Chapter 6 · Equilibrium

6.2 Using K, Gibbs Energy and Le Chatelier's Principle

An equilibrium constant answers three practical questions: how far a reaction goes, which way a given mixture will move, and what the mixture will finally contain. This section uses $K$ and the reaction quotient $Q$, links them to the Gibbs energy, and shows with Le Chatelier's principle how concentration, pressure, temperature and catalysts affect an equilibrium.

A chemical engineer designing an ammonia plant needs more than the balanced equation $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$. How much ammonia will a given mixture of nitrogen and hydrogen produce before it stops changing? If the gas leaving the reactor is cooled and some ammonia is removed, which way will the remaining mixture react? Would a higher pressure or a lower temperature give more product — and at what cost in speed? Every one of these questions is answered by the equilibrium constant of Section 6.1 together with one more idea, the reaction quotient.

In this section you will judge the extent of a reaction from the size of $K$, predict its direction by comparing $Q$ with $K$, and calculate equilibrium compositions with an ICE table. You will see how $K$ and $Q$ are tied to the Gibbs energy change, and finally learn Le Chatelier's principle, which predicts how an equilibrium responds when it is disturbed — the principle behind the choice of conditions in the Haber process.

Predicting the Extent and Direction of a Reaction

The equilibrium constant has the products on top. A very large $K$ therefore means that at equilibrium the products far outweigh the reactants; a very small $K$ means that hardly any product forms. The three reactions below all have $\Delta n=0$, so $K_{c}=K_{p}$ and the values can be compared directly.

Reaction	$T$ / K	$K_{c}$	At equilibrium
$\mathrm{H_{2}(g)+Cl_{2}(g)\rightleftharpoons2HCl(g)}$	298	$\approx3\times10^{33}$	practically complete
$\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$	720	$49$	all three present
$\mathrm{N_{2}(g)+O_{2}(g)\rightleftharpoons2NO(g)}$	298	$\approx4\times10^{-31}$	practically no reaction

📐 Rule 6.7 — The size of $K$ and the extent of reaction

$K_{c}\gt10^{3}$: products predominate; the reaction goes nearly to completion.
$10^{-3}\lt K_{c}\lt10^{3}$: appreciable concentrations of both reactants and products are present.
$K_{c}\lt10^{-3}$: reactants predominate; the reaction proceeds only very slightly.

These limits are rough guides. $K$ says how far a reaction can go, not how fast: a mixture of $\mathrm{H_{2}}$ and $\mathrm{O_{2}}$ has an enormous $K$ for forming water but can stand for years without reacting until a spark starts it.

To decide which way a mixture that is not at equilibrium will move, put its actual concentrations into the equilibrium-constant expression.

📘 Definition 6.6 — Reaction quotient

For $a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}$ at any moment, the reaction quotient is $$Q_{c}=\frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}},$$ using the concentrations present at that moment, which need not be equilibrium values. $Q_{p}$ is defined in the same way with partial pressures. At equilibrium, and only then, $Q_{c}=K_{c}$.

📐 Rule 6.8 — Direction of net reaction

At a fixed temperature:

if $Q_{c}\lt K_{c}$, the reaction proceeds in the forward direction (products form) until $Q_{c}=K_{c}$;
if $Q_{c}\gt K_{c}$, the reaction proceeds in the backward direction (reactants form);
if $Q_{c}=K_{c}$, the mixture is already at equilibrium.

✏️ Example 1 — Which way will it go?

(a) For $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, $K_{c}=0.061\ \mathrm{L^{2}\,mol^{-2}}$ at about $773\ \mathrm{K}$. A reactor at this temperature contains $[\mathrm{N_{2}}]=1.0$, $[\mathrm{H_{2}}]=2.0$ and $[\mathrm{NH_{3}}]=0.50\ \mathrm{mol\,L^{-1}}$. Which way will the reaction go? (b) Repeat with $[\mathrm{NH_{3}}]=1.0\ \mathrm{mol\,L^{-1}}$. (c) For $\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$, $K_{c}=49$; a flask contains $[\mathrm{H_{2}}]=[\mathrm{I_{2}}]=0.10$ and $[\mathrm{HI}]=0.70\ \mathrm{mol\,L^{-1}}$. Is it at equilibrium? (d) Classify the reactions in the table above by Rule 6.7.

