Equilibrium Constant, Reaction Quotient and Gibbs Energy

Where a Reaction Actually Stops

A reaction mixture held at a fixed temperature and pressure has a Gibbs energy $G$ that depends on what is in the vessel. Pure reactants have one value. Pure products have another. Every mixture in between has a value of its own, and that value is not the straight-line average of the two ends, because mixing two substances lowers the Gibbs energy of the collection all by itself.

Plot $G$ for the whole mixture against the extent of reaction — zero on the left when nothing has reacted, one on the right when the reactants are exhausted — and the curve sags below the line joining its two ends. It falls, flattens and rises again. Somewhere between the two extremes it passes through a minimum.

A system at constant temperature and pressure moves in whichever direction lowers its Gibbs energy. Starting from pure reactants the mixture slides to the right, downhill. Starting from pure products it slides to the left, downhill. Both journeys end at the same place — the bottom of the valley. There is nowhere lower to go, so the net movement stops.

Key Point: Equilibrium is the composition at which the Gibbs energy of the mixture is a minimum for the given temperature and pressure. The system settles there not because the molecules run out of anything, but because any movement away from that composition would raise $G$.

The reaction itself has not stopped. Forward and reverse processes continue at equal rates, exactly as the earlier sections described. What has stopped is the net change, and the thermodynamic reason it stopped is that the mixture has reached the floor of the valley.

The slope of that curve is the quantity written $\Delta G$ — the Gibbs energy change for the reaction as written, evaluated for the mixture as it stands at that moment. A downward slope means $\Delta G$ is negative and the forward reaction runs. An upward slope means $\Delta G$ is positive and the reverse reaction runs. At the bottom the slope is zero, and so is $\Delta G$.

• $\Delta G < 0$ — forward reaction is spontaneous from this composition.
• $\Delta G > 0$ — reverse reaction is spontaneous from this composition.
• $\Delta G = 0$ — no net drive in either direction; the mixture is at equilibrium.

Every equilibrium constant in this chapter is a report on where the bottom of that valley lies.

Why the Curve Has a Valley

At constant temperature and pressure the Gibbs energy of the mixture is $G = H - TS$, and two separate things change as reactants turn into products.

The first is the chemistry itself. Bonds break and form, which changes the enthalpy, and the number and kind of particles changes, which changes the entropy. Taken alone, this contribution varies smoothly from the pure-reactant value to the pure-product value — a straight line joining the two ends of the plot.

The second is the simple fact that the vessel holds a mixture. Two kinds of particle sharing one container can be arranged in far more ways than either kind alone, so mixing raises the entropy of the collection. Since $G = H - TS$, a raised $S$ lowers $G$. This mixing contribution is zero at both ends, where only one substance is present, and largest somewhere in between.

The full curve is the straight line pulled downward in the middle by the mixing term. A straight line pulled down in the middle is a valley, and a valley has a bottom.

Key Point: A reaction stops short of completion because a mixture has a lower Gibbs energy than either pure end. Equilibrium is the compromise between the chemical drive and the entropy of mixing.

Two consequences follow. Every reaction has a genuine equilibrium composition containing at least a trace of everything, which is why $K$ is never exactly zero and never infinite. And the position of the bottom, not the depth of either end, is what the equilibrium constant reports.

The Equation That Joins Them

The position of the minimum is fixed by one equation, which links the instantaneous drive $\Delta G$ to the composition through the reaction quotient $Q$.

Key Point: $$\Delta G = \Delta G^{\circ} + RT \ln Q$$

Four symbols, and each one has to be read correctly.

$\Delta G$ is the Gibbs energy change for the reaction with the mixture at its actual present composition. It changes continuously as the reaction proceeds. This is the term that decides whether the mixture moves forward or backward at this instant.

$\Delta G^{\circ}$ is the standard Gibbs energy change — the value for the same balanced equation when every reactant and product is in its standard state: a gas at $1\ \mathrm{bar}$, a solute at $1\ \mathrm{mol\,L^{-1}}$, a pure solid or pure liquid as itself. For one balanced equation at one temperature it is a single fixed number, read off a table. Nothing that happens inside the vessel alters it.

$RT$ carries the temperature. With $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$ and $T$ in kelvin, $RT$ comes out in $\mathrm{J\,mol^{-1}}$, which fixes the units of the whole equation.

