From Gibbs Energy to Equilibrium
The Gibbs energy of a system is defined as $G = H - TS$, an extensive state function. For a chemical change carried out at constant temperature and pressure, the change in Gibbs energy is given by the Gibbs equation $\Delta G = \Delta H - T\Delta S$. NCERT establishes this as the working criterion of spontaneity: a process for which $\Delta G < 0$ is spontaneous, one for which $\Delta G > 0$ is non-spontaneous, and the borderline case $\Delta G = 0$ corresponds to equilibrium.
All the spontaneity analysis up to Section 5.6 concerns reactions driven in one direction. A reversible reaction, in the strict thermodynamic sense, is one carried out so that the system stays in perfect equilibrium with its surroundings at every step. When applied to a chemical reaction, the term means the reaction can proceed in either direction simultaneously and a dynamic equilibrium is established. Since both the forward and backward directions would each have to lower the free energy, the only consistent possibility is that at equilibrium the free energy of the system is at a minimum. Were it not, the system would spontaneously slide toward a configuration of lower free energy.
This gives the criterion of equilibrium directly. For a general reaction such as $\ce{A + B <=> C + D}$, equilibrium is reached when
This is the hinge of the whole section. The actual Gibbs energy change of the reaction mixture falls toward zero as the mixture approaches equilibrium, and it is exactly zero once equilibrium is reached. It is worth emphasising what this does not say. Equilibrium is not the cessation of reaction: the forward and reverse reactions both continue, but at equal rates, so the macroscopic composition stops changing while molecular activity persists. The statement $\Delta_r G = 0$ is the thermodynamic signature of this dynamic balance, and it holds regardless of the path by which the mixture arrived there, because $G$ is a state function.
The Master Relation ΔG° = −RT ln K
To convert the equilibrium criterion into a usable equation, NCERT relates the actual change $\Delta_r G$ to the standard change $\Delta_r G^\circ$ and the reaction quotient $Q$:
At equilibrium two things happen simultaneously: $\Delta_r G = 0$, and the reaction quotient $Q$ becomes the equilibrium constant $K$. Substituting both conditions:
Rearranging gives the relation that NCERT prints as equation 5.23, the single most examined formula in this part of the chapter:
The factor 2.303 simply converts the natural logarithm to base-10 logarithm ($\ln x = 2.303 \log x$), which is convenient when $K$ spans many orders of magnitude. Here $\Delta_r G^\circ$ is the standard Gibbs energy change — the value when all reactants and products are present in their standard states — and $K$ is the thermodynamic equilibrium constant at the same temperature $T$. Two features of this equation deserve attention before any calculation. First, the equilibrium constant that appears is dimensionless, formed from activities (approximated by concentrations or partial pressures relative to the standard state); this is why a logarithm of it is meaningful. Second, $\Delta_r G^\circ$ and $K$ both belong to the same temperature, so a value of $K$ measured at one temperature cannot be paired with $\Delta_r G^\circ$ at another without first correcting for the temperature dependence.
The two faces of the master equation — read it either way.
If $\Delta_r G^\circ$ is known, $K$ follows; if $K$ is measured in the laboratory, $\Delta_r G^\circ$ follows. NCERT lists exactly these two practical uses of equation 5.23.
ΔG versus ΔG°: The Quotient Q
The most frequent conceptual error in this topic is treating $\Delta G$ and $\Delta G^\circ$ as the same quantity. They are not. The relation $\Delta_r G = \Delta_r G^\circ + RT \ln Q$ keeps them distinct and shows precisely how they connect.
$\Delta_r G$ is the actual Gibbs energy change of the reaction mixture at its current composition. As the reaction advances, $Q$ changes, so $\Delta_r G$ changes too; it sweeps toward zero and arrives at zero only at equilibrium. By contrast, $\Delta_r G^\circ$ is a fixed number at a given temperature — the change for converting reactants in their standard states entirely to products in their standard states. It does not change as the mixture equilibrates.
| Property | ΔG (actual) | ΔG° (standard) |
|---|---|---|
| Depends on composition? | Yes, via $Q$ | No, fixed at a given $T$ |
| Defining equation | ΔG = ΔG° + RT ln Q | ΔG° = −RT ln K |
| Value at equilibrium | Exactly $0$ | Generally non-zero (zero only if $K = 1$) |
| Predicts spontaneity of mixture? | Yes (sign of $\Delta G$) | No, only the position of equilibrium ($K$) |
| Quantity it pairs with | $Q$ (reaction quotient) | $K$ (equilibrium constant) |
The pairing in the last row is worth memorising as a discipline: $\Delta G$ goes with $Q$, and $\Delta G^\circ$ goes with $K$. Substituting equilibrium into $\Delta G = \Delta G^\circ + RT \ln Q$ collapses $Q \to K$ and $\Delta G \to 0$, which is exactly why $\Delta G^\circ = -RT \ln K$ contains $K$, not $Q$.