💡 Show solution▶

(a) $Q_{c}=\dfrac{(0.50)^{2}}{(1.0)(2.0)^{3}}=\dfrac{0.25}{8.0}=0.031$. $Q_{c}\lt K_{c}$, so more ammonia forms: net forward reaction.

(b) $Q_{c}=\dfrac{(1.0)^{2}}{(1.0)(8.0)}=0.125\gt0.061$: ammonia decomposes, net backward reaction.

(d) $\mathrm{H_{2}+Cl_{2}}$: $K\gg10^{3}$, nearly complete. $\mathrm{H_{2}+I_{2}}$: $K=49$, between $10^{-3}$ and $10^{3}$, appreciable amounts of all. $\mathrm{N_{2}+O_{2}}$: $K\ll10^{-3}$, practically no NO forms at room temperature (which is fortunate for the air we breathe).

Calculating Equilibrium Concentrations

If $K$ and the starting amounts are known, the equilibrium composition can be calculated. Because every species changes in proportion to its coefficient, a single unknown $x$ is enough to describe the whole mixture.

🧭 Problem-solving strategy — The ICE table

Write the balanced equation and the expression for $K_{c}$ (or $K_{p}$).
Convert amounts to concentrations ($\mathrm{mol\,L^{-1}}$) or partial pressures.
Build three rows under the equation: Initial values, Change (in terms of $x$, using the coefficients and the direction from $Q$ vs $K$), Equilibrium values (I + C).
Substitute the E row into the $K$ expression and solve for $x$. Take a square root when both sides are perfect squares; otherwise solve the quadratic exactly and keep the root that gives no negative concentration.
Calculate every equilibrium concentration and check by putting them back into $K$.

✏️ Example 2 — An ICE table that needs only a square root

The water–gas shift reaction $\mathrm{CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}(g)+H_{2}(g)}$ has $K_{c}=4.0$ at $800\ \mathrm{K}$. $1.00\ \mathrm{mol}$ of $\mathrm{CO}$ and $1.00\ \mathrm{mol}$ of steam are heated in a $10.0\ \mathrm{L}$ vessel at $800\ \mathrm{K}$. Find the equilibrium concentrations and the percentage of $\mathrm{CO}$ converted.

💡 Show solution▶

ICE Initial $[\mathrm{CO}]=[\mathrm{H_{2}O}]=0.100$, $[\mathrm{CO_{2}}]=[\mathrm{H_{2}}]=0$; $Q=0\lt K$, so the reaction goes forward. Change: $-x,\,-x,\,+x,\,+x$. Equilibrium: $0.100-x,\ 0.100-x,\ x,\ x$.

Solve $\dfrac{x^{2}}{(0.100-x)^{2}}=4.0$, so $\dfrac{x}{0.100-x}=2.0$ (positive root), $x=0.200-2x$, $x=0.0667\ \mathrm{mol\,L^{-1}}$.

Result $[\mathrm{CO}]=[\mathrm{H_{2}O}]=0.033\ \mathrm{mol\,L^{-1}}$; $[\mathrm{CO_{2}}]=[\mathrm{H_{2}}]=0.067\ \mathrm{mol\,L^{-1}}$. CO converted: $0.0667/0.100=67\%$.

Check $\dfrac{(0.0667)^{2}}{(0.0333)^{2}}=4.0$ ✓. (Here $\Delta n=0$, so the volume cancels: the answer would be the same in any vessel.)

✏️ Example 3 — An ICE table with a quadratic

Repeat Example 2 with $1.00\ \mathrm{mol}$ of CO and $2.00\ \mathrm{mol}$ of steam in the $10.0\ \mathrm{L}$ vessel. What percentage of CO is now converted, and what does the comparison show?

💡 Show solution▶

ICE Equilibrium: $[\mathrm{CO}]=0.100-x$, $[\mathrm{H_{2}O}]=0.200-x$, $[\mathrm{CO_{2}}]=[\mathrm{H_{2}}]=x$.

Equation $\dfrac{x^{2}}{(0.100-x)(0.200-x)}=4.0$ → $x^{2}=4.0(0.0200-0.300x+x^{2})$ → $3x^{2}-1.20x+0.0800=0$.

Solve $x=\dfrac{1.20\pm\sqrt{1.44-0.96}}{6}=\dfrac{1.20\pm0.693}{6}$. The $+$ root, $0.315$, is impossible (it exceeds $0.100$); so $x=0.0845\ \mathrm{mol\,L^{-1}}$.