$\ln Q$ is the correction for the mixture not being in its standard state. $Q$ is the reaction quotient of the previous section — the equilibrium expression with the concentrations or partial pressures present right now substituted into it. Each term in $Q$ is understood as a ratio to its standard value, which is why $Q$ is a pure number and the logarithm is legal.

The shape of the correction term is worth holding on to.

• A reactant-rich mixture has $Q < 1$, so $\ln Q$ is negative and $RT\ln Q$ pulls $\Delta G$ downward. This is why a reaction with an unfavourable $\Delta G^{\circ}$ still runs forward at the start.
• A product-rich mixture has $Q > 1$, so $RT\ln Q$ is positive and pushes $\Delta G$ upward, opposing further forward reaction.
• When $Q = 1$ the correction vanishes and $\Delta G = \Delta G^{\circ}$. The standard Gibbs energy change is the drive of the reaction at the one special composition where every species stands at its standard state.

The equation therefore does exactly what the sagging curve did: it makes the drive depend on where the mixture currently sits.

Setting the Slope to Zero

At the bottom of the valley two things are true at once. The drive has vanished, so $\Delta G = 0$. And the composition is the equilibrium composition, so the reaction quotient has become the equilibrium constant, $Q = K$.

Substituting both into the working equation,

$$0 = \Delta G^{\circ} + RT \ln K$$

which rearranges to the single most useful result in this section.

Key Point: $$\Delta G^{\circ} = -RT \ln K \qquad \text{or} \qquad \ln K = -\frac{\Delta G^{\circ}}{RT}$$

Taking antilogarithms gives the exponential form,

$$K = e^{-\Delta G^{\circ}/RT}$$

and converting the natural logarithm to base ten with $\ln x = 2.303 \log x$ gives the form most calculations actually use,

$$\Delta G^{\circ} = -2.303\,RT \log K$$

Three consequences follow immediately.

1. $K$ is fixed by $\Delta G^{\circ}$ and $T$, and by nothing else. The right-hand side contains no concentration and no pressure, which is the thermodynamic reason $K$ is independent of the initial amounts taken.
2. $K$ changes with temperature. $T$ appears explicitly, and $\Delta G^{\circ}$ itself depends on temperature. This is the root of everything the next section says about heating an equilibrium.
3. The relation is exponential. $\Delta G^{\circ}$ sits inside an exponent, so a small change in $\Delta G^{\circ}$ produces an enormous change in $K$. A few kilojoules per mole move the equilibrium constant by orders of magnitude.

A catalyst changes neither $\Delta G^{\circ}$ nor $T$, so it cannot change $K$. That statement, usually justified by rate arguments, drops straight out of this equation. [JEE/NEET]

Reading the Sign of the Standard Gibbs Energy

Because $\ln K = -\Delta G^{\circ}/RT$ and $RT$ is always positive, the sign of $\Delta G^{\circ}$ alone decides which side of $1$ the equilibrium constant falls on.

$\Delta G^{\circ}$	$-\Delta G^{\circ}/RT$	$K = e^{-\Delta G^{\circ}/RT}$	Equilibrium mixture
Negative	Positive	$K > 1$	Products predominate; the forward reaction proceeds to a large extent
Zero	Zero	$K = 1$	Neither side is favoured; comparable amounts of both
Positive	Negative	$K < 1$	Reactants predominate; only a small quantity of product forms

The exponential makes the scale steep. At $298\ \mathrm{K}$ the numbers work out as follows.

$\Delta G^{\circ}$ at $298\ \mathrm{K}$	$\log K$	$K$
$-34.2\ \mathrm{kJ\,mol^{-1}}$	$+6$	$10^{6}$
$-11.4\ \mathrm{kJ\,mol^{-1}}$	$+2$	$100$
$-5.7\ \mathrm{kJ\,mol^{-1}}$	$+1$	$10$
$0$	$0$	$1$
$+5.7\ \mathrm{kJ\,mol^{-1}}$	$-1$	$0.1$
$+11.4\ \mathrm{kJ\,mol^{-1}}$	$-2$	$0.01$

Under $6\ \mathrm{kJ\,mol^{-1}}$ — less than the energy of a single weak hydrogen bond — swings $K$ by a factor of ten. Around $34\ \mathrm{kJ\,mol^{-1}}$ swings it by a million.

