Why can a reaction with a positive ΔG°′ still run forward in a cell?

ΔG°′ describes standard conditions (1 M concentrations at pH 7), but the actual free energy change is ΔG = ΔG°′ + RT ln Q, which depends on the real concentrations of reactants and products. If a cell keeps product concentration low (Q small, so ln Q is a large negative number), ΔG can be negative and the reaction proceeds even when ΔG°′ is positive. This is how many 'unfavorable' biochemical reactions are pulled forward in vivo.

What is the difference between ΔG° and ΔG°′?

ΔG° is the standard free energy change under standard chemical conditions (1 M concentrations, pH 0). ΔG°′ is the biochemical standard state: pH 7, and by convention water and, for reactions involving it, [H+] are treated as constants rather than 1 M. Biochemistry problems almost always use ΔG°′.

What does the crossover temperature actually mean?

It's the temperature where ΔG = 0, i.e. T = ΔH/ΔS, the point where a reaction switches from spontaneous to non-spontaneous (or vice versa) as temperature changes. It only exists as a real physical flip when ΔH and ΔS have the same sign; if they have opposite signs, the reaction is spontaneous at every temperature (or non-spontaneous at every temperature).

Why does entropy need to be converted from J to kJ before using ΔG = ΔH − TΔS?

ΔH is conventionally reported in kJ/mol while ΔS is conventionally reported in J/mol·K, mixing units without converting one of them would make the subtraction meaningless. Dividing ΔS by 1000 puts both terms in kJ/mol before combining them.

Gibbs Free Energy Calculator (ΔG=ΔH−TΔS): Free

ΔH, enthalpy (kJ/mol)

ΔS, entropy (J/mol·K)

Temperature

Temp units

How to tell if a reaction is spontaneous

Spontaneity is set by the sign of ΔG, not by ΔH or ΔS alone. Use ΔG = ΔH − TΔS (with T in kelvin and ΔS converted from J to kJ). If ΔG is negative the reaction is spontaneous (exergonic) and releases free energy; if positive it is non-spontaneous (endergonic) and needs an input of energy; if zero the system is at equilibrium. Because temperature multiplies the entropy term, some reactions switch spontaneity at a crossover temperature of T = ΔH / ΔS. In biochemistry, the actual ΔG also depends on concentrations through ΔG = ΔG°′ + RT ln Q, which is why a reaction with a positive ΔG°′ can still run forward when products are kept low.

Constants: R = 8.314 J/mol·K = 0.008314 kJ/mol·K. 0 °C = 273.15 K. Standard biochemical conditions use ΔG°′ (pH 7).

Worked example 1: from ΔH, ΔS, and T (the default)

ΔH = −40 kJ/mol, ΔS = −100 J/mol·K, at 25 °C.

T = 25 + 273.15 = 298.15 K ΔS = −100 J/mol·K = −0.100 kJ/mol·K ΔG = ΔH − TΔS = (−40) − (298.15)(−0.100) = −40 + 29.815 = −10.19 kJ/mol → ΔG < 0: spontaneous (exergonic) Crossover: T = ΔH/ΔS = (−40)/(−0.100) = 400.0 K (126.9 °C) Both ΔH and ΔS are negative, so below 400 K the reaction is spontaneous; above 400 K it becomes non-spontaneous.

This matches the calculator's default output exactly. Notice the sign logic: a negative ΔH (heat-releasing) and negative ΔS (more ordered products) compete, at low temperature the enthalpy term wins and the reaction is spontaneous; push the temperature high enough and the entropy penalty (−TΔS becomes a large positive number) takes over.

Worked example 2: ATP hydrolysis in a cell (ΔG°′ and Q)

ATP hydrolysis has a standard biochemical free energy of ΔG°′ = −30.5 kJ/mol, the tool's default. Inside a cell, product concentrations are kept low relative to substrate, giving a reaction quotient around Q = 0.01, at 37 °C body temperature.

T = 37 + 273.15 = 310.15 K R = 0.008314 kJ/mol·K RT ln Q = (0.008314)(310.15) × ln(0.01) = 2.5786 × (−4.6052) = −11.87 kJ/mol ΔG = ΔG°′ + RT ln Q = (−30.5) + (−11.87) = −42.37 kJ/mol

The cell's actual ΔG (−42.4 kJ/mol) is considerably more negative than the textbook standard value (−30.5 kJ/mol). This is a real illustration of how cells keep ATP hydrolysis strongly favorable by controlling concentrations, not just by relying on the standard free energy value.

FAQ

Why can a reaction with positive ΔG°′ still run forward in a cell?
Actual ΔG = ΔG°′ + RT ln Q depends on real concentrations, not the standard 1 M state. If a cell keeps products low, Q is small, ln Q is a large negative number, and ΔG can go negative even when ΔG°′ is positive.

What's the difference between ΔG° and ΔG°′?
ΔG° is standard conditions (1 M, pH 0). ΔG°′ is the biochemical standard state, pH 7, with water and (where relevant) H⁺ treated as constants rather than 1 M. Biochemistry problems almost always use ΔG°′.

What does the crossover temperature mean?
The temperature where ΔG = 0, T = ΔH/ΔS, where the reaction flips between spontaneous and non-spontaneous. It's only a real physical flip when ΔH and ΔS share the same sign; opposite signs mean the reaction is spontaneous (or non-spontaneous) at every temperature.

Why convert ΔS from J to kJ?
ΔH is conventionally in kJ/mol, ΔS in J/mol·K, mixing units without converting would make ΔH − TΔS meaningless. Dividing ΔS by 1000 puts both terms in kJ/mol before combining.

Practice problems

1. A reaction has ΔH = +50 kJ/mol and ΔS = +120 J/mol·K at 25°C. Is it spontaneous? Find the crossover temperature.

Show answer

T = 25 + 273.15 = 298.15 K ΔS = 120 J/mol·K = 0.120 kJ/mol·K ΔG = ΔH − TΔS = 50 − (298.15)(0.120) = 50 − 35.78 = 14.22 kJ/mol → ΔG > 0: non-spontaneous at 25°C Crossover: T = ΔH/ΔS = 50/0.120 = 416.7 K (143.5°C) Both ΔH and ΔS are positive, so above 416.7 K the reaction becomes spontaneous; below it, non-spontaneous, the opposite temperature dependence from a reaction with both negative.

2. Why does a reaction with positive ΔH and positive ΔS behave oppositely (temperature-wise) to one with both negative?

Show answer

ΔG = ΔH − TΔS. The −TΔS term's sign depends on ΔS's sign: if ΔS is positive, increasing T makes −TΔS more negative, pulling ΔG down (toward spontaneous) as temperature rises. If ΔS is negative instead, increasing T makes −TΔS more positive, pushing ΔG up (toward non-spontaneous) as temperature rises, the opposite direction. So a positive-ΔH/positive-ΔS reaction is spontaneous only above its crossover temperature (entropy needs heat to "win"), while a negative-ΔH/negative-ΔS reaction is spontaneous only below its crossover temperature (enthalpy needs to stay dominant).