Q.A spontaneous reaction is not possible if

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 $R = 8.314 \text{ J/mol·K}$:

$-33,300 = -(8.314)(298) \ln K$ …