One warning about language. A positive $\Delta G^{\circ}$ is often described as making the reaction non-spontaneous, and the description is fair only if it is understood correctly: it means $K < 1$, so the forward reaction proceeds to a small extent and stalls with reactants predominating. It does not mean that nothing at all happens. Starting from pure reactants, $Q = 0$, and $RT\ln Q$ is hugely negative, so $\Delta G$ is negative and product does form. The reaction simply runs out of drive early.

Key Point: $\Delta G^{\circ} > 0$ means a small $K$, not a forbidden reaction. Every reaction with a finite $K$ produces some product before it reaches equilibrium.

The Distinction That Costs Marks

$\Delta G$ and $\Delta G^{\circ}$ differ by one superscript and by everything else.

	$\Delta G^{\circ}$	$\Delta G$
Belongs to	the balanced equation	the mixture at this instant
Composition assumed	every species at standard state	whatever is actually present
Value during a reaction	fixed, does not move	varies continuously
Value at equilibrium	unchanged, generally not zero	exactly zero
Decides spontaneity now	no	yes
Source	tables, or $-RT\ln K$	calculated from $\Delta G^{\circ}$ and $Q$

The row that matters most is the one headed Value at equilibrium. $\Delta G^{\circ}$ is not zero at equilibrium. It keeps whatever value it had before the reaction started, because it refers to a fixed hypothetical composition, not to the vessel. The quantity that goes to zero at equilibrium is $\Delta G$.

Eliminating $\Delta G^{\circ}$ between the two equations makes the connection to the previous section exact. Substituting $\Delta G^{\circ} = -RT\ln K$ into $\Delta G = \Delta G^{\circ} + RT\ln Q$,

$$\Delta G = RT \ln Q - RT \ln K = RT \ln\!\left(\frac{Q}{K}\right)$$

Key Point: $\Delta G = RT\ln(Q/K)$. The direction of a reaction depends only on how $Q$ compares with $K$, and the Gibbs energy statement and the $Q$-against-$K$ statement are the same statement written twice.

Composition	$Q/K$	$\ln(Q/K)$	$\Delta G$	Net change
Too much reactant	$<1$	Negative	Negative	Forward
Equilibrium	$=1$	$0$	$0$	None
Too much product	$>1$	Positive	Positive	Reverse

A reaction with a strongly positive $\Delta G^{\circ}$ can still have a negative $\Delta G$ at a given moment, provided $Q$ is small enough. A reaction with a strongly negative $\Delta G^{\circ}$ can have a positive $\Delta G$ if the vessel has been loaded with product. Deciding direction from the sign of $\Delta G^{\circ}$ alone is wrong whenever the mixture is not near standard conditions. [JEE Main]

Handling the Numbers

Most errors in this section are arithmetic, and nearly all of them come from three places.

Units. $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$, so $RT\ln K$ comes out in $\mathrm{J\,mol^{-1}}$ and so does $\Delta G^{\circ}$. Tabulated values of $\Delta G^{\circ}$ are quoted in $\mathrm{kJ\,mol^{-1}}$. Multiply a tabulated value by $10^{3}$ before dividing by $RT$, and divide a calculated value by $10^{3}$ before reporting it. A missing factor of a thousand is the single commonest mistake here, and it is easy to spot: it gives a $K$ absurdly close to $1$.

The two constants at $298\ \mathrm{K}$. These are worth memorising.

$$RT = 8.314 \times 298 = 2478\ \mathrm{J\,mol^{-1}}$$ $$2.303\,RT = 2.303 \times 8.314 \times 298 = 5705\ \mathrm{J\,mol^{-1}} = 5.705\ \mathrm{kJ\,mol^{-1}}$$

With the second one, the whole calculation at $298\ \mathrm{K}$ collapses to a single division.

$$\Delta G^{\circ}\ (\mathrm{in\ kJ\,mol^{-1}}) = -5.705 \log K \qquad \log K = -\frac{\Delta G^{\circ}\ (\mathrm{in\ kJ\,mol^{-1}})}{5.705}$$

Neither shortcut survives a change of temperature. At $300\ \mathrm{K}$, $2.303\,RT = 5744\ \mathrm{J\,mol^{-1}}$; at $500\ \mathrm{K}$ it is $9574\ \mathrm{J\,mol^{-1}}$. Recompute rather than reuse.

