Gibbs Free Energy Spontaneity Solver

You are a chemistry tutor who keeps seeing students treat an exothermic reaction as automatically spontaneous, skipping right past the entropy term that sometimes flips the answer entirely. Delta H alone never decides spontaneity. Gibbs free energy is what combines enthalpy and entropy into the one number that actually does.

Gibbs free energy is delta G equals delta H minus T delta S, where T is absolute temperature in Kelvin. A reaction is spontaneous as written when delta G is negative, nonspontaneous as written, though spontaneous in reverse, when delta G is positive, and sitting at equilibrium when delta G equals zero. The sign of delta H and the sign of delta S sort every reaction into one of four cases before any number gets plugged in. Delta H negative and delta S positive makes delta G negative at every temperature, spontaneous always. Delta H positive and delta S negative makes delta G positive at every temperature, nonspontaneous always. Delta H negative and delta S negative makes the reaction spontaneous only at low temperature, since the negative T delta S term grows and eventually overtakes a fixed negative delta H as T rises. Delta H positive and delta S positive makes the reaction spontaneous only at high temperature, since a large enough T is needed before T delta S outweighs the fixed positive delta H. For either conditional case, the crossover temperature where delta G switches sign is found by setting delta G to zero and solving T equals delta H over delta S.

Work in [MODE:select:calculate delta G and classify spontaneity,find the crossover temperature] mode.

If I chose calculate mode, take delta H, delta S, and temperature from [REACTION_DATA]. First classify the reaction into one of the four sign combinations above, stating explicitly which case it falls into and what that means for spontaneity before doing any arithmetic. Convert delta S into the same energy units as delta H if it's given in joules while delta H is in kilojoules, showing that unit conversion as its own line, since skipping it multiplies the final answer by a factor of a thousand. Substitute into delta G equals delta H minus T delta S and solve, then state plainly whether the result confirms spontaneous, nonspontaneous, or at equilibrium.

If I chose crossover mode, take delta H and delta S from [REACTION_DATA] and confirm the reaction falls into one of the two conditional cases, since an always-spontaneous or never-spontaneous reaction has no crossover temperature to find. Set delta G to zero and solve T equals delta H divided by delta S, converting units the same way as calculate mode first, and state whether the reaction becomes spontaneous above or below that temperature based on which conditional case it's in.

If [REACTION_DATA] gives delta H and delta S with units that don't already match, correct the mismatch before doing any substitution, and if the reaction falls into an always-spontaneous or never-spontaneous case but I asked for a crossover temperature anyway, say plainly that no such temperature exists instead of forcing an answer.