Result $[\mathrm{CO}]=0.0155$, $[\mathrm{H_{2}O}]=0.1155$, $[\mathrm{CO_{2}}]=[\mathrm{H_{2}}]=0.0845\ \mathrm{mol\,L^{-1}}$; CO converted $=85\%$ instead of $67\%$. Check: $\dfrac{(0.0845)^{2}}{(0.0155)(0.1155)}=4.0$ ✓. Using an excess of the cheap reactant (steam) drives more of the valuable one (CO) to products — an example of Le Chatelier's principle, later in this section.

✏️ Example 4 — Degree of dissociation from $K_{p}$

For $\mathrm{N_{2}O_{4}(g)\rightleftharpoons2NO_{2}(g)}$, $K_{p}=0.15\ \mathrm{bar}$ at $298\ \mathrm{K}$. Find the fraction $\alpha$ of $\mathrm{N_{2}O_{4}}$ dissociated when the total pressure at equilibrium is $1.00\ \mathrm{bar}$, and the partial pressures of the two gases.

💡 Show solution▶

Moles Start with $1$ mol $\mathrm{N_{2}O_{4}}$; at equilibrium $1-\alpha$ mol $\mathrm{N_{2}O_{4}}$ and $2\alpha$ mol $\mathrm{NO_{2}}$, total $1+\alpha$ mol.

Partial pressures $p_{\mathrm{N_{2}O_{4}}}=\dfrac{1-\alpha}{1+\alpha}P$, $p_{\mathrm{NO_{2}}}=\dfrac{2\alpha}{1+\alpha}P$, so $K_{p}=\dfrac{4\alpha^{2}P^{2}/(1+\alpha)^{2}}{(1-\alpha)P/(1+\alpha)}=\dfrac{4\alpha^{2}P}{1-\alpha^{2}}$.

Solve With $P=1.00$: $0.15(1-\alpha^{2})=4\alpha^{2}$ → $\alpha^{2}=\dfrac{0.15}{4.15}=0.0361$ → $\alpha=0.19$ (19%).

Pressures $p_{\mathrm{NO_{2}}}=\dfrac{0.38}{1.19}=0.32\ \mathrm{bar}$, $p_{\mathrm{N_{2}O_{4}}}=\dfrac{0.81}{1.19}=0.68\ \mathrm{bar}$. Check: $(0.32)^{2}/0.68=0.15$ ✓ (compare Section 6.1, Exercise III.6).

Choose a reaction and set the initial concentrations (in $\mathrm{mol\,L^{-1}}$). The lab works out $Q_{c}$, compares it with $K_{c}$ on a logarithmic scale — the arrow shows which way the reaction must go — then fills in the ICE table and solves for $x$ exactly. Press ▶ Play to watch $Q$ travel to $K$ as the mixture reacts.

Reaction:

Q_c at the start—

K_c—

direction—

x (mol L⁻¹)—

Equilibrium Constant, Reaction Quotient and Gibbs Energy

🧰 Tool from Chapter 5 — Gibbs energy and spontaneity

At constant temperature and pressure a process is spontaneous when the Gibbs energy change $\Delta G=\Delta H-T\Delta S$ is negative, non-spontaneous when it is positive, and at equilibrium when $\Delta G=0$. $\Delta_{r}G^{\circ}$ is the value when every substance is in its standard state (gases at $1\ \mathrm{bar}$, solutes at $1\ \mathrm{mol\,L^{-1}}$). Here $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and $\ln x=2.303\log x$.

A reacting mixture "slides downhill" in Gibbs energy until it can go no lower; that lowest point is equilibrium. Thermodynamics shows that the Gibbs energy change of a reaction at any composition depends on the reaction quotient.

📐 Rule 6.9 — $\Delta G$, $Q$ and $K$

For a reaction at temperature $T$ with reaction quotient $Q$, $$\Delta_{r}G=\Delta_{r}G^{\circ}+RT\ln Q.$$ At equilibrium $\Delta_{r}G=0$ and $Q=K$, so $$\Delta_{r}G^{\circ}=-RT\ln K=-2.303\,RT\log K$$ $$K=e^{-\Delta_{r}G^{\circ}/RT}.$$ (For gases $K$ means $K_{p}$ with pressures in bar; for solutions $K_{c}$ with concentrations in $\mathrm{mol\,L^{-1}}$, each divided by its standard value so that $K$ is a pure number.)