Logarithms. Use $\ln$ with $-RT\ln K$ and $\log$ with $-2.303\,RT\log K$. Mixing them scales the answer by $2.303$ in one direction or the other, which is why an answer that is $2.3$ times too large or too small almost always means the wrong logarithm was taken. The bridge is $\ln K = 2.303 \log K$.

Two sign checks close the loop, and both take a second.

• A negative $\Delta G^{\circ}$ must give $K > 1$; a positive $\Delta G^{\circ}$ must give $K < 1$.
• A $K$ greater than $1$ must give a negative $\Delta G^{\circ}$; a $K$ less than $1$ must give a positive one.

$Q$ and $K$ go into the logarithm as pure numbers. At this level, whatever numerical value of $K_c$ or $K_p$ is quoted in the question is the number that goes in.

Choosing the Form to Use

Four forms of one relation cover every question in this section. The data given decides which one to reach for.

Given	Asked	Form to use
$\Delta G^{\circ}$ and $T$	$K$	$\ln K = -\dfrac{\Delta G^{\circ}}{RT}$, then $K = e^{\ln K}$
$K$ and $T$	$\Delta G^{\circ}$	$\Delta G^{\circ} = -RT\ln K$
$\Delta G^{\circ}$ in $\mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$	$K$, quickly	$\log K = -\dfrac{\Delta G^{\circ}}{5.705}$
$\Delta G^{\circ}$, $T$ and a composition	$\Delta G$ and the direction	$\Delta G = \Delta G^{\circ} + RT\ln Q$
$K$ and a composition	the direction only	$\Delta G = RT\ln\!\left(\dfrac{Q}{K}\right)$, or compare $Q$ with $K$

Order of work in a numerical question. Write the balanced equation first, because both $\Delta G^{\circ}$ and $K$ belong to a particular equation with particular coefficients, and a halved or doubled equation carries a different pair of values. Decide next whether the question concerns the standard state or the vessel, since that choice alone selects between $\Delta G^{\circ}$ and $\Delta G$. Fix the units before touching the calculator. Then compute, and finish with the sign check.

Reporting conventions are worth following. Quote $\Delta G^{\circ}$ in $\mathrm{kJ\,mol^{-1}}$ with its sign written explicitly, and quote $K$ as a pure number in scientific notation. A $K$ written with a sign, or a $\Delta G^{\circ}$ written without one, loses marks that the arithmetic had already earned. [Board]

Question 1: Standard Gibbs energy to equilibrium constant

The value of $\Delta G^{\circ}$ for the phosphorylation of glucose in glycolysis is $13.8\ \mathrm{kJ\,mol^{-1}}$. Find the value of $K_c$ at $298\ \mathrm{K}$.

Answer:

First I convert the standard Gibbs energy change into joules, because $R$ is in joules.

$$\Delta G^{\circ} = 13.8\ \mathrm{kJ\,mol^{-1}} = 13.8 \times 10^{3}\ \mathrm{J\,mol^{-1}}$$

Now I use $\Delta G^{\circ} = -RT \ln K_c$, rearranged for $\ln K_c$.

$$\ln K_c = -\frac{\Delta G^{\circ}}{RT} = -\frac{13.8 \times 10^{3}}{8.314 \times 298} = -\frac{13800}{2478} = -5.569$$

Taking the antilogarithm,

$$K_c = e^{-5.569} = 3.81 \times 10^{-3}$$

The sign check works: $\Delta G^{\circ}$ is positive, so $K_c$ must come out below $1$, and it does.

Ans: $K_c = 3.81 \times 10^{-3}$ at $298\ \mathrm{K}$ Watch out: Using $13.8$ instead of $13.8 \times 10^{3}$ gives $\ln K_c = -0.00557$ and $K_c = 0.994$, a value suspiciously close to $1$. That is the signature of a forgotten kilo.

Question 2: Equilibrium constant to standard Gibbs energy

Hydrolysis of sucrose gives $\mathrm{sucrose} + \mathrm{H_2O} \rightleftharpoons \mathrm{glucose} + \mathrm{fructose}$, with $K_c = 2 \times 10^{13}$ at $300\ \mathrm{K}$. Calculate $\Delta G^{\circ}$ at $300\ \mathrm{K}$.

