What does it mean for a reaction to be "thermodynamically favored?"

[Adriana Rangel 1A]

I saw somewhere that reactions with a negative enthalpy change are exothermic and thus more favorable than reactions with a positive enthalpy change (endothermic).
But according to the 2nd Law of Thermodynamics (states that reactions tend to go from low to high entropy) then why is a reaction with a negative enthalpy change favorable?

[Ishita Monga 1B]

Whether or not a reaction is favored is determined by deltaG. You have to use the relation deltaG=deltaH - TdeltaS in order to determine favorability. If a reaction has a negative enthalpy and entropy increases over the course of the reaction, then a negative minus a positive is still negative. Thus, deltaG would be negative which means that the reaction is favorable.

[Cristina Sarmiento 1E]

Reactions that do not require energy are seen as more favorable. Since exothermic reactions release energy and endothermic reactions require energy, exothermic reactions are more favorable.

[Angela G 2K]

Enthalpy tends to go from high enthalpy to low enthalpy, whereas entropy tends to go from low entropy to high entropy.
Delta H -, Delta S + = favored
Delta H -, Delta S - = favored at low temperatures only
Delta H +, Delta S + = favored at high temperatures only
Delta H +, Delta S - = not favored (reverse reaction is thermodynamically favored)

[Shreya Ramineni 2L]

Thermodynamically favorable means spontaneous, or that the reaction does not require energy in order to happen.

[Phillip Winters 2F]

A reaction if thermodynamically favorable if delta G is negative, which means that the reaction is spontaneous and thus doesn't require energy to occur

[Kevin Tabibian 1A]

It means that it is energetically favorable and naturally occurring ( spontaneous). This is represented by a negative delta G