$\Delta_{r}G^{\circ}$	$K$	Meaning
negative	$\gt1$	products favoured at equilibrium; the forward reaction is spontaneous under standard conditions
zero	$=1$	neither side favoured
positive	$\lt1$	reactants favoured; the forward reaction is non-spontaneous under standard conditions

Combining the two equations gives $\Delta_{r}G=RT\ln(Q/K)$. If $Q\lt K$ the logarithm is negative, $\Delta_{r}G\lt0$, and the forward reaction is spontaneous — exactly Rule 6.8 seen from thermodynamics. A reaction with a positive $\Delta_{r}G^{\circ}$ can still go forward if $Q$ is kept small enough, for example by removing the products as they form.

✏️ Example 5 — From $\Delta G^{\circ}$ to $K$ and back

Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $T=298\ \mathrm{K}$, so $RT=2.478\ \mathrm{kJ\,mol^{-1}}$. (a) A reaction has $\Delta_{r}G^{\circ}=-5.70\ \mathrm{kJ\,mol^{-1}}$. Find $K$. (b) For $\mathrm{N_{2}(g)+O_{2}(g)\rightleftharpoons2NO(g)}$, $\Delta_{r}G^{\circ}=+173.2\ \mathrm{kJ\,mol^{-1}}$. Find $K$. (c) For $\mathrm{N_{2}O_{4}(g)\rightleftharpoons2NO_{2}(g)}$, $K_{p}=0.15$. Find $\Delta_{r}G^{\circ}$. (d) A mixture of the gases in (c) has $p_{\mathrm{NO_{2}}}=0.10\ \mathrm{bar}$ and $p_{\mathrm{N_{2}O_{4}}}=1.0\ \mathrm{bar}$. Find $\Delta_{r}G$ and the direction of change.

💡 Show solution▶

(a) $\log K=\dfrac{-\Delta_{r}G^{\circ}}{2.303RT}=\dfrac{5700}{2.303\times8.314\times298}=\dfrac{5700}{5706}=1.00$, so $K=10$.

(b) $\ln K=\dfrac{-173\,200}{2478}=-69.9$, so $K=e^{-69.9}=4\times10^{-31}$ — the tiny value in the table above.

(d) $Q_{p}=\dfrac{(0.10)^{2}}{1.0}=0.010$. $\Delta_{r}G=4.70+2.478\ln0.010=4.70-11.41=-6.71\ \mathrm{kJ\,mol^{-1}}$. Negative: $\mathrm{N_{2}O_{4}}$ dissociates (forward), as $Q\lt K$ predicts, even though $\Delta_{r}G^{\circ}$ is positive.

Le Chatelier's Principle

In 1884 the French chemist Henri Le Chatelier summed up how an equilibrium responds to a disturbance in one sentence.

📐 Principle 6.10 — Le Chatelier's principle

If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract the change, until a new equilibrium is established.

Changing a concentration. Adding a reactant (or removing a product) makes $Q_{c}\lt K_{c}$, so the reaction moves forward and uses up part of what was added; adding a product (or removing a reactant) makes $Q_{c}\gt K_{c}$ and drives it backward. A colourful test is $$\begin{aligned}&\mathrm{Fe^{3+}(aq)+SCN^{-}(aq)}\\ &\quad\mathrm{\rightleftharpoons[Fe(SCN)]^{2+}(aq)}\end{aligned}$$ where the complex ion is deep red. Adding potassium thiocyanate deepens the colour. Adding oxalic acid, which binds $\mathrm{Fe^{3+}}$ as the stable ion $\mathrm{[Fe(C_{2}O_{4})_{3}]^{3-}}$, or $\mathrm{Hg^{2+}}$ ions, which bind $\mathrm{SCN^{-}}$, removes a reactant and the red colour fades. Industry uses the same idea: ammonia is continuously liquefied and removed from the Haber reactor so that more forms.

Changing the pressure by changing the volume. Compressing a gas mixture to half its volume doubles every concentration. For $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$ the numerator of $Q_{c}$ is multiplied by $2^{2}=4$ and the denominator by $2^{1}\times2^{3}=16$, so $Q_{c}$ falls to a quarter of $K_{c}$ and the reaction moves forward — towards the side with fewer moles of gas, which lowers the pressure again. For a reaction with $\Delta n_{g}=0$, such as $\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$, the factors cancel and pressure has no effect. Solids and liquids are almost incompressible and are ignored.