Answer:

I use the same relation, this time in the forward direction.

$$\Delta G^{\circ} = -RT \ln K_c$$

I need the natural logarithm of $2 \times 10^{13}$.

$$\ln(2 \times 10^{13}) = \ln 2 + 13 \ln 10 = 0.693 + 29.934 = 30.63$$

Substituting,

$$\Delta G^{\circ} = -(8.314\ \mathrm{J\,K^{-1}\,mol^{-1}})(300\ \mathrm{K})(30.63) = -7.64 \times 10^{4}\ \mathrm{J\,mol^{-1}}$$

Ans: $\Delta G^{\circ} = -7.64 \times 10^{4}\ \mathrm{J\,mol^{-1}} = -76.4\ \mathrm{kJ\,mol^{-1}}$ Watch out: $K_c$ is far above $1$, so $\Delta G^{\circ}$ must be negative. Dropping the leading minus sign is the most frequent error in this direction, and the sign check catches it at once.

Question 3: The log-ten route at 298 K

A reaction has $\Delta G^{\circ} = -13.6\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. Find $K$.

Answer:

At $298\ \mathrm{K}$ the constant $2.303\,RT = 5705\ \mathrm{J\,mol^{-1}}$, so I use the base-ten form.

$$\Delta G^{\circ} = -2.303\,RT \log K$$ $$\log K = -\frac{\Delta G^{\circ}}{2.303\,RT} = -\frac{-13600}{5705} = 2.384$$

Taking the antilogarithm,

$$K = 10^{2.384} = 2.4 \times 10^{2}$$

Ans: $K = 2.4 \times 10^{2}$ (about $242$) Watch out: Dividing by $RT = 2478$ instead of $2.303\,RT = 5705$ gives $5.49$ and a $K$ of about $3 \times 10^{5}$. That is the natural logarithm being treated as a common logarithm — a factor of $2.303$ in the exponent, which is a huge error in $K$.

Question 4: Working back from a small K

The equilibrium constant for a reaction is $1.0 \times 10^{-3}$ at $298\ \mathrm{K}$. Calculate $\Delta G^{\circ}$.

Answer:

$\log K$ is exactly $-3$ here, so the base-ten form is the quick route.

$$\Delta G^{\circ} = -2.303\,RT \log K = -(5705\ \mathrm{J\,mol^{-1}})(-3) = +17115\ \mathrm{J\,mol^{-1}}$$

Converting,

$$\Delta G^{\circ} = +17.1\ \mathrm{kJ\,mol^{-1}}$$

$K$ is below $1$, so a positive $\Delta G^{\circ}$ is what I expect.

Ans: $\Delta G^{\circ} = +17.1\ \mathrm{kJ\,mol^{-1}}$

Question 5: A temperature that is not 298 K

For a gas-phase reaction, $K_p = 4.0 \times 10^{-2}$ at $500\ \mathrm{K}$. Calculate $\Delta G^{\circ}$ at this temperature.

Answer:

The shortcut constant $5705$ belongs to $298\ \mathrm{K}$ only, so I rebuild it for $500\ \mathrm{K}$.

$$2.303\,RT = 2.303 \times 8.314 \times 500 = 9574\ \mathrm{J\,mol^{-1}}$$

Now,

$$\log K_p = \log(4.0 \times 10^{-2}) = -1.398$$ $$\Delta G^{\circ} = -(9574)(-1.398) = +1.34 \times 10^{4}\ \mathrm{J\,mol^{-1}}$$

Ans: $\Delta G^{\circ} = +13.4\ \mathrm{kJ\,mol^{-1}}$ at $500\ \mathrm{K}$ Watch out: Reusing $5705$ at $500\ \mathrm{K}$ gives $+7.97\ \mathrm{kJ\,mol^{-1}}$. The shortcut constant is temperature-specific; only $R$ and $2.303$ are fixed.

Question 6: The drive at a non-equilibrium instant

For $\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$, $\Delta G^{\circ} = -33.0\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. A vessel contains $\mathrm{N_2}$ at $1.0\ \mathrm{bar}$, $\mathrm{H_2}$ at $3.0\ \mathrm{bar}$ and $\mathrm{NH_3}$ at $0.50\ \mathrm{bar}$. Find $\Delta G$ and state the direction of net change.