ΔG vs ΔG°, and Q vs K — never swap them
Examiners exploit two confusions. First, candidates write "$\Delta G = -RT \ln K$", which is wrong: the standard symbol $\Delta G^\circ$ belongs in that equation. The actual $\Delta G$ obeys $\Delta G = \Delta G^\circ + RT \ln Q$. Second, candidates put $K$ inside the quotient term while away from equilibrium; the variable composition term always carries $Q$, and $Q$ becomes $K$ only at equilibrium.
Rule: $\Delta G^\circ \leftrightarrow K$ (a constant pair); $\Delta G \leftrightarrow Q$ (a variable pair). $\Delta G = 0$ at equilibrium, but $\Delta G^\circ$ is zero only when $K = 1$.
Sign of ΔG° and Magnitude of K
Because $\Delta_r G^\circ = -2.303\,RT \log K$ and $RT$ is always positive, the sign of $\Delta_r G^\circ$ fixes whether $\log K$ is positive or negative, and therefore whether $K$ is greater or less than unity. This is the qualitative payoff of the relation: a single number tells you which side of the reaction is favoured at equilibrium.
| Sign of ΔrG° | log K | K relative to 1 | Position / extent of reaction |
|---|---|---|---|
| $\Delta_r G^\circ < 0$ (negative) | $> 0$ | $K > 1$ | Products favoured; reaction proceeds far forward |
| $\Delta_r G^\circ = 0$ | $= 0$ | $K = 1$ | Comparable amounts of reactants and products |
| $\Delta_r G^\circ > 0$ (positive) | $< 0$ | $K < 1$ | Reactants favoured; little product forms |
NCERT links this directly to enthalpy. For strongly exothermic reactions $\Delta_r H^\circ$ is large and negative, so $\Delta_r G^\circ$ is likely to be large and negative too, $K$ is much larger than 1, and the reaction can go nearly to completion. For strongly endothermic reactions $\Delta_r H^\circ$ is large and positive, $K$ is much smaller than 1, and little product is formed. The entropy term $\Delta_r S^\circ$ modifies this picture depending on its sign.
The dependence of $K$ on $\Delta_r G^\circ$ is exponential, not linear, which is why modest changes in $\Delta_r G^\circ$ produce dramatic swings in equilibrium yield. At 298 K, every $5.7\ \text{kJ mol}^{-1}$ of additional negative $\Delta_r G^\circ$ multiplies $K$ by a factor of ten, since $2.303\,RT \approx 5.7\ \text{kJ mol}^{-1}$. A reaction with $\Delta_r G^\circ = -57\ \text{kJ mol}^{-1}$ therefore has $K \approx 10^{10}$ and is effectively irreversible, while one with $\Delta_r G^\circ = +57\ \text{kJ mol}^{-1}$ has $K \approx 10^{-10}$ and barely proceeds. This is the quantitative reason chemists treat a large negative $\Delta_r G^\circ$ as the marker of a "feasible" reaction.
The Gibbs equation $\Delta G = \Delta H - T\Delta S$ and the spontaneity sign rules underpin everything here. Revise them in Spontaneity, Entropy & Gibbs Energy.
The Free-Energy Minimum
The geometric statement of equilibrium is that the Gibbs energy of the reacting system, plotted against the extent of reaction, passes through a minimum. Approaching that minimum from the reactant side, the slope is negative ($\Delta G < 0$, forward reaction spontaneous); approaching from the product side, the slope is positive ($\Delta G > 0$, reverse reaction spontaneous). Exactly at the minimum the slope is zero, which is the analytical meaning of $\Delta_r G = 0$.
The position of that minimum along the extent-of-reaction axis is what $K$ encodes. A reaction with $K \gg 1$ has its minimum displaced far toward the product end, so equilibrium is reached only after most reactant has been consumed; a reaction with $K \ll 1$ has its minimum near the reactant end, and equilibrium arrives after only a trace of product forms. Crucially, the curve dips below both pure-reactant and pure-product extremes: mixing reactants and products always lowers $G$ through the entropy of mixing, which is why no real reaction goes fully to completion and a genuine equilibrium mixture always contains some of every species.
Gibbs energy of the system versus extent of reaction, with the minimum marking equilibrium.
Because both directions must lower $G$, the only stable configuration is the minimum. Any displacement away from it raises $G$, so the system returns — the thermodynamic basis of Le Chatelier-type restoring behaviour.
Worked Examples
The two interconversions below mirror NCERT Problems 5.13 and 5.12 and the type of arithmetic NEET sets. Keep units consistent: with $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $\Delta_r G^\circ$ comes out in joules per mole.
