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Chemical Equilibrium question

2019 · 12 Apr · Shift 2 · Q13
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Chemical Equilibrium question

2019 · 12 Apr · Shift 2 · Q13

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The INCORRECT match in the following is :
  1. A
    Δ\DeltaΔ Go = 0, K = 1
  2. B
    Δ\DeltaΔ Go < 0, K < 1
  3. C
    Δ\DeltaΔ Go > 0, K < 1
  4. D
    Δ\DeltaΔ Go < 0, K > 1
View written solutionFree

Correct answer: B

  1. The relation between standard Gibbs free energy change and equilibrium constant is
ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln KΔG∘=−RTlnK

where R>0R>0R>0 and T>0T>0T>0.

So, the sign of ΔG∘\Delta G^\circΔG∘ depends on the value of ln⁡K\ln KlnK.

  1. Analyze the possible cases:

  • If K=1K=1K=1, then

    ln⁡1=0⇒ΔG∘=−RT(0)=0\ln 1 = 0 \Rightarrow \Delta G^\circ = -RT(0)=0ln1=0⇒ΔG∘=−RT(0)=0

    So, ΔG∘=0  ⟺  K=1\Delta G^\circ=0 \iff K=1ΔG∘=0⟺K=1.

  • If K>1K>1K>1, then

    ln⁡K>0⇒ΔG∘=−RTln⁡K<0\ln K>0 \Rightarrow \Delta G^\circ = -RT\ln K<0lnK>0⇒ΔG∘=−RTlnK<0

    So, K>1  ⟺  ΔG∘<0K>1 \iff \Delta G^\circ<0K>1⟺ΔG∘<0.

  • If K<1K<1K<1, then

    ln⁡K<0⇒ΔG∘=−RTln⁡K>0\ln K<0 \Rightarrow \Delta G^\circ = -RT\ln K>0lnK<0⇒ΔG∘=−RTlnK>0

    So, K<1  ⟺  ΔG∘>0K<1 \iff \Delta G^\circ>0K<1⟺ΔG∘>0.

  1. Now check each option:

  • A: ΔG∘=0, K=1\Delta G^\circ = 0,\ K=1ΔG∘=0, K=1
    Correct.

  • B: ΔG∘<0, K<1\Delta G^\circ < 0,\ K<1ΔG∘<0, K<1
    Incorrect, because if ΔG∘<0\Delta G^\circ<0ΔG∘<0, then necessarily K>1K>1K>1.

  • C: ΔG∘>0, K<1\Delta G^\circ > 0,\ K<1ΔG∘>0, K<1
    Correct.

  • D: Interpreting the intended match as ΔG∘<0, K>1\Delta G^\circ<0,\ K>1ΔG∘<0, K>1
    Correct.

  1. Therefore, the incorrect match is
B\boxed{\text{B}}B​
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