Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2016 · 9 Apr · Shift 1 · Q16
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Thermodynamics
  5. /2016 · 9 Apr · Shift 1 · Q16

Thermodynamics question

2016 · 9 Apr · Shift 1 · Q16

JEE MainChemistryThermodynamicsMCQ+4 / −1
A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following :
  1. A
    Both Δ\DeltaΔ H and Δ\DeltaΔ S are negative.
  2. B
    Both Δ\DeltaΔ H and Δ\DeltaΔ S are positive.
  3. C
    Δ\DeltaΔ H is positive while Δ\DeltaΔ S is negative.
  4. D
    Δ\DeltaΔ H is negative while Δ\DeltaΔ S is positive.
View written solutionFree

Correct answer: B

  1. For a reaction at constant pressure, spontaneity is determined by Gibbs free energy:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

A reaction is spontaneous when:

ΔG<0\Delta G < 0ΔG<0

  1. Given condition:
  • At low temperature: reaction is non-spontaneous ⇒ΔG>0\Rightarrow \Delta G > 0⇒ΔG>0
  • At high temperature: reaction becomes spontaneous ⇒ΔG<0\Rightarrow \Delta G < 0⇒ΔG<0

So, increasing temperature makes ΔG\Delta GΔG decrease from positive to negative.

  1. Analyze the expression:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

For ΔG\Delta GΔG to become smaller as TTT increases, the term −TΔS-T\Delta S−TΔS must become more negative with increasing TTT.

This happens when:

ΔS>0\Delta S > 0ΔS>0

  1. Now check the low-temperature condition. If ΔS>0\Delta S > 0ΔS>0, then at low TTT:

ΔG≈ΔH\Delta G \approx \Delta HΔG≈ΔH

Since the reaction is non-spontaneous at low temperature, we need:

ΔH>0\Delta H > 0ΔH>0

  1. Therefore,

ΔH>0,ΔS>0\Delta H > 0, \quad \Delta S > 0ΔH>0,ΔS>0

  1. Check options:
  • A: ΔH<0,ΔS<0\Delta H<0, \Delta S<0ΔH<0,ΔS<0 ❌
  • B: ΔH>0,ΔS>0\Delta H>0, \Delta S>0ΔH>0,ΔS>0 ✅
  • C: ΔH>0,ΔS<0\Delta H>0, \Delta S<0ΔH>0,ΔS<0 ❌
  • D: ΔH<0,ΔS>0\Delta H<0, \Delta S>0ΔH<0,ΔS>0 ❌

Hence the correct option is B.

PreviousNext

More from Thermodynamics

  • The plot shows the variation of −ln Kp versus temperature for the two reactions. M(s) +21​ O2​(g) → MO(s) and C(s) +21​ O2​(g) → CO(s) Identify the correct statement : Includes diagram2016 · MCQ
  • If 100 mole of H2​O2​ decompose at 1 bar and 300 K, the work done (kJ) by one mole of O2​(g) as it expands against 1 bar pressure is : 2H2​O2​(l) ⇌ 2H2​O(l) + O2​(g) (R = 8.3 J K − 1 mol − 1)2016 · MCQ
  • The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJ mol–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :2016 · MCQ
  • The following reaction is performed at 298 K 2NO(g) + O2​ (g) ⇋ 2NO2​ (g) The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2​(g) at…2015 · MCQ
  • For complete combustion of ethanol, C2​H5​OH(l) + 3O2​(g) → 2CO2​(g) + 3H2​O(l) the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol–1 at 25oC. Assuming ideality the Enthalpy of combustion, $\Delta…2014 · MCQ
  • A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be : (R = 8.314 J/mol K) ( l n…2013 · MCQ
  • The incorrect expression among the following is :2012 · MCQ
  • The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27oC is :2011 · MCQ