ΔG°reaction = ΣΔG°f(products) − ΣΔG°f(reactants) is the products-minus-reactants equation that calculates a reaction's standard Gibbs free energy change from tabulated standard free energies of formation; if the result is negative, the process is thermodynamically favored (AP Chem Topic 9.3).
This equation lets you calculate ΔG° for any reaction using a table of standard Gibbs free energies of formation (ΔG°f). You add up the ΔG°f values of the products (each multiplied by its coefficient), do the same for the reactants, and subtract reactants from products. It's the same products-minus-reactants logic you used with ΔH°f values back in Unit 6, just with free energy instead of enthalpy.
The "standard" part matters. ΔG° only applies when everything is in its standard state, meaning pure substances, 1.0 M solutions, or gases at 1.0 atm (or 1.0 bar), per essential knowledge 9.3.A.1. The payoff is one number that tells you whether the reaction is thermodynamically favored. ΔG° < 0 means favored, ΔG° > 0 means not favored. The CED prefers "thermodynamically favored" over "spontaneous" because spontaneous wrongly suggests the reaction happens fast. ΔG° says nothing about speed; that's kinetics' job. Also remember that ΔG°f for any element in its most stable form is zero by definition, which saves you a step in calculations.
This equation lives in Topic 9.3 (Gibbs Free Energy and Thermodynamic Favorability) in Unit 9: Thermodynamics and Electrochemistry. It directly supports learning objective 9.3.A, which asks you to explain whether a physical or chemical process is thermodynamically favored based on an evaluation of ΔG°. Before you can evaluate ΔG°, you usually have to calculate it, and this equation is one of the two main routes (the other is ΔG° = ΔH° − TΔS°). It's also where AP Chem cashes in the big idea that thermodynamic data is composable. Because G is a state function, you can build the ΔG° of any reaction from formation values without ever running the reaction.
Keep studying AP® Chemistry Unit 9
ΔG° = ΔH° − TΔS° (Unit 9)
This is the other route to the same answer. The formation-values equation works when you're handed a ΔG°f table; the ΔH° − TΔS° equation works when you have enthalpy and entropy data, and it's the one to use when temperature changes the verdict. Both should give matching ΔG° values at 298 K, which makes them a great self-check.
Enthalpy Change and ΔH°f calculations (Unit 6)
You've already done this exact math with enthalpies of formation. ΔG°rxn = ΣΔG°f(products) − ΣΔG°f(reactants) is the Hess's-law pattern copied over to free energy, because G, like H, is a state function. If you can do a ΔH°f problem, you can do this one. Just swap the table you're reading from.
Thermodynamic favorability (Unit 9, Topic 9.3)
The number this equation spits out is the whole point of Topic 9.3. A negative ΔG° means thermodynamically favored; a positive ΔG° means not favored. But favored does not mean fast. A reaction with ΔG° = −500 kJ/mol can still sit there for years if the activation energy is huge.
Expect this as a calculation step inside a larger thermodynamics question. A typical setup gives you a table of ΔG°f values and a balanced equation, then asks for ΔG°rxn and whether the reaction is thermodynamically favored, with justification. The two classic point-losers are forgetting to multiply each ΔG°f by its stoichiometric coefficient and subtracting in the wrong order (it's always products minus reactants). On FRQs, the favorability claim has to be tied to the sign of ΔG°, not vibes. Write something like "ΔG° = −86 kJ/mol < 0, so the reaction is thermodynamically favored." Also watch the language trap baked into the CED. Saying a reaction with negative ΔG° will "happen quickly" can cost you, because ΔG° measures favorability, not rate.
Both calculate the same quantity, ΔG°, but from different inputs. The formation equation (ΣΔG°f products − ΣΔG°f reactants) uses a table of free energies of formation and only works at the temperature the table was made for, usually 298 K. The ΔH° − TΔS° equation builds ΔG° from enthalpy and entropy, so it lets you plug in different temperatures and figure out when a reaction flips between favored and not favored. On the exam, the data you're given tells you which one to use.
ΔG°rxn = ΣΔG°f(products) − ΣΔG°f(reactants), and every ΔG°f value must be multiplied by its coefficient from the balanced equation.
If the calculated ΔG° is negative, the reaction is thermodynamically favored; if positive, it is not favored under standard conditions.
The ΔG°f of any element in its most stable standard-state form is zero, so those species drop out of the sum.
Standard state means pure substances, 1.0 M solutions, and gases at 1.0 atm (or 1.0 bar); that's what the ° symbol guarantees (EK 9.3.A.1).
A negative ΔG° tells you nothing about speed. The CED uses "thermodynamically favored" instead of "spontaneous" specifically to avoid that mix-up.
This equation and ΔG° = ΔH° − TΔS° are two routes to the same ΔG°; use formation values when given a table, use ΔH° − TΔS° when temperature matters.
It's the formula for finding a reaction's standard Gibbs free energy change by summing the standard free energies of formation of the products, summing those of the reactants, and subtracting reactants from products. A negative result means the reaction is thermodynamically favored (Topic 9.3).
No. ΔG° < 0 only means the reaction is thermodynamically favored, not that it's fast. Diamond converting to graphite has a negative ΔG° and takes essentially forever. Rate comes from kinetics (Unit 5), not thermodynamics.
Both calculate the same ΔG°. The formation-values equation needs a ΔG°f data table and is locked to the table's temperature (usually 298 K), while ΔG° = ΔH° − TΔS° lets you change T and find where favorability flips. Pick based on what data the problem gives you.
By definition, forming an element in its most stable standard-state form from itself involves no change, so its ΔG°f is 0 kJ/mol. That means O2(g), N2(g), Fe(s), and similar species contribute nothing to the sum.
Yes, always. ΔG°f values are per mole, so 2H2O(g) contributes 2 × ΔG°f of water vapor. Skipping coefficients is one of the most common ways to lose the calculation point on an FRQ.
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