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

2024 · 27 Jan · Shift 2 · Q30
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. /2024 · 27 Jan · Shift 2 · Q30

Thermodynamics question

2024 · 27 Jan · Shift 2 · Q30

JEE MainChemistryThermodynamicsNumerical+4 / −1
For a certain thermochemical reaction M→N\mathrm{M} \rightarrow \mathrm{N}M→N at T=400 K,ΔH⊖=77.2 kJ mol−1,ΔS=122 JK−1,log⁡\mathrm{T}=400 \mathrm{~K}, \Delta \mathrm{H}^{\ominus}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}=122 \mathrm{~JK}^{-1}, \logT=400 K,ΔH⊖=77.2 kJ mol−1,ΔS=122 JK−1,log equilibrium constant (log⁡K)(\log K)(logK) is ‾\underline{\hspace{2cm}}​×10−1\times 10^{-1}×10−1.
Numerical answer
View written solutionFree

Correct answer: -37

  1. Use the relation between Gibbs free energy and equilibrium constant

For a reaction at temperature TTT,

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circΔG∘=ΔH∘−TΔS∘

and also,

ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln KΔG∘=−RTlnK

So first we calculate ΔG∘\Delta G^\circΔG∘.

  1. Convert units properly

Given:

ΔH∘=77.2 kJ mol−1=77200 J mol−1\Delta H^\circ = 77.2\,\text{kJ mol}^{-1} = 77200\,\text{J mol}^{-1}ΔH∘=77.2kJ mol−1=77200J mol−1 ΔS∘=122 J K−1mol−1\Delta S^\circ = 122\,\text{J K}^{-1}\text{mol}^{-1}ΔS∘=122J K−1mol−1 T=400 KT = 400\,\text{K}T=400K
  1. Calculate ΔG∘\Delta G^\circΔG∘
ΔG∘=77200−(400)(122)\Delta G^\circ = 77200 - (400)(122)ΔG∘=77200−(400)(122) ΔG∘=77200−48800=28400 J mol−1\Delta G^\circ = 77200 - 48800 = 28400\,\text{J mol}^{-1}ΔG∘=77200−48800=28400J mol−1
  1. Use ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln KΔG∘=−RTlnK
ln⁡K=−ΔG∘RT\ln K = -\frac{\Delta G^\circ}{RT}lnK=−RTΔG∘​

Taking R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}R=8.314J mol−1K−1,

ln⁡K=−28400(8.314)(400)\ln K = -\frac{28400}{(8.314)(400)}lnK=−(8.314)(400)28400​ ln⁡K=−284003325.6≈−8.54\ln K = -\frac{28400}{3325.6} \approx -8.54lnK=−3325.628400​≈−8.54
  1. Convert to common logarithm

Since

log⁡K=ln⁡K2.303\log K = \frac{\ln K}{2.303}logK=2.303lnK​ log⁡K=−8.542.303≈−3.71\log K = \frac{-8.54}{2.303} \approx -3.71logK=2.303−8.54​≈−3.71

Thus,

log⁡K≈−3.7=−37×10−1\log K \approx -3.7 = -37 \times 10^{-1}logK≈−3.7=−37×10−1
  1. Final integer

The blank in

log⁡K=‾×10−1\log K = \underline{\hspace{2cm}} \times 10^{-1}logK=​×10−1

is therefore

−37-37−37
  1. Comparison with stored answer

Stored correct answer = 373737

But the actual value is negative:

log⁡K≈−37×10−1\log K \approx -37 \times 10^{-1}logK≈−37×10−1

So the sign is missing in the stored answer.

PreviousNext

More from Thermodynamics

  • Which of the following is not correct?2024 · MCQ
  • Standard enthalpy of vapourisation for CCl4​ is 30.5 kJ mol−1. Heat required for vapourisation of 284 g of CCl4​ at constant temperature is ​kJ.…2024 · Numerical
  • An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→A as shown in the diagram… Includes diagram2024 · Numerical
  • Two reactions are given below: ​2Fe(s)​+23​O2( g)​→Fe2​O3( s)​,ΔH∘=−822 kJ/molC(s)​+21​O2( g)​→CO(g)​,ΔH∘=−110 kJ/mol​…2024 · Numerical
  • Consider the following reaction at 298 K⋅23​O2(g)​⇌O3(g)​⋅KP​=2.47×10−29. Δr​G⊖ for the reaction is ​…2024 · Numerical
  • If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, w, is −x J. The value of x is ​…2024 · Numerical
  • At 25∘C, the enthalpy of the following processes are given : What would be the value of X for the following reaction ? ​ (Nearest integer) H2​O(g)→H(g)+OH(g) ΔH∘=X kJ mol−1… Includes table2023 · Numerical
  • 0.3 g of ethane undergoes combustion at 27∘C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5∘C. The heat evolved during combustion…2023 · Numerical