Q.The value of ∆G ° for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

Gibbs Free Energy from $K$ — The Bridge Between Thermodynamics and Equilibrium

Imagine you're pushing a heavy box across a rough floor. You push hard, but the box barely moves. The potential to move is there — you're applying force — but the actual motion is tiny. That's the difference between thermodynamic spontaneity (the push) and equilibrium (where the box sits, barely budging).

Gibbs Free Energy ($\Delta G$) tells you the push — whether a reaction can happen. The equilibrium constant $K$ tells you how far it actually goes before stopping. The equation that links them is one of the most powerful in chemistry:

Let's unpack this from the ground up.

Step 1: What is $\Delta G$?

Gibbs Free Energy change ($\Delta G$) measures the maximum useful work a reaction can do at constant temperature and pressure. More practically:

• If $\Delta G < 0$: the reaction is spontaneous (it can happen on its own).
• If $\Delta G > 0$: the reaction is non-spontaneous (it needs energy input).
• If $\Delta G = 0$: the system is at equilibrium — no net change.

But here's the catch: $\Delta G$ depends on how much reactant and product you have at any moment. It's not a fixed number.

Step 2: Standard vs. Non-standard Conditions

Chemists define a standard state (pure substances at 1 bar, 1 M concentration for solutions, 25°C usually). Under those conditions, the free energy change is called $\Delta G^\circ$ (standard Gibbs free energy change).

But real reactions rarely start at standard conditions. So we have:

where $Q$ is the reaction quotient (ratio of products to reactants at that instant, raised to their stoichiometric coefficients).

Step 3: At Equilibrium — The Key Insight

At equilibrium, the reaction has no net tendency to go forward or backward. That means:

And the reaction quotient $Q$ becomes exactly the equilibrium constant $K$.

So plug into the equation:

Rearrange:

That's it. This single equation connects a thermodynamic property ($\Delta G^\circ$) with a concentration-based constant ($K$).

Step 4: What This Tells You

Step 5: A Concrete Example

Consider the reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

At 25°C, $\Delta G^\circ = -33.3 \text{ kJ/mol}$. Using $R = 8.314 \text{ J/mol·K}$:

This huge $K$ tells you that at equilibrium, ammonia dominates — the reaction strongly favours products. That matches the negative $\Delta G^\circ$.

Step 6: Why This Matters for Exams

You'll use this equation in three main ways:

1. Calculate $K$ from $\Delta G^\circ$ (as above)
2. Calculate $\Delta G^\circ$ from $K$ (just rearrange)
3. Find temperature where $K=1$ (set $\Delta G^\circ = 0$ and solve for $T$)

The Big Picture

Gibbs Free Energy from $K$ is the thermodynamic compass that tells you where a reaction will settle. $\Delta G^\circ$ gives the direction and magnitude of the drive, while $K$ gives the destination. They're two sides of the same coin — and this equation is the mint that stamps them together.

A quick search for "Gibbs Free Energy from K class 11 chemistry" or "NCERT chemistry syllabus gibbs free energy from k" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.