
309 of 560
Understanding +4 | -1
The correct relation between electron gain enthalpy and electron affinity is
View answer and explanation
Correct option: C
ΔegH=−E.A−52RT\Delta_{e g} H=-E . A-\frac{5}{2} R T
Option C (ΔegH=−E.A−52RT\Delta_{e g} H=-E . A-\frac{5}{2} R T) is correct because it represents the relationship between electron gain enthalpy and electron affinity, considering the conversion from electron affinity to enthalpy change.
310 Medium
310 of 560
Analyzing +4 | -1
Given that
C+O2→CO2:ΔH∘=−xkJ\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2: \Delta \mathrm{H}^{\circ}=-\mathrm{x} \mathrm{kJ}

2CO+O2→2CO2:ΔH∘=−ykJ2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2: \Delta \mathrm{H}^{\circ}=-\mathrm{y} \mathrm{kJ}

the enthalpy of formation of carbon monoxide will be
View answer and explanation
Correct option: B
y−2x2\frac{y-2 x}{2}
Option B (y−2x2\frac{{y-2x}}{{2}}) is correct because it is derived from Hess's Law by manipulating the given equations to obtain the formation equation for CO.
311 Easy
311 of 560
Understanding +4 | -1
All the naturally occurring processes, i.e., spontaneous proceed spontaneously in a direction which leads to:
View answer and explanation
Correct option: A
Decrease of free energy
Option A (Decrease of free energy) is correct because spontaneous processes always proceed in a direction that lowers the Gibbs free energy of the system.
312 Medium
312 of 560
Understanding +4 | -1
Absorption of gasses on solid surface is generally exothermic because
View answer and explanation
Correct option: B
Entropy decreases
Option B (Entropy decreases) is correct because when gases are adsorbed onto a solid surface, their freedom of movement is restricted, leading to a decrease in entropy.
313 Medium
313 of 560
Understanding +4 | -1
The work done during the process when 1 mole of gas is allowed to expand freely into vacuum is:
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because when a gas expands freely into a vacuum, the external pressure is zero, resulting in no work being done.
314 Medium
314 of 560
Remembering +4 | -1
Which one among the following is the correct expression for isothermal reversible change?
View answer and explanation
Correct option: B
w=−2.303  nRT  log⁡VfViw = - 2.303\,\,nRT\,\,\log \frac{{{V_f}}}{{{V_i}}}
Option B is correct because the correct expression for isothermal reversible work is w = -2.303nRTlog(Vf/Vi).
315 Medium
315 of 560
Analyzing +4 | -1
ΔS∘\Delta {S^ \circ } will be highest for the reaction:
View answer and explanation
Correct option: B
CaCO3(s)→CaO(s)+CO2(g)CaC{O_3}(s) \to CaO(s) + C{O_2}(g)
Option B (CaCO3(s)→CaO(s)+CO2(g)CaC{O_3}(s) \to CaO(s) + C{O_2}(g)) is correct because it involves the decomposition of a solid into a solid and a gas, leading to a significant increase in entropy.
316 Medium
316 of 560
Remembering +4 | -1
Entropy change of fusion at constant pressure is given by:
View answer and explanation
Correct option: A
ΔS(f)=ΔHfT\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}
Option A (ΔS(f)=ΔHfT\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}) is correct because the entropy change of fusion at constant pressure is defined as the enthalpy of fusion divided by the temperature at the melting point.
317 Medium
317 of 560
Remembering +4 | -1
The enthalpy change accompanying a reaction
is called
View answer and explanation
Correct option: A
reaction enthalpy
Option A (reaction enthalpy) is correct because the enthalpy change accompanying a reaction is specifically termed the reaction enthalpy.
318 Medium
318 of 560
Application +4 | -1
The enthalpy and entropy change for the reaction,Br2(l)+Cl2(g)→2BrCl(g)\mathrm{B}{\mathrm{r}_2}\left( l \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{BrCl}\left( \mathrm{g} \right)Are 30 kJmol−1\mathrm{mo}{\mathrm{l}^{ - 1}}and105JK−1mol−1105\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}respectively. The temperature at which the reaction will be in equilibrium is:
View answer and explanation
Correct option: C
285.7 K
Option C (285.7 K) is correct because the equilibrium temperature is calculated using T = ∆H/∆S.
319 Hard
319 of 560
Application +4 | -1
An ideal gas undergoes isothermal compression from 5 m35\,{m^3} to 1 m31\,{m^3} against a constant external pressure of 4 Nm−24\,N{m^{ - 2}}. Heat released in this process is used to increase the temperature of 1 mole of AlAl. If molar heat capacity of AlAl is 24 J mol−1 K−124\,J\,mo{l^{ - 1}}\,{K^{ - 1}}, the temperature of AlAl increases by
View answer and explanation
Correct option: D
23K\frac{2}{3}K
Option D (2/3 K) is correct because the heat released during isothermal compression is used to increase the temperature of Aluminum, and the temperature increase is calculated using the formula q = nCΔT.
320 Medium
320 of 560
Application +4 | -1
The value of ΔS\Delta S for freezing of 10g of H2O(ℓ){H_2}O\left( \ell \right) (enthalpy of fusion is 80 cal/g) at 0∘C0^\circ C and 1 atm pressure is:
View answer and explanation
Correct option: D
-12.25 J/K
Option D (-12.25 J/K) is correct because the change in entropy during freezing is calculated using the formula ΔS=−QT\Delta S = \frac{-Q}{T}, where Q is the heat released during freezing and T is the temperature in Kelvin.
