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Chemical Equilibrium question

2020 · 6 Sep · Shift 1 · Q7
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Chemical Equilibrium question

2020 · 6 Sep · Shift 1 · Q7

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The variation of equilibrium constant with temperature is given below :

Temperature Equilibrium Constant
T1 = 25oC K1 = 10
T2 = 100oC K2 = 100

The values of Δ\DeltaΔ Ho, Δ\DeltaΔ Go at
T1 and Δ\DeltaΔ Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
  1. A
    28.4, –5.71 and –14.29
  2. B
    0.64, –7.14 and –5.71
  3. C
    28.4, –7.14 and –5.71
  4. D
    0.64, –5.71 and –14.29
View written solutionFree

Correct answer: A

  1. Use van’t Hoff equation to find ΔH∘\Delta H^\circΔH∘

For two temperatures,

ln⁡(K2K1)=ΔH∘R(1T1−1T2)\ln\left(\frac{K_2}{K_1}\right)=\frac{\Delta H^\circ}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)ln(K1​K2​​)=RΔH∘​(T1​1​−T2​1​)

Given:

T1=25∘C=298 K,K1=10T_1=25^\circ C=298\,K, \quad K_1=10T1​=25∘C=298K,K1​=10 T2=100∘C=373 K,K2=100T_2=100^\circ C=373\,K, \quad K_2=100T2​=100∘C=373K,K2​=100

So,

ln⁡(K2K1)=ln⁡(10)=2.303\ln\left(\frac{K_2}{K_1}\right)=\ln(10)=2.303ln(K1​K2​​)=ln(10)=2.303

Also,

1T1−1T2=1298−1373\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{298}-\frac{1}{373}T1​1​−T2​1​=2981​−3731​ =373−298298×373=75111154≈6.747×10−4 K−1=\frac{373-298}{298\times 373}=\frac{75}{111154}\approx 6.747\times 10^{-4}\,K^{-1}=298×373373−298​=11115475​≈6.747×10−4K−1

Hence,

ΔH∘=Rln⁡(K2/K1)(1/T1−1/T2)\Delta H^\circ=\frac{R\ln(K_2/K_1)}{(1/T_1-1/T_2)}ΔH∘=(1/T1​−1/T2​)Rln(K2​/K1​)​ =8.314×2.3036.747×10−4=\frac{8.314\times 2.303}{6.747\times 10^{-4}}=6.747×10−48.314×2.303​ ≈2.84×104 J mol−1=28.4 kJ mol−1\approx 2.84\times 10^4\,J\,mol^{-1}=28.4\,kJ\,mol^{-1}≈2.84×104Jmol−1=28.4kJmol−1

So,

ΔH∘≈28.4 kJ mol−1\boxed{\Delta H^\circ \approx 28.4\,kJ\,mol^{-1}}ΔH∘≈28.4kJmol−1​
  1. Find ΔG∘\Delta G^\circΔG∘ at T1T_1T1​

Relation:

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

At T1=298 KT_1=298\,KT1​=298K, K1=10K_1=10K1​=10:

ΔG1∘=−8.314×298×ln⁡10\Delta G_1^\circ=-8.314\times 298\times \ln 10ΔG1∘​=−8.314×298×ln10 =−8.314×298×2.303=-8.314\times 298\times 2.303=−8.314×298×2.303 ≈−5698 J mol−1\approx -5698\,J\,mol^{-1}≈−5698Jmol−1 ≈−5.71 kJ mol−1\approx -5.71\,kJ\,mol^{-1}≈−5.71kJmol−1

Thus,

ΔG∘(T1)≈−5.71 kJ mol−1\boxed{\Delta G^\circ(T_1)\approx -5.71\,kJ\,mol^{-1}}ΔG∘(T1​)≈−5.71kJmol−1​
  1. Find ΔG∘\Delta G^\circΔG∘ at T2T_2T2​

At T2=373 KT_2=373\,KT2​=373K, K2=100K_2=100K2​=100:

ΔG2∘=−RT2ln⁡K2\Delta G_2^\circ=-RT_2\ln K_2ΔG2∘​=−RT2​lnK2​

Now,

ln⁡100=2ln⁡10=4.606\ln 100=2\ln 10=4.606ln100=2ln10=4.606

So,

ΔG2∘=−8.314×373×4.606\Delta G_2^\circ=-8.314\times 373\times 4.606ΔG2∘​=−8.314×373×4.606 ≈−14279 J mol−1\approx -14279\,J\,mol^{-1}≈−14279Jmol−1 ≈−14.29 kJ mol−1\approx -14.29\,kJ\,mol^{-1}≈−14.29kJmol−1

Thus,

ΔG∘(T2)≈−14.29 kJ mol−1\boxed{\Delta G^\circ(T_2)\approx -14.29\,kJ\,mol^{-1}}ΔG∘(T2​)≈−14.29kJmol−1​
  1. Match with options

The values are:

ΔH∘=28.4,ΔG∘(T1)=−5.71,ΔG∘(T2)=−14.29\Delta H^\circ=28.4,\quad \Delta G^\circ(T_1)=-5.71,\quad \Delta G^\circ(T_2)=-14.29ΔH∘=28.4,ΔG∘(T1​)=−5.71,ΔG∘(T2​)=−14.29

This matches Option A.

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