Free Energy and Equilibrium: Study Guide (Chemistry 9.5)

Two ways of saying “products are favoured”

You now have two separate tools that describe where a reaction ends up.

• From Unit 7, the equilibrium constant K. Products are on the top of the K expression, so a K greater than 1 means the equilibrium mixture contains more products than reactants (in the sense of the expression). See Topic 7.5.
• From Topic 9.3, the standard Gibbs free energy change ΔG°. A negative ΔG° means the reaction is thermodynamically favoured under standard conditions.

These are not two different facts. They are the same fact in two languages. When a chemist says a process is “thermodynamically favoured” (ΔG° < 0), they mean that, under standard conditions, products are favoured at equilibrium (K > 1).

ΔG°	K	What the equilibrium mixture looks like
Negative (ΔG° < 0)	Greater than 1	Products favoured; the reaction is thermodynamically favoured
Zero (ΔG° = 0)	Exactly 1	Neither side favoured; significant amounts of both
Positive (ΔG° > 0)	Less than 1 (but above 0)	Reactants favoured; only a small amount of product forms

Notice the last row. A positive ΔG° does not mean “no reaction”. It means K is small. A small K still gives some product, as you saw in Topic 7.5.

The equation that links ΔG° and K

The exact relationship is:

• ΔG° = −RT ln K
• or, rearranged, K = e^(−ΔG°/RT)

What each symbol means:

• ΔG° is the standard free energy change for the balanced equation, in J mol⁻¹.
• R is the gas constant, 8.314 J mol⁻¹ K⁻¹.
• T is the absolute temperature in kelvin.
• ln is the natural logarithm, and e is its base (about 2.718).

Two unit traps catch many students:

1. ΔG° values are usually given in kJ mol⁻¹, but R is in J mol⁻¹ K⁻¹. Multiply ΔG° by 1000 before you divide by RT. If you forget, ΔG°/RT comes out about a thousand times too small and K looks almost exactly 1.
2. T must be in kelvin. 25 °C is 298 K.

Why the minus sign? A negative ΔG° makes −ΔG°/RT positive, and e raised to a positive power is greater than 1. So the minus sign is exactly what makes “negative ΔG°” match “K greater than 1”.

The quantity ΔG°/RT has no unit: J mol⁻¹ divided by (J mol⁻¹ K⁻¹ × K) leaves nothing. That is a good check. An exponent must always be a pure number.

Estimating K without a full calculation

You do not always need a calculator. The CED expects you to make the connection by estimation too. The trick is to compare ΔG° with RT.

At 298 K, RT = 8.314 × 298 = 2478 J mol⁻¹, which is about 2.5 kJ mol⁻¹.

• If ΔG° is close to zero (a few kJ mol⁻¹ at most, similar in size to RT), then ΔG°/RT is small, and K is close to 1. Both reactants and products are present in significant amounts.
• If ΔG° is much larger than RT in size (tens of kJ mol⁻¹ or more), then K is very far from 1: huge if ΔG° is negative, tiny if ΔG° is positive.

A handy rule at 298 K: RT ln 10 = 2.478 × 2.303 ≈ 5.7 kJ mol⁻¹. So every 5.7 kJ mol⁻¹ of ΔG° changes K by a factor of 10.

ΔG° at 298 K (kJ mol⁻¹)	ΔG° ÷ RT	K	Interpretation
−40	−16.1	about 1 × 10⁷	Essentially all products
−20	−8.1	about 3 × 10³	Strongly product-favoured
−5.7	−2.3	about 10	Product-favoured
0	0	1	Neither side favoured
+5.7	+2.3	about 0.1	Reactant-favoured
+20	+8.1	about 3 × 10⁻⁴	Strongly reactant-favoured
+40	+16.1	about 1 × 10⁻⁷	Essentially no products

Look at how quickly K changes. A ΔG° of −40 kJ mol⁻¹ is not an especially large energy for a chemical reaction, yet it gives a K of about ten million. This is why many reactions are described as “going to completion”: a ΔG° of a few tens of kJ mol⁻¹ is enough.

Where temperature comes in

T appears twice in this topic.

1. Inside the equation. For the same ΔG°, a higher T makes ΔG°/RT smaller, so K is a little closer to 1.
2. Inside ΔG° itself. From Topic 9.3, ΔG° = ΔH° − TΔS°. When T changes, ΔG° changes, and so does K. This is usually the bigger effect.

So to predict K at a new temperature, first find ΔG° at that temperature, then convert to K. In this course you may assume that ΔH° and ΔS° stay roughly constant over a moderate temperature range.

A useful consequence: when ΔH° and ΔS° have the same sign, there is a crossover temperature T = ΔH° ÷ ΔS° at which ΔG° = 0 and K = 1. On one side of it the reaction is product-favoured; on the other side it is reactant-favoured.

Worked example 1: K from ΔG°

Question. For the invented isomerisation P(aq) ⇌ Q(aq), ΔG° = −12.6 kJ mol⁻¹ at 298 K. Calculate K and describe the equilibrium mixture.

