What is the difference between ΔG and ΔG°?

ΔG is the Gibbs energy change of a reaction mixture under the actual, instantaneous composition, and it depends on the reaction quotient Q. ΔG° is the standard Gibbs energy change, measured when every reactant and product is in its standard state (unit activity, that is Q = 1), so ΔG° is a fixed constant for a reaction at a given temperature. They are linked by ΔG = ΔG° + RT ln Q. Crucially, the sign of ΔG decides spontaneity at the current composition, whereas the sign of ΔG° only fixes the value and direction of the equilibrium constant K through ΔG° = −RT ln K.

How is ΔG° related to the equilibrium constant K?

At equilibrium ΔG = 0 and Q = K, so the equation ΔG = ΔG° + RT ln Q reduces to 0 = ΔG° + RT ln K, giving ΔG° = −RT ln K, or equivalently K = e^(−ΔG°/RT). A negative ΔG° makes the exponent positive, so K > 1 and products dominate at equilibrium; a positive ΔG° makes K < 1, so reactants dominate; and ΔG° = 0 gives K = 1.

How do you predict the direction of a reaction using Q and K?

Compute the reaction quotient Q from the current concentrations or partial pressures, then compare it with K. If Q K it proceeds in the reverse direction toward reactants; if Q = K the system is already at equilibrium and no net reaction occurs. The same comparison can be read through Gibbs energy: Q K gives ΔG > 0 (reverse), and Q = K gives ΔG = 0.

Does a negative ΔG° guarantee that a reaction goes to completion?

No. A negative ΔG° guarantees only that K > 1, meaning products predominate at equilibrium, not that the reaction goes fully to completion. Some reactant always remains because equilibrium is dynamic. A large negative ΔG° gives a very large K and pushes the position of equilibrium far toward products, so the reaction is nearly complete; a small negative ΔG° gives a modest K. ΔG° fixes only the equilibrium position, not the rate.

Why does ΔG become zero at equilibrium?

ΔG is the driving force for a reaction; it measures the free energy still available to push the system toward products or reactants. As the reaction proceeds, Q changes and ΔG = ΔG° + RT ln Q steadily moves toward zero. At equilibrium the forward and reverse tendencies exactly balance, so there is no net free energy left to drive change in either direction. ΔG = 0 marks the minimum of the Gibbs energy versus reaction-extent curve.

What temperature must be used in ΔG° = −RT ln K, and in what units is R taken?

Temperature T must be the absolute (kelvin) temperature, and R is the gas constant 8.314 J K⁻¹ mol⁻¹ when ΔG° is expressed in joules per mole. Always convert any ΔG° given in kJ mol⁻¹ to J mol⁻¹ before substituting, and remember that ln K is the natural logarithm; if you use log₁₀ you must write ΔG° = −2.303 RT log K instead.

Relationship Between K, Q and ΔG — NEET Chemistry

Q: How do you predict the direction of a reaction using Q and K?

Compute the reaction quotient Q from the current concentrations or partial pressures, then compare it with K. If Q K it proceeds in the reverse direction toward reactants; if Q = K the system is already at equilibrium and no net reaction occurs. The same comparison can be read through Gibbs energy: Q K gives ΔG > 0 (reverse), and Q = K gives ΔG = 0.

Why thermodynamics fixes K

The value of $K_c$ for a reaction does not depend on how fast the reaction runs. As established in the chapter on Thermodynamics, it is instead governed by the change in Gibbs energy, $\Delta G$. Gibbs energy is the thermodynamic quantity whose sign tells us the spontaneous direction of a process at constant temperature and pressure. A reaction proceeds in the direction that lowers the Gibbs energy of the system, and it stops — reaches equilibrium — at the composition where the Gibbs energy is at its minimum.

NCERT states the qualitative rule first. If $\Delta G$ is negative, the reaction is spontaneous and proceeds in the forward direction. If $\Delta G$ is positive, the forward reaction is non-spontaneous; the reverse reaction has a negative $\Delta G$, so products convert back to reactants. If $\Delta G$ is zero, the reaction has reached equilibrium and there is no longer any free energy left to drive it in either direction.

Sign of ΔG	Thermodynamic meaning	What the system does
`ΔG < 0`	Forward reaction spontaneous	Net reaction proceeds forward (toward products)
`ΔG > 0`	Forward reaction non-spontaneous	Net reaction proceeds in reverse (toward reactants)
`ΔG = 0`	No net driving force	System is at equilibrium; no net change

The master equation: ΔG = ΔG° + RT ln Q

The qualitative picture is made quantitative by relating $\Delta G$ at any instant to the standard Gibbs energy change $\Delta G^\circ$ and the reaction quotient $Q$. NCERT writes this as equation 6.21:

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

Here $\Delta G^\circ$ is the standard Gibbs energy change, $R$ is the gas constant ($8.314\ \text{J K}^{-1}\text{mol}^{-1}$), $T$ is the absolute temperature, and $Q$ is the reaction quotient evaluated from the actual, instantaneous composition. For a general reaction $a\,\ce{A} + b\,\ce{B} \rightleftharpoons c\,\ce{C} + d\,\ce{D}$, the quotient has the same algebraic form as the equilibrium constant, $Q_c = \dfrac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$, but uses concentrations that are not necessarily equilibrium values.