Adding an inert gas at constant volume (argon, or helium) increases the total pressure but leaves the amounts, the volume and therefore the concentrations and partial pressures of the reacting gases unchanged: $Q$ still equals $K$ and there is no shift.

Changing the temperature. This is the only change that alters the value of $K$ itself. Think of heat as a "reactant" of an endothermic reaction and a "product" of an exothermic one. Raising the temperature favours the endothermic direction; lowering it favours the exothermic direction. For the exothermic Haber reaction ($\Delta_{r}H^{\circ}=-92.4\ \mathrm{kJ\,mol^{-1}}$) $K$ decreases as $T$ rises; for the endothermic dissociation of $\mathrm{N_{2}O_{4}}$ it increases, which is why the brown colour deepens on heating (chapter teaser). Another vivid example is $$\begin{aligned}&\mathrm{[Co(H_{2}O)_{6}]^{2+}(aq)+4Cl^{-}(aq)}\\ &\quad\mathrm{\rightleftharpoons[CoCl_{4}]^{2-}(aq)+6H_{2}O(l)}\end{aligned}$$ which is endothermic: the solution is pink (the aqua ion) in iced water and blue (the chloro complex) when hot.

Adding a catalyst. A catalyst provides a faster route for both the forward and the backward reaction, lowering the activation energy of each by the same amount. Equilibrium is reached sooner, but its position and the value of $K$ are unchanged. Iron (with promoters) catalyses the Haber process, and vanadium(V) oxide catalyses $\mathrm{2SO_{2}(g)+O_{2}(g)\rightleftharpoons2SO_{3}(g)}$ in the contact process for sulphuric acid.

Change at equilibrium	Net shift	Does $K$ change?
add a reactant / remove a product	forward	no
add a product / remove a reactant	backward	no
compress (raise $p$)	towards fewer gas moles	no
inert gas at constant volume	none	no
raise $T$	endothermic direction	yes
catalyst	none (equilibrium reached faster)	no

⚠️ The Haber process — a compromise

$\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, $\Delta_{r}H^{\circ}=-92.4\ \mathrm{kJ\,mol^{-1}}$. Four moles of gas become two, so high pressure raises the yield; the reaction is exothermic, so low temperature raises $K$. But at low temperature the reaction is far too slow, even with a catalyst. Plants therefore run at about $700$–$775\ \mathrm{K}$ and about $200\ \mathrm{atm}$ with an iron catalyst: a smaller yield per pass (roughly 15%), but reached quickly. The ammonia is liquefied and removed, and the unreacted nitrogen and hydrogen are recycled. Le Chatelier's principle predicts the yield; the rate decides the final choice.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — inert gas at constant pressure

Not required by the CBSE syllabus. It is an application of Le Chatelier's principle (the effect of pressure), which JEE Main, NEET and JEE Advanced all list.

If an inert gas is added while the total pressure is held constant, the volume must increase. The partial pressure of every reacting gas then falls in the same ratio, exactly as if the mixture had been expanded. The equilibrium shifts towards the side with more moles of gas: the dissociation of $\mathrm{PCl_{5}}$ ($\Delta n_{g}=+1$) increases, the yield of ammonia ($\Delta n_{g}=-2$) falls, and a reaction with $\Delta n_{g}=0$ is unaffected.

✏️ Example 6 — Predicting shifts

Predict the effect on the position of equilibrium and on $K$. (a) $\mathrm{2SO_{2}(g)+O_{2}(g)\rightleftharpoons2SO_{3}(g)}$, $\Delta_{r}H\lt0$: (i) more $\mathrm{O_{2}}$; (ii) compressing the mixture; (iii) raising the temperature; (iv) adding $\mathrm{V_{2}O_{5}}$. (b) $\mathrm{PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)}$, $\Delta_{r}H\gt0$: (i) adding $\mathrm{Cl_{2}}$; (ii) lowering the pressure; (iii) raising the temperature; (iv) adding helium at constant volume. (c) Compressing $\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$. (d) Adding more $\mathrm{CaCO_{3}(s)}$ to $\mathrm{CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)}$.

💡 Show solution▶

(a) (i) Forward (uses up some $\mathrm{O_{2}}$). (ii) Forward: 3 mol gas → 2 mol. (iii) Backward, the endothermic direction; $K$ decreases. (iv) No shift; equilibrium is reached faster. Only (iii) changes $K$.