Answer:

First I build the reaction quotient in partial pressures, with the coefficients as powers.

$$Q_p = \frac{p_{\mathrm{NH_3}}^{2}}{p_{\mathrm{N_2}} \times p_{\mathrm{H_2}}^{3}} = \frac{(0.50)^{2}}{(1.0)(3.0)^{3}} = \frac{0.25}{27} = 9.26 \times 10^{-3}$$

Next the correction term.

$$RT \ln Q_p = (8.314)(298)\ln(9.26 \times 10^{-3}) = (2478)(-4.682) = -11.6 \times 10^{3}\ \mathrm{J\,mol^{-1}}$$

Adding it to the standard value,

$$\Delta G = \Delta G^{\circ} + RT\ln Q_p = -33.0 + (-11.6) = -44.6\ \mathrm{kJ\,mol^{-1}}$$

$\Delta G$ is negative, so the mixture moves forward and makes more ammonia.

Ans: $\Delta G = -44.6\ \mathrm{kJ\,mol^{-1}}$; net reaction goes forward Watch out: Cubing the hydrogen pressure is not optional. Using $3.0$ instead of $27$ gives $Q_p = 0.083$, then $\Delta G = -39.2\ \mathrm{kJ\,mol^{-1}}$ — right sign, wrong number.

Question 7: Positive standard value, forward-running mixture

For $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$, $\Delta G^{\circ} = +5.0\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. A mixture has $Q = 0.020$. Find $K$, find $\Delta G$, and say which way the mixture moves.

Answer:

First the equilibrium constant.

$$\ln K = -\frac{5.0 \times 10^{3}}{2478} = -2.018 \qquad K = e^{-2.018} = 0.133$$

Now the drive at this composition.

$$RT \ln Q = (2478)\ln(0.020) = (2478)(-3.912) = -9.69 \times 10^{3}\ \mathrm{J\,mol^{-1}}$$ $$\Delta G = +5.0 + (-9.69) = -4.69\ \mathrm{kJ\,mol^{-1}}$$

$\Delta G$ is negative, so the forward reaction runs. The same conclusion comes from $Q = 0.020$ being smaller than $K = 0.133$.

Ans: $K = 0.133$, $\Delta G = -4.69\ \mathrm{kJ\,mol^{-1}}$, net change forward Watch out: A positive $\Delta G^{\circ}$ does not forbid the forward reaction. It fixes $K$ below $1$, and any mixture with $Q$ below that $K$ still moves forward until $Q$ climbs to $0.133$.

Question 8: Scaling the equation

Given $\Delta G^{\circ} = -33.0\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$ for $\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$, find $K$ for this equation and then $\Delta G^{\circ}$ and $K$ for $\tfrac{1}{2}\mathrm{N_2}(g) + \tfrac{3}{2}\mathrm{H_2}(g) \rightleftharpoons \mathrm{NH_3}(g)$.

Answer:

For the equation as written,

$$\ln K = -\frac{-33.0 \times 10^{3}}{2478} = 13.32 \qquad K = e^{13.32} = 6.1 \times 10^{5}$$

Halving every coefficient halves the amount of reaction the equation describes, so $\Delta G^{\circ}$ is halved.

$$\Delta G^{\circ}_{\text{half}} = \tfrac{1}{2}(-33.0) = -16.5\ \mathrm{kJ\,mol^{-1}}$$

Halving $\Delta G^{\circ}$ halves $\ln K$, which takes the square root of $K$.

$$\ln K_{\text{half}} = \frac{16.5 \times 10^{3}}{2478} = 6.66 \qquad K_{\text{half}} = e^{6.66} = 7.8 \times 10^{2}$$

And $\sqrt{6.1 \times 10^{5}} = 7.8 \times 10^{2}$, which confirms it.

Ans: $K = 6.1 \times 10^{5}$; for the halved equation $\Delta G^{\circ} = -16.5\ \mathrm{kJ\,mol^{-1}}$ and $K = 7.8 \times 10^{2}$ Watch out: $\Delta G^{\circ}$ scales linearly with the equation while $K$ scales as a power. Halving the equation halves $\Delta G^{\circ}$ but square-roots $K$ — the two never change in the same way.

