Organic Chemistry · An Overview of Organic Reactions
Describing a Reaction: Equilibria, Rates, and Energy Changes
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In 30 seconds
A chemical reaction can be described by three distinct questions: Does it happen? (equilibrium), How fast? (kinetics), and Does it release or absorb energy? (energy changes). Confusing them is one of the most common errors in organic chemistry. The equilibrium constant Keq measures the product/reactant ratio at equilibrium and connects to the free-energy change through ΔG°= -RTlnKeq. The rate depends on the activation energy Ea and on concentrations, not on thermodynamic favorability. And ΔH°, related to the free energy by ΔG°= ΔH°- TΔS°, tells whether a reaction is exothermic or endothermic. A reaction can be thermodynamically favorable yet extremely slow — understanding why is essential for predicting whether it will occur on a useful timescale.
Why this matters
Every practical decision in chemistry — which reaction to run, at what temperature, and whether a product can be isolated — comes down to these three descriptors. A drug may be thermodynamically able to react with water yet kinetically stable enough to survive in the body for hours; that kinetic stability is what makes many pharmaceuticals work. Industrial synthesis balances thermodynamics (does equilibrium lie far enough toward product?) against kinetics (is it fast enough to be economical?) — which is why catalysts exist: they lower activation energy without changing equilibrium. Enzymes accelerate biological reactions by lowering Ea, while the equilibrium position of a metabolic step sets the direction of a pathway. On exams, the "favorable vs. fast" distinction is a favorite trap.
The college version
Core Concepts
Equilibrium constants and the direction of a reaction
For a reaction A + B ⇌ C + D, the equilibrium constant is
Keq = [C][D][A][B]
where the brackets mean molar concentrations at equilibrium. If Keq > 1, products dominate; if Keq < 1, reactants dominate. The equilibrium constant is a ratio of concentrations — never a rate, and it says nothing about how long the reaction takes.
Free energy and equilibrium
The connection between equilibrium and energy is
ΔG°= -RTlnKeq
where ΔG° is the standard Gibbs free-energy change, R is the gas constant (8.314 J mol⁻¹ K⁻¹), T is the Kelvin temperature, and ln is the natural logarithm. A large Keq gives negative ΔG° (favorable); a small Keq gives positive ΔG° (unfavorable). The free energy is composed of enthalpy and entropy:
ΔG°= ΔH°- TΔS°
An exothermic reaction (ΔH°< 0) and one that increases disorder (ΔS°> 0) both push ΔG° negative; a reaction can be favorable through either term or both.
Rates: how fast a reaction proceeds
The rate of a reaction tells how quickly reactants convert to products. For a simple reaction A → B, the rate law is often rate = k[A], where k is the rate constant. The rate constant depends on temperature and the activation energy Ea, the barrier reactants must climb; higher temperature and lower Ea both make k larger. Importantly, k and the rate law are set by the mechanism — the pathway — while ΔG° and Keq depend only on the starting and ending states. That is why a reaction can have a favorable equilibrium yet a high barrier: thermodynamics describes the destination, kinetics describes the journey.
Exothermic vs. endothermic
A reaction that releases heat is exothermic (ΔH°< 0); one that absorbs heat is endothermic (ΔH°> 0). You can estimate ΔH° from bond dissociation energies — add bonds broken, subtract bonds formed (the next topic develops this fully). Being exothermic does not make a reaction fast — burning paper is highly exothermic but needs a spark to start. Nor is an exothermic reaction automatically spontaneous if entropy opposes it; the full story is always in ΔG°.
Putting the three descriptors together
A complete description answers three questions: where does it end (equilibrium, Keq, ΔG°)? how fast does it get there (rate, k, Ea)? what is the heat of the journey (ΔH°, with ΔS° filling out ΔG°)? Two reactions can share thermodynamics yet differ wildly in rate, or vice versa. Keeping the three questions separate — and knowing which equation answers which — is the core skill.
How It Works / Step-by-Step Process
- Ask the thermodynamic question: write Keq from the balanced reaction, decide which side dominates, and compute ΔG°= -RTlnKeq if a value is available.
- Ask the energy question: identify bonds broken and formed to estimate ΔH°, and note whether entropy (ΔS°) helps or opposes; combine them in ΔG°= ΔH°- TΔS°.
- Ask the kinetic question: check the mechanism, activation energy, and temperature; higher T and lower Ea mean faster.