The two terms play different roles. $\Delta G^\circ$ is a fixed constant for the reaction at a given temperature — it depends only on the standard states of reactants and products. The term $RT\ln Q$ is the composition-dependent correction: as the mixture evolves, $Q$ changes, and so does $\Delta G$.

From equilibrium to ΔG° = −RT ln K

At equilibrium two facts hold simultaneously: the driving force has vanished, so $\Delta G = 0$, and the reaction quotient has reached its equilibrium value, so $Q = K$. Substituting both into the master equation collapses it (NCERT equation 6.22):

$$\Delta G = \Delta G^\circ + RT\ln K = 0 \quad\Longrightarrow\quad \Delta G^\circ = -RT\ln K$$

Rearranging gives $\ln K = -\dfrac{\Delta G^\circ}{RT}$, and taking the antilogarithm of both sides yields the exponential form (equation 6.23):

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

This is the bridge the chapter is built around. It means the equilibrium constant is completely determined once $\Delta G^\circ$ and $T$ are known, and conversely $\Delta G^\circ$ can be obtained by measuring $K$. Note that the logarithm here is the natural logarithm; if you prefer base-10, use $\Delta G^\circ = -2.303\,RT\log K$.

Sign of ΔG° and the magnitude of K

Because $K = e^{-\Delta G^\circ / RT}$, the sign of $\Delta G^\circ$ controls whether $K$ exceeds or falls below unity, and therefore whether products or reactants predominate at equilibrium. NCERT spells out the two cases explicitly.

If $\Delta G^\circ < 0$, then $-\Delta G^\circ / RT$ is positive, so $e^{-\Delta G^\circ / RT} > 1$, making $K > 1$. This implies a reaction that proceeds in the forward direction to such an extent that products are present predominantly. If $\Delta G^\circ > 0$, then $-\Delta G^\circ / RT$ is negative, so $e^{-\Delta G^\circ / RT} < 1$, making $K < 1$; the forward reaction proceeds to such a small degree that only a minute quantity of product forms.

Sign of ΔG°	Value of K	Extent of reaction at equilibrium
`ΔG° < 0` (negative)	`K > 1`	Products predominate; forward reaction proceeds far
`ΔG° = 0`	`K = 1`	Comparable amounts of reactants and products
`ΔG° > 0` (positive)	`K < 1`	Reactants predominate; very little product forms

The relationship is steep because of the exponential. A modest $\Delta G^\circ$ of only a few tens of kJ mol⁻¹ moves $K$ across many orders of magnitude. This is why some reactions with a favourable $\Delta G^\circ$ have enormous equilibrium constants, as in the hydrolysis of sucrose where $K_c$ reaches $2\times10^{13}$.

Worked example: K from ΔG°

Worked Example 1

The value of $\Delta G^\circ$ for the phosphorylation of glucose in glycolysis is $13.8\ \text{kJ mol}^{-1}$. Find the value of $K_c$ at $298\ \text{K}$. (NCERT Problem 6.10)

Convert to joules: $\Delta G^\circ = 13.8\times10^{3}\ \text{J mol}^{-1}$ (note the sign is positive). Apply $\Delta G^\circ = -RT\ln K_c$:

$$\ln K_c = \frac{-\Delta G^\circ}{RT} = \frac{-13.8\times10^{3}}{8.314 \times 298} = -5.569$$

Hence $K_c = e^{-5.569} = 3.81\times10^{-3}$.

Because $\Delta G^\circ$ is positive, $K_c$ comes out far below 1, consistent with a reaction that barely proceeds on its own — exactly why phosphorylation in the cell is coupled to ATP hydrolysis.

The numerical recipe is the same every time: convert $\Delta G^\circ$ to joules, divide by $-RT$ to get $\ln K$, then exponentiate. The only place students lose marks is dropping the negative sign or forgetting the kJ-to-J conversion.

Build the foundation first

Shaky on how $K_c$ and $K_p$ are written and connected? Revisit Equilibrium Constant: Kc and Kp before applying the Gibbs-energy relation.

Q versus K: predicting direction

The reaction quotient $Q$ is the practical tool for deciding which way a non-equilibrium mixture will move. Computed exactly like $K$ but from the current composition, $Q$ is then compared with the (fixed) equilibrium constant. NCERT gives the generalisations directly, and the same comparison maps onto the sign of $\Delta G$ through the master equation.