1. Convert to joules: ΔG° = −12 600 J mol⁻¹.
2. RT = 8.314 J mol⁻¹ K⁻¹ × 298 K = 2477.6 J mol⁻¹.
3. Exponent: −ΔG°/RT = −(−12 600) ÷ 2477.6 = +5.086 (no unit).
4. K = e^5.086 = 162 (3 significant figures).

Interpretation. ΔG° is negative, so K > 1: products are favoured. Because Q and P appear to the first power, K = [Q]/[P] = 162, so at equilibrium there are about 162 Q molecules for every P molecule. The reaction is thermodynamically favoured.

Estimation check. 12.6 kJ mol⁻¹ is a little more than 2 × 5.7 kJ mol⁻¹, so K should be a little more than 10² = 100. 162 fits.

Trap. If you leave ΔG° in kJ, you get e^(12.6/2477.6) = 1.005, which wrongly suggests that neither side is favoured.

Worked example 2: ΔG° from K

Question. A different invented reaction has K = 3.2 × 10⁻⁴ at 350 K. Calculate ΔG° at 350 K. Is the reaction thermodynamically favoured at this temperature?

1. RT = 8.314 × 350 = 2909.9 J mol⁻¹.
2. ln K = ln(3.2 × 10⁻⁴) = −8.047.
3. ΔG° = −RT ln K = −(2909.9)(−8.047) = +23 417 J mol⁻¹ = +23.4 kJ mol⁻¹.

Interpretation. ΔG° is positive and K is less than 1, so the reaction is not thermodynamically favoured at 350 K: reactants are favoured at equilibrium. The two signs agree, as they must.

Check the sign. A K below 1 always has a negative ln K. The minus sign in front of RT then turns it into a positive ΔG°. If your ΔG° came out negative for a K below 1, you dropped a minus sign.

Worked example 3: how temperature changes K

Question. For an invented gas-phase reaction, ΔH° = +45.0 kJ mol⁻¹ and ΔS° = +150 J mol⁻¹ K⁻¹. Assume both stay constant. (a) Calculate ΔG° and K at 298 K and at 400 K. (b) At what temperature is K = 1?

(a) At 298 K.

1. ΔG° = ΔH° − TΔS° = 45.0 kJ mol⁻¹ − (298 K)(0.150 kJ mol⁻¹ K⁻¹) = 45.0 − 44.7 = +0.3 kJ mol⁻¹.
2. K = e^(−300 / 2477.6) = e^(−0.121) = 0.89.

ΔG° is tiny compared with RT, so K is close to 1: both reactants and products are present in significant amounts, with reactants very slightly favoured.

At 400 K.

1. ΔG° = 45.0 − (400)(0.150) = 45.0 − 60.0 = −15.0 kJ mol⁻¹.
2. RT = 8.314 × 400 = 3325.6 J mol⁻¹, so −ΔG°/RT = 15 000 ÷ 3325.6 = 4.51.
3. K = e^4.51 = 91.

Now products are clearly favoured. Raising the temperature made the −TΔS° term larger, and that outweighed the positive ΔH°.

(b) K = 1 when ΔG° = 0, so T = ΔH° ÷ ΔS° = 45.0 kJ mol⁻¹ ÷ 0.150 kJ mol⁻¹ K⁻¹ = 300 K. Below 300 K reactants are favoured; above it, products are favoured.

Check. The answers fit the estimation rule. At 298 K, ΔG° is almost zero, so K ≈ 1. At 400 K, ΔG° is about 4.5 times RT, so K is far enough from 1 to be about 10².

Common misconceptions

• “ΔG° > 0 means the reaction does not happen.” It means K < 1. Some product still forms; the equilibrium mixture is mostly reactants.
• “ΔG° = 0 means nothing happens.” It means K = 1. Forward and reverse reactions still run, and both sides are present in significant amounts at equilibrium.
• “Thermodynamically favoured means fast.” ΔG° and K say where the reaction ends up, not how quickly it gets there. A favoured reaction can be too slow to observe (Topic 9.4).
• Mixing kJ and J. Convert ΔG° to J mol⁻¹ before using R = 8.314 J mol⁻¹ K⁻¹. A K that comes out as 1.00something for a ΔG° of several kJ is the warning sign.
• Using log instead of ln. The equation needs the natural logarithm. Using log gives a ΔG° that is too small in size by a factor of 2.303.
• Using degrees Celsius. T in the equation is always in kelvin.
• “A negative ΔG° means K is negative.” K is always positive. ΔG° sets whether K is above or below 1, never its sign.
• “Doubling ΔG° doubles K.” The link is exponential. At 298 K, ΔG° = −5.7 kJ mol⁻¹ gives K ≈ 10, but ΔG° = −11.4 kJ mol⁻¹ gives K ≈ 100, not 20. Doubling ΔG° squares K.

Where this leads

This topic joins Unit 7 and Unit 9: equilibrium constants and free energy describe the same thing. Next, Topic 9.6, Free Energy of Dissolution, applies these ideas to dissolving a salt, where ΔG° links to K_sp. Later, Topic 9.9 adds a third language, the cell potential E°, which is also tied to ΔG°. Try the practice questions now, then use the revision notes and the checklist to consolidate.