- Synthesize: a reaction is "favorable" if ΔG°< 0 and "fast" only if Ea is low enough — report both separately.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| "Favorable" (thermodynamics) | "Fast" (kinetics) | ΔG°< 0 says products are favored at equilibrium; Ea says how fast. Favorable reactions can be essentially instantaneous or take millennia. |
| Keq | Rate constant k | Keq is a ratio of equilibrium concentrations (no time); k sets the speed (time-dependent). |
| Exothermic | Spontaneous | Exothermic (ΔH°< 0) is only part of ΔG°= ΔH°- TΔS°; a strongly unfavorable entropy term can make an exothermic reaction nonspontaneous. |
| lnK | log10 K | The equation ΔG°= -RTlnK uses the natural logarithm; using base 10 changes the answer by a factor of ln10 ≈ 2.303. |
| Catalysts change the equilibrium | Catalysts change only the rate | Catalysts lower Ea; they do not change Keq, ΔG°, or the equilibrium position. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Think of a reaction like sledding down a hill into a valley. Equilibrium asks: is the valley lower than the start, and how many kids end up at the bottom versus the top? Rate asks: is there a fence in between that slows the sled, even though the valley is lower? Energy asks: does the ride give off warmth or need it? A hill can be lower at the bottom yet have a huge fence at the top — a favorable but slow reaction.
Worked example
Example 1: Converting an equilibrium constant to a free-energy change
A reaction at 298 K has Keq = 1.0 × 104. What is ΔG°?
Start with the formula before substituting numbers:
ΔG°= -RTlnKeq
Substitute R = 8.314 J mol-1K-1, T = 298 K, Keq = 1.0 × 104:
ΔG°= -(8.314)(298)ln(1.0 × 104)
Compute ln(1.0 × 104) = 9.21:
ΔG°= -(8.314)(298)(9.21) = -2.28 × 104 J mol-1 = -22.8 kJ mol-1
Dimensional analysis: (J mol⁻¹ K⁻¹)(K) = J mol⁻¹, then 1000 J = 1 kJ. The negative sign matches Keq > 1: products favored.
Example 2: A reactant-favored equilibrium
At 298 K, a reaction has Keq = 2.5 × 10-3. Is the reaction favorable?
Use the same formula:
ΔG°= -RTlnKeq = -(8.314)(298)ln(2.5 × 10-3)
ln(2.5 × 10-3) = -5.99:
ΔG°= -(8.314)(298)(-5.99) = +1.48 × 104 J mol-1 = +14.8 kJ mol-1
A positive ΔG° means reactants are favored — the reaction needs an external push (removing product, coupling to another reaction) to proceed. Units check: J mol⁻¹ K⁻¹ × K = J mol⁻¹, converted to kJ mol⁻¹.
Key takeaways
- Keq = [C][D][A][B]: products over reactants at equilibrium; Keq > 1 favors products.
- ΔG°= -RTlnKeq with R = 8.314 J mol⁻¹ K⁻¹; negative ΔG° ⟺ Keq > 1.
- ΔG°= ΔH°- TΔS°: enthalpy, entropy, and temperature set spontaneity.
- Rate is controlled by Ea and temperature, not by ΔG° — favorable ≠ fast.
- Exothermic: ΔH°< 0; endothermic: ΔH°> 0.
- Catalysts lower Ea and speed reactions without changing Keq or ΔG°.
- Equilibrium describes the endpoint; kinetics the path; energy the heat. Never swap the three questions.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Write the equation connecting ΔG° to Keq, and state the sign of ΔG° when Keq > 1.
Show answer
ΔG°= -RTlnKeq. When Keq > 1, lnKeq > 0, so ΔG°< 0 — the reaction favors products.
A reaction has Keq = 1.0 × 10-5 at 298 K. Roughly, is ΔG° positive or negative, and what does that mean?
Show answer
Positive. Keq < 1 means lnKeq < 0, so ΔG°> 0; reactants dominate at equilibrium and the reaction is unfavorable as written.
What is the difference between the equilibrium constant and the rate constant?
Show answer
Keq is a unitless ratio of equilibrium concentrations (where the reaction ends); k is a rate constant with units such as s⁻¹ (how fast). They are unrelated.
A reaction is exothermic but extremely slow at room temperature. Explain this in terms of ΔH°, ΔG°, and Ea.
Show answer
Exothermic (ΔH°< 0) tends to make ΔG° negative, but the reaction is slow because Ea is high — molecules lack the energy to climb the barrier even though the products are lower.
For ΔH°= -50 kJ mol-1 and ΔS°= -100 J mol-1K-1 at 298 K, compute ΔG°.
Show answer
ΔS°= -0.100 kJ mol-1K-1; TΔS°= (298)(-0.100) = -29.8 kJ mol-1; ΔG°= -50 - (-29.8) = -20.2 kJ mol-1 (favorable at 298 K).
Does adding a catalyst change Keq? Why or why not?
Show answer
No. A catalyst lowers Ea and speeds up forward and reverse reactions equally; Keq (the ratio of the two rate constants) and ΔG° are unchanged.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- equilibrium constant Kₑq
- Ratio of product concentrations to reactant concentrations at equilibrium.
- Gibbs free energy Δ G°
- Energy available to do work under standard conditions; sign predicts spontaneity.
- enthalpy change Δ H°
- Heat absorbed or released at constant pressure.
- entropy change Δ S°
- Change in disorder or randomness.
- rate constant k
- A temperature-dependent proportionality constant in the rate law.
- activation energy Eₐ
- The energy barrier between reactants and products.
- activation energy (Eₐ)
- Energy difference between reactants and the transition state.
Sources & references
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