The standard Gibbs energy change for a reaction at 298 K is $\Delta_r G^\circ = -13.6\ \text{kJ mol}^{-1}$. Find the equilibrium constant $K$. (NCERT Problem 5.13)
Start from $\Delta_r G^\circ = -2.303\,RT \log K$ and solve for $\log K$:
$$\log K = \frac{-\Delta_r G^\circ}{2.303\,RT} = \frac{-(-13.6 \times 10^{3}\ \text{J mol}^{-1})}{2.303 \times (8.314\ \text{J K}^{-1}\text{mol}^{-1}) \times (298\ \text{K})}$$
$$\log K = \frac{13600}{5705} \approx 2.38 \;\Rightarrow\; K = \text{antilog}(2.38) = 2.4 \times 10^{2}$$
A negative $\Delta_r G^\circ$ has produced $K \gg 1$, confirming a product-favoured reaction, exactly as the sign table predicts.
For the conversion of oxygen to ozone, $\ce{3/2 O2(g) -> O3(g)}$ at 298 K, the equilibrium constant is $K_p = 2.47 \times 10^{-29}$. Calculate $\Delta_r G^\circ$. (NCERT Problem 5.12)
Apply $\Delta_r G^\circ = -2.303\,RT \log K_p$ with $\log(2.47 \times 10^{-29}) \approx -28.61$:
$$\Delta_r G^\circ = -2.303 \times (8.314\ \text{J K}^{-1}\text{mol}^{-1}) \times (298\ \text{K}) \times (-28.61)$$
$$\Delta_r G^\circ \approx +163000\ \text{J mol}^{-1} = +163\ \text{kJ mol}^{-1}$$
A vanishingly small $K_p$ has produced a large positive $\Delta_r G^\circ$: the formation of ozone from oxygen is strongly non-spontaneous under standard conditions, consistent with ozone's negligible equilibrium yield.
A reaction has equilibrium constant $K = 10$ at a temperature $T$. State the sign of $\Delta_r G^\circ$ and explain what $\Delta_r G^\circ$ would be if $K$ were exactly 1.
Since $K = 10 > 1$, $\log K = 1 > 0$, so $\Delta_r G^\circ = -2.303\,RT(1) < 0$: the reaction is product-favoured. If instead $K = 1$, then $\log K = 0$ and $\Delta_r G^\circ = 0$ — the only case in which the standard change vanishes. This is the precise boundary separating $K > 1$ from $K < 1$.
Temperature, Enthalpy and K
Substituting the Gibbs equation $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$ into the master relation lets temperature enter the value of $K$. Combining the two,
This shows the two practical routes NCERT highlights. If $\Delta_r H^\circ$ and $\Delta_r S^\circ$ are known, $\Delta_r G^\circ$ and hence $K$ can be estimated at any temperature for predicting economic yields; conversely, if $K$ is measured at one temperature, $\Delta_r G^\circ$ at another can be inferred. Many reactions that are non-spontaneous at low temperature ($\Delta_r G^\circ > 0$, $K < 1$) become spontaneous at high temperature ($\Delta_r G^\circ < 0$, $K > 1$) when $\Delta_r S^\circ$ is positive, because the $-T\Delta_r S^\circ$ term grows.
Calorie vs joule values of R
Some NEET items quote $R = 2\ \text{cal K}^{-1}\text{mol}^{-1}$ instead of $8.314\ \text{J K}^{-1}\text{mol}^{-1}$. The whole calculation then returns $\Delta G^\circ$ in calories, not joules. Watch the units stated in the question and carry them through; the form $\Delta G^\circ = -2.303\,RT \log K$ is identical in either unit system.
Rule: Pick $R$ to match the answer's required unit — $8.314\ \text{J K}^{-1}\text{mol}^{-1}$ for joules, $2\ \text{cal K}^{-1}\text{mol}^{-1}$ for calories.
Gibbs energy & equilibrium in one screen
- Equilibrium criterion: $\Delta_r G = 0$, the minimum of the system's free-energy curve.
- Away from equilibrium: $\Delta_r G = \Delta_r G^\circ + RT \ln Q$; $\Delta G$ pairs with $Q$.
- Master relation: $\Delta_r G^\circ = -RT \ln K = -2.303\,RT \log K$; $\Delta G^\circ$ pairs with $K$.
- $\Delta_r G^\circ < 0 \Rightarrow K > 1$ (products favoured); $\Delta_r G^\circ > 0 \Rightarrow K < 1$; $\Delta_r G^\circ = 0 \Rightarrow K = 1$.
- Strongly exothermic ($\Delta_r H^\circ \ll 0$) typically gives $K \gg 1$ and near-complete reaction.
- The equation runs both ways: compute $K$ from $\Delta_r G^\circ$, or $\Delta_r G^\circ$ from a measured $K$.