What happens to the free energy (∆G) if both ∆H is negative and ∆S is positive?

In thermodynamics, the change in free energy (∆G) is determined by the equation:

[ \Delta G = \Delta H - T \Delta S ]

where:

• ∆G is the change in free energy,
• ∆H is the change in enthalpy,
• T is the temperature in Kelvin,
• ∆S is the change in entropy.

When ∆H is negative, it indicates that the reaction is exothermic, releasing heat. A negative ∆H is generally favorable for spontaneity.

When ∆S is positive, it suggests that the disorder of the system increases, which is also favorable for spontaneity.

If both ∆H is negative and ∆S is positive, the equation indicates that the first term (∆H) will contribute a negative value, while the second term (-T∆S) will be subtracted from it. Since T is positive (as temperature in Kelvin cannot be zero or negative), the entropy term will always be negative. Combining these effects leads to a situation where ∆G will be negative at all relevant temperatures, thus favoring spontaneity.

Therefore, if both ∆H is negative and ∆S is positive,