Comparison	Resulting ΔG	Net direction of reaction
`Q < K`	`ΔG < 0`	Forward — net reaction goes from left to right (toward products)
`Q = K`	`ΔG = 0`	At equilibrium — no net reaction occurs
`Q > K`	`ΔG > 0`	Reverse — net reaction goes from right to left (toward reactants)

The logic behind the $\Delta G$ column follows from $\Delta G = \Delta G^\circ + RT\ln Q = -RT\ln K + RT\ln Q = RT\ln(Q/K)$. When $Q < K$ the ratio $Q/K < 1$, its logarithm is negative, and $\Delta G < 0$, so the forward reaction is spontaneous. When $Q > K$ the logarithm is positive and $\Delta G > 0$, so the system runs backward. The two viewpoints — comparing $Q$ with $K$, or reading the sign of $\Delta G$ — are one and the same statement.

Figure 1. The reaction quotient is read against the fixed value of K. To the left of equilibrium ($Q<K$) the system has too little product, ΔG is negative, and the net reaction runs forward; to the right ($Q>K$) it runs in reverse until $Q$ climbs or falls back to $K$.

Worked example: predicting direction

Worked Example 2

The value of $K_c$ for the reaction $\ce{2A <=> B + C}$ is $2\times10^{-3}$. At a given time the composition of the reaction mixture is $[\ce{A}] = [\ce{B}] = [\ce{C}] = 3\times10^{-4}\ \text{M}$. In which direction will the reaction proceed? (NCERT Problem 6.7)

Write the reaction quotient for this stoichiometry: $Q_c = \dfrac{[\ce{B}][\ce{C}]}{[\ce{A}]^2}$.

$$Q_c = \frac{(3\times10^{-4})(3\times10^{-4})}{(3\times10^{-4})^2} = 1$$

Compare with $K_c = 2\times10^{-3}$. Since $Q_c\,(1) > K_c\,(2\times10^{-3})$, the system has too much product relative to equilibrium. Therefore $\Delta G > 0$ and the reaction proceeds in the reverse direction, consuming B and C to re-form A until $Q_c$ falls to $2\times10^{-3}$.

Notice that the answer hinged on a comparison, not on solving an equilibrium expression. Whenever a question hands you a set of instantaneous concentrations and an equilibrium constant, the fastest route is to evaluate $Q$, compare, and read the direction off the table.

The Gibbs energy versus extent curve

The whole framework is captured by a single curve: the Gibbs energy of the system plotted against the extent of reaction. Starting from pure reactants, $G$ falls as the reaction advances, reaches a minimum, then would rise again toward pure products. The minimum is the equilibrium composition, and the slope of the curve at any point is $\Delta G$.

Figure 2. Gibbs energy passes through a minimum as a reaction advances. Left of the minimum the system rolls forward ($\Delta G<0$); right of it the system rolls back ($\Delta G>0$); the trough itself is equilibrium, where $\Delta G=0$ and $Q=K$. A deeper minimum shifted toward products corresponds to a more negative $\Delta G^\circ$ and a larger $K$.

ΔG versus ΔG°: the core distinction

The single most frequent error in this topic is treating $\Delta G$ and $\Delta G^\circ$ as interchangeable. They are not. $\Delta G^\circ$ is a constant tied to standard states and to $K$; $\Delta G$ is a variable that tracks the actual composition and decides spontaneity here and now.

NEET Trap

Do not confuse ΔG with ΔG°, or Q with K

A reaction with $\Delta G^\circ > 0$ (hence $K < 1$) can still run forward at a given instant if the mixture happens to have $Q < K$ — because then $\Delta G < 0$. The sign of $\Delta G^\circ$ tells you only where equilibrium lies; the sign of $\Delta G$ tells you which way the mixture moves right now. Likewise, "$\Delta G = 0$" means equilibrium, but "$\Delta G^\circ = 0$" merely means $K = 1$.

Use $\Delta G^\circ = -RT\ln K$ only at equilibrium (or to find $K$). Use $\Delta G = \Delta G^\circ + RT\ln Q = RT\ln(Q/K)$ for any instantaneous composition.

Feature	ΔG (Gibbs energy change)	ΔG° (standard Gibbs energy change)
Depends on composition?	Yes — varies with Q	No — fixed for the reaction at given T
Decides spontaneity?	Yes — its sign gives the net direction	No — only sets the value of K
Value at equilibrium	`0`	`−RT ln K` (generally non-zero)
Governing equation	`ΔG = ΔG° + RT ln Q`	`ΔG° = −RT ln K`

Quick Recap

One relationship, three quantities

Master equation: $\Delta G = \Delta G^\circ + RT\ln Q$, valid at any instant.
At equilibrium $\Delta G = 0$ and $Q = K$, giving $\Delta G^\circ = -RT\ln K$, equivalently $K = e^{-\Delta G^\circ/RT}$.
$\Delta G^\circ < 0 \Rightarrow K > 1$ (products predominate); $\Delta G^\circ > 0 \Rightarrow K < 1$ (reactants predominate); $\Delta G^\circ = 0 \Rightarrow K = 1$.
Direction from $Q$ vs $K$: $QK\Rightarrow\Delta G>0$ (reverse); $Q=K\Rightarrow\Delta G=0$ (equilibrium).
$\Delta G$ varies with composition and decides direction; $\Delta G^\circ$ is fixed and only sets $K$. Always use $T$ in kelvin and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.

Relationship Between K, Q and ΔG