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

2008 · Shift 0 · Q9
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. /2008 · Shift 0 · Q9

Thermodynamics question

2008 · Shift 0 · Q9

JEE MainChemistryThermodynamicsMCQ+4 / −1
Standard entropy of X2X_2X2​, Y2Y_2Y2​ and XY3XY_3XY3​ are 60, 40 and 50 JK−1 mol−1 , respectively. For the reaction, 12X2{1 \over 2} X_221​X2​+32Y2→{3 \over 2} Y_2 \to23​Y2​→ XY3XY_3XY3​, ΔH\Delta HΔH = -30 kJ, to be at equilibrium, the temperature will be :
  1. A
    1250 K
  2. B
    500 K
  3. C
    750 K
  4. D
    1000 K
View written solutionFree

Correct answer: C

  1. Condition for equilibrium

At equilibrium for the reaction under standard conditions,

ΔG=ΔH−TΔS=0\Delta G = \Delta H - T\Delta S = 0ΔG=ΔH−TΔS=0

So,

T=ΔHΔST = \frac{\Delta H}{\Delta S}T=ΔSΔH​
  1. Calculate standard entropy change

Given:

  • S∘(X2)=60 J K−1mol−1S^\circ(X_2)=60\ \text{J K}^{-1}\text{mol}^{-1}S∘(X2​)=60 J K−1mol−1
  • S∘(Y2)=40 J K−1mol−1S^\circ(Y_2)=40\ \text{J K}^{-1}\text{mol}^{-1}S∘(Y2​)=40 J K−1mol−1
  • S∘(XY3)=50 J K−1mol−1S^\circ(XY_3)=50\ \text{J K}^{-1}\text{mol}^{-1}S∘(XY3​)=50 J K−1mol−1

Reaction:

12X2+32Y2→XY3\frac{1}{2}X_2+\frac{3}{2}Y_2 \to XY_321​X2​+23​Y2​→XY3​

Thus,

ΔS∘=S∘(products)−S∘(reactants)\Delta S^\circ = S^\circ(\text{products}) - S^\circ(\text{reactants})ΔS∘=S∘(products)−S∘(reactants) ΔS∘=50−(12×60+32×40)\Delta S^\circ = 50 - \left(\frac{1}{2}\times 60 + \frac{3}{2}\times 40\right)ΔS∘=50−(21​×60+23​×40) ΔS∘=50−(30+60)=50−90=−40 J K−1mol−1\Delta S^\circ = 50 - (30+60) = 50-90 = -40\ \text{J K}^{-1}\text{mol}^{-1}ΔS∘=50−(30+60)=50−90=−40 J K−1mol−1
  1. Use given enthalpy change

Given:

ΔH=−30 kJ mol−1=−30000 J mol−1\Delta H = -30\ \text{kJ mol}^{-1} = -30000\ \text{J mol}^{-1}ΔH=−30 kJ mol−1=−30000 J mol−1

At equilibrium,

T=ΔHΔS=−30000−40=750 KT = \frac{\Delta H}{\Delta S} = \frac{-30000}{-40} = 750\ \text{K}T=ΔSΔH​=−40−30000​=750 K
  1. Match with options
T=750 KT = 750\ \text{K}T=750 K

So the correct option is C.

PreviousNext

More from Thermodynamics

  • Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100∘C(Given : molar enthalpy of vapourisation of water at 1 bar…2007 · MCQ
  • Identify the correct statement regarding a spontaneous process :2007 · MCQ
  • In conversion of lime-stone to lime, CaCO3​(s) → CaO(s) + CO2​ (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above…2007 · MCQ
  • The standard enthalpy of formation Δf​Ho at 298 K for methane, CH4​(g), is –74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be :2006 · MCQ
  • An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct?2006 · MCQ
  • (ΔH−ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is : (R = 8.314 J K–1 mol–1)2006 · MCQ
  • The enthalpy changes for the following processes are listed below : Cl2​(g) = 2Cl(g), 242.3 kJ mol–1 I2​(g) = 2I(g), 151.0 kJ mol–1 ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1 I2​(s) = I2​(g), 62.76 kJ mol–1 Given that the…2006 · MCQ
  • Consider the reaction: N2​ + 3H2​ → 2NH3​ carried out at constant temperature and pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?2005 · MCQ