Equilibrium constant, reaction quotient and Gibbs energy

How are K, Q and Gibbs energy related?

In plain words: Δ_rG is the “push” on a reaction mixture. It has two parts: a fixed part, Δ_rG^⊖, set by the reaction and the temperature, and a part RT ln Q that depends on how far the mixture has already gone. As products build up, Q grows and the push shrinks, until it vanishes at Q = K. That is the same rule as comparing Q with K, written in energy terms.

Here Δ_rG is the instantaneous slope of the mixture’s Gibbs energy with reaction extent, per mole of reaction as written; it is not the finite total Gibbs energy change between two arbitrary mixtures. Its sign predicts a favoured direction, not a reaction rate.

Gibbs energy and the direction of reaction

NCERT states (page 184), recalling Unit 5:

• if ΔG is negative, the reaction is spontaneous and proceeds forward;
• if ΔG is positive, the forward reaction is non-spontaneous; the reverse reaction has a negative ΔG, so products are converted to reactants;
• if ΔG is 0, the reaction has reached equilibrium:

∆G is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

NCERT Chemistry Part I (Class 11), Section 6.7, page 184

NCERT opens Section 6.7 by noting that the value of K_c does not depend on the rate of the reaction; it is fixed by the thermodynamics of the reaction. The relation between them (NCERT Eqs. 6.21–6.22):

At equilibrium Δ_rG = 0 and Q = K, so 0 = Δ_rG^⊖ + RT ln K:

Here Δ_rG^⊖ refers to specified standard states and Q and K are dimensionless activity products. For ideal gases use p/p^⊖, usually with p^⊖ = 1 bar; for dilute solutions use approximately c/c^⊖, usually with c^⊖ = 1 mol L⁻¹. A real 1 M solution need not have unit activity. Present pure bulk solids and liquids have activity approximately 1. A gas concentration-standard K can also be used, but its standard Gibbs energy must match that convention; mixing concentration-based K with a pressure-standard Gibbs energy gives the wrong result when Δn ≠ 0.

Putting the two together gives a compact form, beyond the NCERT text but useful in JEE problems:

So Δ_rG < 0 exactly when Q < K, and the mixture moves forward; Δ_rG > 0 when Q > K, and it moves backward (compare the standard Gibbs energy and K in Unit 5).

What the sign of ΔG^⊖ tells us

From K = e^{−Δ_rG⊖/RT} (NCERT Eq. 6.23, page 184):

• Δ_rG^⊖ < 0 gives K > 1: the products are favoured at equilibrium;
• Δ_rG^⊖ > 0 gives K < 1: the reactants are favoured; the actual product fraction also depends on stoichiometry and initial amounts.

NCERT calls the first case “a spontaneous reaction” and the second “a non-spontaneous reaction”. That is a shorthand. The sign of Δ_rG^⊖ tells us whether K is above or below 1, and which way a mixture would go if every substance were in its standard state (Q = 1); “spontaneous” here means spontaneous under standard conditions. Whether a particular mixture reacts forward depends on Δ_rG, that is on Q. A reaction with Δ_rG^⊖ > 0 still goes forward from pure reactants, when an absent gaseous or dissolved product makes Q approach zero, until Q = K. This is a limiting argument for ln Q, not a finite evaluation of ln 0; an absent pure solid product does not contribute a zero activity. (Two print slips on the same NCERT page: “G^⊖ is standard Gibbs energy” should read ΔG^⊖, the standard Gibbs energy change; and its two inequalities for e^−ΔG⊖/RT are misprinted: they mean > 1 and < 1.)

Solved examples

Common mistakes to avoid

Quick revision summary

• Δ_rG = Δ_rG^⊖ + RT ln Q; at equilibrium Δ_rG = 0 and Q = K.
• Δ_rG^⊖ = −RT ln K = −2.303 RT log K, so K = e^{−Δ_rG⊖/RT}.
• Δ_rG = RT ln(Q/K): negative when Q < K (forward), positive when Q > K (backward).
• Δ_rG^⊖ < 0: K > 1, products favoured; Δ_rG^⊖ > 0: K < 1.

Questions students ask

What is the difference between ΔG and ΔG^⊖?

Δ_rG^⊖ is the Gibbs energy change for the specified standard states (unit activity, approximated by ideal gas at 1 bar or ideal solute at 1 mol L^⁻¹), so it has one fixed value for a reaction at a given temperature, while Δ_rG is the Gibbs energy change for the mixture actually present and changes as the reaction proceeds. The two are linked by Δ_rG = Δ_rG^⊖ + RT ln Q and are equal only when Q = 1. Δ_rG^⊖ fixes the position of equilibrium through K, while the sign of Δ_rG tells which way a particular mixture will react.

Is the Q in ΔG = ΔG^⊖ + RT ln Q the same as the heat q?

No, the Q in Δ_rG = Δ_rG^⊖ + RT ln Q is the reaction quotient, not the heat q of the first law of thermodynamics, ΔU = q + w. It is written like the equilibrium constant but with the partial pressures (in bar) and concentrations (in mol L^⁻¹) of the mixture at that moment, each divided by its standard value so that Q is a pure number, and with pure solids and liquids left out. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with p_N2 = 1.0 bar, p_H2 = 3.0 bar and p_NH3 = 0.50 bar, Q = 0.50^²/(1.0 × 3.0^³) ≈ 9.3 × 10^⁻³.

What happens to ΔG^⊖ and K when an equation is reversed or doubled?

Δ_rG^⊖ is quoted per mole of reaction as the equation is written, so reversing the equation changes its sign and doubling the equation doubles it; since Δ_rG^⊖ = −RT ln K, K is inverted on reversal and squared on doubling. For a reaction with Δ_rG^⊖ = −6.00 kJ mol^⁻¹ at 298 K, K = e^6000/2477.6 ≈ 11.3; the doubled equation has −12.0 kJ mol^⁻¹ and K ≈ 127 (11.3^²), the reversed one +6.00 kJ mol^⁻¹ and K ≈ 0.089. This is the thermodynamic reason behind the rules for changing the equation.

How does Le Chatelier’s principle follow from ΔG = ΔG^⊖ + RT ln Q?

Le Chatelier’s principle for changes of concentration and pressure follows from Δ_rG = RT ln(Q/K): selectively increasing a reactant activity or decreasing a product activity while holding the other activities fixed makes Q smaller than K, so Δ_rG turns negative and the mixture reacts forward until Q = K again, while adding a product does the reverse. At 298 K, a change that halves Q in an equilibrium mixture gives Δ_rG = 8.314 × 298 × ln 0.5 ≈ −1.7 kJ mol^⁻¹. A change of temperature is different: it changes Δ_rG^⊖ = −RT ln K, and so K itself.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A reaction has K = 1.0 × 10⁻⁵ at 298 K. Find Δ_rG^⊖.

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   Δ_rG^⊖ = −2.303 RT log K = −2.303 × 8.314 × 298 × (−5) ≈ +28.5 kJ mol⁻¹.

2. Q2. For a reaction at 300 K, Δ_rG^⊖ = −20.0 kJ mol⁻¹. Find K.

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   ln K = 20 000/(8.314 × 300) = 20 000/2494.2 = 8.02, so K = e^8.02 ≈ 3.0 × 10³.

3. Q3. A reaction has K = 10 at 298 K. A mixture has Q = 100. Find Δ_rG and say which way the mixture moves.

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   Δ_rG = RT ln(Q/K) = 8.314 × 298 × ln 10 ≈ +5.7 kJ mol⁻¹. It is positive, so the mixture moves backward, towards reactants, until Q falls to 10.