321 Hard
321 of 560
Application +4 | -1
To raise the temperature of 1 mole gas from 400 K400 \mathrm{~K} to 500 K500 \mathrm{~K}, how much heat is require while Cp=3+1.2×10−3 T cal/molC_p=3+1.2 \times 10^{-3} \mathrm{~T} \mathrm{~cal} / \mathrm{mol}
View answer and explanation
Correct option: B
354 cal354 \mathrm{~cal}
Option B (354 cal) is correct because it is calculated by integrating the given Cp function over the temperature range.
322 Easy
322 of 560
Understanding +4 | -1
At constant pressure and temperature, the direction of any chemical reaction is one where, the … decrease.
View answer and explanation
Correct option: C
Gibbs energy
Option C (Gibbs energy) is correct because at constant pressure and temperature, the direction of a spontaneous chemical reaction is towards a decrease in Gibbs free energy.
323 Hard
323 of 560
Application +4 | -1
xx g sample of NH4NO3\mathrm{NH}_4 \mathrm{NO}_3 is decomposed in a Bomb calorimeter. The temperature of calorimeter increase by 4∘C4^{\circ} \mathrm{C}. The heat capacity of the system is 125 kJ/∘C125 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}. The value of xx is
(Given molar heat of decomposition of NH4NO3\mathrm{NH}_4 \mathrm{NO}_3 at constant volume is 400 kJ mol−1400 \mathrm{~kJ} \mathrm{~mol}^{-1} )
View answer and explanation
Correct option: A
1
Option A (1) is correct because it represents the mass of NH4NO3 required to produce the given temperature change in the calorimeter, calculated using the heat capacity and molar heat of decomposition.
324 Easy
324 of 560
Application +4 | -1
Heat of formation of COCO and CO2C{O_2} are -26.4, -94.0 Kcal/mole respectively. What is the heat of combustion of CO in K cals?
View answer and explanation
Correct option: C
−67.6- 67.6
Option C (−67.6- 67.6) is correct because the heat of combustion of CO is calculated using the heats of formation of CO and CO2 and applying Hess's Law.
325 Medium
325 of 560
Application +4 | -1
The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}. The heat released when 0.5 mole of HNO3\mathrm{HN}{\mathrm{O}_3} solution is mixed with 0.2 mole of KOH is
View answer and explanation
Correct option: B
11.4 kJ
Option B (11.4 kJ) is correct because the heat released is determined by the limiting reactant (KOH) and the heat of neutralization.
326 Easy
326 of 560
Application +4 | -1
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25∘C25^{\circ} \mathrm{C} is
View answer and explanation
Correct option: B
298×107×8.31×2.303log⁡2298 \times 10^7 \times 8.31 \times 2.303 \log 2
Option B (298×107×8.31×2.303log⁡2298 \times 10^7 \times 8.31 \times 2.303 \log 2) is correct because it represents the work done for a reversible expansion, calculated using the formula W=−2.303nRTlog⁡(V2/V1)W = -2.303nRT\log(V_2/V_1) with R in ergs.
327 Easy
327 of 560
Understanding +4 | -1
Standard state Gibbs free energy changes for the isomerization reaction,cis-2-pentene⇌trans-2-pentene is -3.67 kJ/mol at 400K. If more trans-2-pentene is added to the reaction vessel, then:
View answer and explanation
Correct option: C
More cis-2-pentene is formed
Option C (More cis-2-pentene is formed) is correct because adding more trans-2-pentene will shift the equilibrium towards the reactants (cis-2-pentene) to counteract the change, according to Le Chatelier's principle.
328 Easy
328 of 560
Application +4 | -1
What is the melting point of benzene if ΔHfusion =9.95 kJ/mol\Delta H_{\text {fusion }}=9.95 \mathrm{~kJ} / \mathrm{mol} and ΔSfusion =35.7 J/K−mol\Delta S_{\text {fusion }}=35.7 \mathrm{~J} / \mathrm{K}-\mathrm{mol} ?
View answer and explanation
Correct option: B
278.7 K278.7 \mathrm{~K}
Option B (278.7 K278.7 \mathrm{~K}) is correct because the melting point is calculated using the formula T=ΔHfusion ΔSfusion T = \frac{\Delta H_{\text {fusion }}}{\Delta S_{\text {fusion }}}.
329 Medium
329 of 560
Understanding +4 | -1
In an irreversible process taking place at constant T and P and in which only pressure volume work is being done, the change in Gibbs free energy (dG) and change in entropy
View answer and explanation
Correct option: B
(dS)V,E>0(dG)T,P<0{\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} < 0
Option B ((dS)V,E>0(dG)T,P<0{\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} < 0) is correct because, in an irreversible process at constant T and P, entropy increases and Gibbs free energy decreases.
330 Hard
330 of 560
Application +4 | -1
When 500 J heat is given to the gas X in an isobaric process its work done comes out as 142.8 J. The gas X is
View answer and explanation
Correct option: B
NH3\mathrm{N}{\mathrm{H}_3}
Option B (NH3) is correct because the calculated value of Cp/Cv is closest to that of NH3.
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