Chemistry · Thermodynamics · NEET
They are not the same. ΔG° (delta G naught) is the standard Gibbs energy change, measured when every reactant and product is in its standard state (1 bar pressure, 1 M concentration). It is a FIXED number for a reaction at a given temperature. ΔG (without the °) is the actual Gibbs energy change at any real moment during the reaction, and it keeps changing as the reaction proceeds. The formula ΔG° = −RT ln K uses ΔG° (the standard one), because K is measured at equilibrium standard conditions. Remember: only ΔG becomes zero at equilibrium, not ΔG°.
If K > 1, then ln K is positive, so −RT ln K is NEGATIVE. That means ΔG° is negative and the forward reaction is spontaneous (products are favoured). If K < 1, ln K is negative, so ΔG° is positive (reactants favoured). If K = 1, ln K = 0, so ΔG° = 0. Simple rule: big K → negative ΔG° → products win. This is a very common NEET one-liner question.
The minus sign makes the signs match up correctly. A spontaneous reaction has a large K (products favoured) and must have a negative ΔG°. Without the minus sign, a large positive ln K would give a positive ΔG°, which would wrongly say the reaction is non-spontaneous. The minus sign flips it so that large K gives negative ΔG°. It comes directly from ΔG = ΔG° + RT ln Q; at equilibrium ΔG = 0 and Q = K, so 0 = ΔG° + RT ln K, giving ΔG° = −RT ln K.
No. ΔG = 0 at equilibrium means the forward and backward reactions happen at the same rate, so there is no NET change. The reaction is still going both ways, just balanced. Note carefully: it is ΔG (actual) that equals zero at equilibrium, NOT ΔG°. ΔG° is a fixed non-zero number (unless K happens to equal 1). Students lose marks by writing ΔG° = 0 at equilibrium — that is wrong.
Both are the same equation. Since ln K = 2.303 × log K, you can write ΔG° = −RT ln K (natural log) or ΔG° = −2.303 RT log K (base-10 log). Use the log-10 version when the numbers give a clean log (like log 10 = 1 or log 100 = 2), and use ln when the question already gives values in ln form. NEET options usually show one form clearly, so match your working to the answer choices.
Use R = 8.314 J mol⁻¹ K⁻¹ when the answer is needed in joules or kilojoules. Use R = 2 cal mol⁻¹ K⁻¹ (approximately) when the question is set in calories. The question always tells you which R to use — read it. Temperature T must always be in kelvin (K), never in °C. Getting units wrong here is the most common NEET calculation mistake.
Hydrolysis of sucrose: Sucrose + H₂O ⇌ Glucose + Fructose. If the equilibrium constant (Kc) is 2 × 10¹³ at 300 K, the value of ΔrG° at the same temperature will be:
The equilibrium concentrations of the species in A + B ⇌ C + D are 2, 3, 10 and 6 mol L⁻¹ respectively at 300 K. ΔG° for the reaction is (Given R = 2 cal mol⁻¹K⁻¹).
If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for ΔG° and K_eq?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It links the standard Gibbs energy change to the equilibrium constant. It lets you find K if you know ΔG°, or find ΔG° if you know K. NEET asks direct one-step plug-in questions from this almost every year, so it is high-value.
ΔG° = 0 only when ln K = 0, which means K = 1. At K = 1, products and reactants are equally favoured at standard conditions. Do not confuse this with ΔG = 0, which happens at equilibrium for any reaction.
No. The equilibrium constant K is always positive (it comes from concentrations or pressures, which cannot be negative). So you can always take ln K. Only ΔG° and ln K can be negative, never K itself.
Use ln K = 2.303 × log K. So ΔG° = −RT ln K is the same as ΔG° = −2.303 RT log K. Pick whichever matches the numbers in the question for easy calculation.
Both equal the same ΔG°. Since ΔG° = −nFE°cell and ΔG° = −RT ln K, you can combine them to get nFE°cell = RT ln K. This connects electrochemistry with equilibrium — a favourite NEET cross-topic idea covered next in Gibbs Energy and Cell EMF.
