Temperature dependence describes how a chemical quantity changes with temperature; in AP Chem Topic 9.5, it means the equilibrium constant K depends on T through K = e^(-ΔG°/RT), so heating or cooling a system actually changes how product-favored it is, not just how fast it gets there.
Temperature dependence is the idea that key chemical quantities are not fixed numbers. They change when T changes. The big one for AP Chem is the equilibrium constant. The CED gives you two versions of the same relationship, ΔG° = -RT ln K and K = e^(-ΔG°/RT), and T is sitting right inside both of them. That means K is only a constant at a given temperature. Change the temperature and you get a new K, which means a genuinely different mix of products and reactants at equilibrium.
Here's the intuition the exam wants. ΔG° itself depends on temperature through ΔG° = ΔH° - TΔS°, and then K depends on ΔG° and T together. For an exothermic reaction with negative ΔS° (like 2A(g) → B(g), where gas molecules combine), raising T makes the -TΔS° term more positive, ΔG° climbs toward zero or above, and K shrinks. The reaction becomes less product-favored when hot. Temperature dependence is the quantitative, thermodynamic reason behind the equilibrium shifts you learned qualitatively with Le Chatelier.
This term lives in Unit 9 (Thermodynamics and Electrochemistry), Topic 9.5, and supports learning objective 9.5.A, explaining whether a process is thermodynamically favored using the relationships between K, ΔG°, and T. The essential knowledge spells out the math (9.5.A.2) and the estimation skill (9.5.A.3). When ΔG° is near zero, K is close to 1. When ΔG° is much bigger or smaller than RT, K deviates strongly from 1. Notice that the comparison point is RT, not just some fixed number. Temperature sets the scale. This is also where Units 7 and 9 finally connect, because the equilibrium constant you treated as a given in Unit 7 turns out to be determined by thermodynamics and temperature.
ΔG° = -RT ln K (Unit 9)
This equation IS the temperature dependence of equilibrium. T appears explicitly, and ΔG° hides another T inside ΔH° - TΔS°. Change T and both pieces move, so K moves too.
Le Chatelier's Principle (Unit 7)
Le Chatelier predicts the direction of the shift when you heat an equilibrium (exothermic reactions shift toward reactants). Topic 9.5 explains why. Heating an exothermic reaction makes ΔG° less negative, which makes K smaller. Same answer, two levels of explanation.
Rate constant (k) (Unit 5)
Rates are temperature dependent too, but for a different reason. Heating raises k because more collisions clear the activation energy barrier. Don't mix these up. Temperature changes k in every reaction, but it changes K based on the sign of ΔH°.
Spontaneity (Unit 9)
Whether ΔG° is negative can flip with temperature whenever ΔH° and ΔS° have the same sign. An exothermic reaction with negative ΔS° is favored when cold and unfavored when hot, because the -TΔS° penalty grows with T.
Multiple-choice questions love pairing temperature dependence with particulate diagrams. A classic stem shows an exothermic reaction like 2A(g) ⇌ B(g) with mostly product molecules at 300 K, then asks what happens to the particle counts and ΔG° at 500 K. You need to reason that heating makes ΔG° less negative, K drops, and the picture shifts toward more reactant particles. Another common setup gives you ΔH° and ΔS° (say, -92 kJ/mol and -198 J/(mol·K)) and asks you to predict the effect of doubling T. Watch the units there, since ΔS° comes in J and ΔH° in kJ. The estimation skill from 9.5.A.3 also shows up, like recognizing that ΔG° = -100 kJ/mol means K is enormous and a particulate model should show almost entirely products. No released FRQ has used the phrase "temperature dependence" verbatim, but FRQs regularly ask you to justify whether a reaction is thermodynamically favored at a given temperature, which is exactly this skill.
Both change with T, but they answer different questions. The rate constant k tells you how fast equilibrium is reached, and heating always increases k (more molecules clear the activation energy barrier). The equilibrium constant K tells you where equilibrium ends up, and heating can increase OR decrease K depending on the sign of ΔH°. Heating an exothermic reaction makes it faster but less product-favored at the same time.
The equilibrium constant K is only constant at a fixed temperature; change T and you get a new K through K = e^(-ΔG°/RT).
For exothermic reactions, raising the temperature makes ΔG° less negative and K smaller, so the equilibrium becomes less product-favored.
Temperature dependence comes from two places at once, the explicit T in ΔG° = -RT ln K and the hidden T in ΔG° = ΔH° - TΔS°.
Use RT as your measuring stick: when ΔG° is near zero, K is near 1, and when ΔG° is much larger than RT in either direction, K is far from 1.
Temperature affects both K (where equilibrium sits) and k (how fast you get there), and these are separate effects you must keep straight.
Le Chatelier's temperature predictions from Unit 7 are the qualitative version of what ΔG° = -RT ln K tells you quantitatively in Unit 9.
It's how quantities like the equilibrium constant K, free energy ΔG°, and the rate constant k change when temperature changes. In Topic 9.5, the key version is that K depends on T through K = e^(-ΔG°/RT), so temperature determines how product-favored a reaction is.
Yes. Temperature is the one condition that actually changes K. Adding reactants, changing pressure, or adding a catalyst shifts concentrations but leaves K alone, while changing T gives you a genuinely new K value.
Heating always increases k because more collisions have enough energy to react, but heating only increases K for endothermic reactions. For an exothermic reaction, raising T makes the reaction faster (bigger k) but less product-favored (smaller K).
Because ΔG° = ΔH° - TΔS° becomes less negative as T grows (especially when ΔS° is also negative), and a less negative ΔG° plugged into K = e^(-ΔG°/RT) gives a smaller K. That's the thermodynamic math behind Le Chatelier's rule that heat shifts an exothermic equilibrium toward reactants.
Yes, whenever ΔH° and ΔS° have the same sign. A reaction with ΔH° = -92 kJ/mol and ΔS° = -198 J/(mol·K) is favored at 300 K but flips to unfavored above roughly 465 K, where TΔS° outweighs ΔH°.
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