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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

Option 4 : ∆G° < 0, K < 1

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
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180 Mins

**Calculation:**

The Gibbs free energy is given by the formula:

∆G = ∆G° + RT ln K_{eq}

∆G° = - RT ln K_{eq}

We know that at equilibrium ∆G° = 0,

If we put K = 1 then ∆G° goes to 0

If ∆G° > 0 means ∆G° is positive and definitely K < 1

If ∆G° < 0 means ∆G° is negative and definitely K > 1

Thus,