Match standard free energy of the reaction with th – Class 11 Chemistry

NCERT Exemplar · Q40

Q. Match standard free energy of the reaction with the corresponding equilibrium constant. Column I

(i) ΔG° > 0

(ii) ΔG° < 0

(iii) ΔG° = 0
Column II

(a) K > 1

(b) K = 1

(d) K < 1

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Concept understanding — Gibbs Free Energy from 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:

\Delta G^\circ = -RT \ln K

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:

\Delta G = \Delta G^\circ + RT \ln Q

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

Note

$R$ is the gas constant (8.314 J/mol·K), $T$ is temperature in Kelvin. The $\ln$ is natural log.

Step 3: At Equilibrium — The Key Insight

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

\Delta G = 0

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

So plug into the equation:

0 = \Delta G^\circ + RT \ln K

Rearrange:

\Delta G^\circ = -RT \ln K

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

$\Delta G^\circ$ value	$K$ value	Meaning
Negative ( $<0$ )	$K > 1$	Products favoured at equilibrium
Zero ( $=0$ )	$K = 1$	Equal amounts at equilibrium
Positive ( $>0$ )	$K < 1$	Reactants favoured at equilibrium

Watch out

A negative $\Delta G^\circ$ does not mean the reaction is fast — only that it's thermodynamically favourable. Kinetics (activation energy) is a separate story.

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 $R = 8.314 \text{ J/mol·K}$ :

$-33,300 = -(8.314)(298) \ln K$ …

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The sign of $\Delta G^\circ$ determines whether products or reactants are favored at equilibrium through the relationship $\Delta G^\circ = -RT \ln K$ . Matching: (i)→(d), (ii)→(a), (iii)→(b).

Why Gibbs Free Energy and the Equilibrium Constant Are Connected

At equilibrium, a reaction has no driving force to proceed in either direction—the system has minimized its free energy. The standard Gibbs free energy change $\Delta G^\circ$ tells us how far the equilibrium position lies from a 1:1 mixture of products and reactants (all at unit activity). The equilibrium constant $K$ quantifies exactly that position: the ratio of product activities to reactant activities when the system settles.

The bridge between them is:

\Delta G^\circ = -RT \ln K

where $R$ is the gas constant, $T$ is temperature, and $K$ is the equilibrium constant. This equation reveals that the sign and magnitude of $\Delta G^\circ$ directly encode whether products ( $K > 1$ ) or reactants ( $K < 1$ ) dominate at equilibrium.

Step-by-Step Matching

1. Start with the fundamental relationship

Rearrange the formula to isolate $K$ :

K = e^{-\Delta G^\circ / RT}

Since $RT$ is always positive (temperature in Kelvin, $R > 0$ ), the sign of $\Delta G^\circ$ controls the exponent and thus the value of $K$ .

2. Case (i): $\Delta G^\circ > 0$

When $\Delta G^\circ$ is positive, the exponent $-\Delta G^\circ / RT$ becomes negative. An exponential with a negative argument gives a number between 0 and 1:

K = e^{-(\text{positive number})} < 1

This means reactants are favored at equilibrium. The reaction is non-spontaneous in the forward direction under standard conditions.

Match: (i) → (d) $K < 1$

3. Case (ii): $\Delta G^\circ < 0$

When $\Delta G^\circ$ is negative, the exponent $-\Delta G^\circ / RT$ becomes positive:

K = e^{-(\text{negative number})} = e^{\text{positive number}} > 1

Products are favored. The reaction proceeds spontaneously forward under standard conditions, and at equilibrium the product concentrations significantly exceed those of reactants.

Match: (ii) → (a) $K > 1$

4. Case (iii): $\Delta G^\circ = 0$

When $\Delta G^\circ$ is exactly zero:

$K = e^{0} = 1$ …

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Q.Match standard free energy of the reaction with the corresponding equilibrium constant. Column I

Why Gibbs Free Energy and the Equilibrium Constant Are Connected

Step-by-Step Matching

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Q.Match standard free energy of the reaction with the corresponding equilibrium constant.
Column I

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