Question 9: Reversing the equation

For the sucrose hydrolysis of Question 2, $\Delta G^{\circ} = -76.4\ \mathrm{kJ\,mol^{-1}}$ and $K_c = 2 \times 10^{13}$ at $300\ \mathrm{K}$. Write $\Delta G^{\circ}$ and $K_c$ for the reverse reaction.

Answer:

Reversing an equation swaps products and reactants, so the equilibrium constant is inverted.

$$K_c(\text{reverse}) = \frac{1}{2 \times 10^{13}} = 5 \times 10^{-14}$$

Inverting $K$ changes the sign of $\ln K$, and so changes the sign of $\Delta G^{\circ}$.

$$\Delta G^{\circ}(\text{reverse}) = -\Delta G^{\circ}(\text{forward}) = +76.4\ \mathrm{kJ\,mol^{-1}}$$

I can check it directly: $-RT\ln(5 \times 10^{-14}) = -(2494)(-30.63) = +7.64 \times 10^{4}\ \mathrm{J\,mol^{-1}}$.

Ans: $K_c = 5 \times 10^{-14}$ and $\Delta G^{\circ} = +76.4\ \mathrm{kJ\,mol^{-1}}$ for the reverse reaction

Question 10: When the standard value is zero

A reaction has $\Delta G^{\circ} = 0$ at some temperature $T$. State $K$, and state whether a vessel containing all species at $1\ \mathrm{bar}$ is at equilibrium.

Answer:

Putting $\Delta G^{\circ} = 0$ into $\ln K = -\Delta G^{\circ}/RT$ gives $\ln K = 0$, so

$$K = e^{0} = 1$$

at every temperature for which $\Delta G^{\circ}$ is zero.

For the vessel, every species is at $1\ \mathrm{bar}$, so $Q = 1$ whatever the stoichiometry. Since $Q = K = 1$, the mixture is at equilibrium, and $\Delta G = RT\ln(Q/K) = 0$ confirms it.

Ans: $K = 1$, and the mixture at $1\ \mathrm{bar}$ throughout is at equilibrium Watch out: This is the one case where a standard-state mixture happens to be an equilibrium mixture. For any other $\Delta G^{\circ}$ the standard mixture has $Q = 1 \neq K$ and will react.

What Goes Wrong

• Treating $\Delta G^{\circ}$ as zero at equilibrium. $\Delta G$ is zero at equilibrium. $\Delta G^{\circ}$ keeps its tabulated value throughout.
• Deciding direction from $\Delta G^{\circ}$. Only $\Delta G$, which contains the $RT\ln Q$ term, reports the direction of net change from the composition actually present.
• Forgetting the factor of $10^{3}$. $\Delta G^{\circ}$ is tabulated in $\mathrm{kJ\,mol^{-1}}$ and $R$ is in $\mathrm{J\,K^{-1}\,mol^{-1}}$. A $K$ that lands near $1$ for a large $\Delta G^{\circ}$ is this error.
• Mixing $\ln$ and $\log$. $-RT\ln K$ and $-2.303\,RT\log K$ are the same quantity; using $RT$ with $\log$, or $2.303\,RT$ with $\ln$, is out by a factor of $2.303$.
• Reusing $5705$ at another temperature. That number is $2.303\,RT$ at $298\ \mathrm{K}$ alone.
• Losing a minus sign. Two minus signs live in $\ln K = -\Delta G^{\circ}/RT$; the sign check comparing $K$ with $1$ costs nothing.
• Calling a positive $\Delta G^{\circ}$ a reaction that cannot happen. It means $K < 1$ and a small yield, not zero yield.
• Assuming a large $K$ means a fast reaction. $\Delta G^{\circ}$ fixes the destination, never the time taken to arrive.
• Scaling $\Delta G^{\circ}$ and $K$ the same way. Doubling the equation doubles $\Delta G^{\circ}$ and squares $K$.
• Using a temperature in degrees Celsius. $T$ in $RT$ is always in kelvin.

Key Point: $\Delta G = \Delta G^{\circ} + RT\ln Q$ describes the drive from wherever the mixture happens to be. Setting $\Delta G = 0$ and $Q = K$ turns it into $\Delta G^{\circ} = -RT\ln K = -2.303\,RT\log K$, which converts a tabulated energy into an equilibrium constant and back. At $298\ \mathrm{K}$, $2.303\,RT = 5705\ \mathrm{J\,mol^{-1}}$.