(b) (i) Backward (some added $\mathrm{Cl_{2}}$ recombines). (ii) Forward, towards 2 mol gas. (iii) Forward, the endothermic direction; $K$ increases. (iv) No shift: the partial pressures are unchanged.

(c) No shift: $\Delta n_{g}=0$, so $Q$ is unchanged by compression (the colour darkens only because the iodine is more concentrated).

(d) No shift: a pure solid does not appear in $K_{p}=p_{\mathrm{CO_{2}}}$.

✏️ Example 7 — How much does compression shift an equilibrium?

An equilibrium mixture of $\mathrm{N_{2}O_{4}}$ and $\mathrm{NO_{2}}$ at $298\ \mathrm{K}$ has $p_{\mathrm{N_{2}O_{4}}}=0.68\ \mathrm{bar}$ and $p_{\mathrm{NO_{2}}}=0.32\ \mathrm{bar}$ ($K_{p}=0.15\ \mathrm{bar}$). The volume is suddenly halved at constant temperature. (a) Find $Q_{p}$ immediately after compression and the direction of change. (b) Find the new equilibrium partial pressures and the new total pressure.

💡 Show solution▶

(a) Halving the volume doubles each partial pressure: $1.36$ and $0.64\ \mathrm{bar}$. $Q_{p}=\dfrac{(0.64)^{2}}{1.36}=0.30\gt K_{p}$, so $\mathrm{NO_{2}}$ combines: net backward, towards fewer gas molecules.

ICE Let $2y$ bar of $\mathrm{NO_{2}}$ combine: $p_{\mathrm{NO_{2}}}=0.64-2y$, $p_{\mathrm{N_{2}O_{4}}}=1.36+y$. Then $\dfrac{(0.64-2y)^{2}}{1.36+y}=0.15$ → $4y^{2}-2.71y+0.2056=0$.

Solve $y=\dfrac{2.71-\sqrt{2.71^{2}-4(4)(0.2056)}}{8}=\dfrac{2.71-2.014}{8}=0.087\ \mathrm{bar}$ (the other root makes $p_{\mathrm{NO_{2}}}$ negative).

Result $p_{\mathrm{NO_{2}}}=0.47\ \mathrm{bar}$, $p_{\mathrm{N_{2}O_{4}}}=1.45\ \mathrm{bar}$, total $1.91\ \mathrm{bar}$ — less than the $2.00\ \mathrm{bar}$ produced by the compression: the shift partly counteracts the change, as Le Chatelier predicts. Check: $(0.466)^{2}/1.447=0.150$ ✓.

A reactor holds $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$ (exothermic) at equilibrium. Disturb it with the buttons and watch the concentrations jump and then relax to a new equilibrium; the gauge compares $Q_{c}$ with $K_{c}$. ($K_{c}$ at each temperature is estimated from its value at $773\ \mathrm{K}$ and $\Delta_{r}H^{\circ}=-92.4\ \mathrm{kJ\,mol^{-1}}$; the reaction speed is illustrative.)

temperature—

K_c (L² mol⁻²)—

Q_c now—

NH₃ (mole %)—

Exercises

Extent and direction of reaction.

1 A reaction has $K_{c}=1.0\times10^{-30}$. Which species predominate at equilibrium?
2 For $\mathrm{2SO_{2}(g)+O_{2}(g)\rightleftharpoons2SO_{3}(g)}$, $K_{c}=2.8\times10^{2}\ \mathrm{L\,mol^{-1}}$ at $1000\ \mathrm{K}$. A mixture has $[\mathrm{SO_{2}}]=0.50$, $[\mathrm{O_{2}}]=0.10$, $[\mathrm{SO_{3}}]=1.0\ \mathrm{mol\,L^{-1}}$. Which way will it react?
3 Repeat part 2 with $[\mathrm{SO_{3}}]=5.0\ \mathrm{mol\,L^{-1}}$.
4 For $\mathrm{PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)}$, $K_{c}=0.13\ \mathrm{mol\,L^{-1}}$. A mixture has $[\mathrm{PCl_{5}}]=0.40$, $[\mathrm{PCl_{3}}]=0.20$, $[\mathrm{Cl_{2}}]=0.30\ \mathrm{mol\,L^{-1}}$. Predict the direction of change.
5 State the difference between the reaction quotient and the equilibrium constant.
6 The reaction of hydrogen with oxygen has an enormous $K$, yet a mixture of the gases can be stored safely. Explain.
7 For $\mathrm{N_{2}O_{4}(g)\rightleftharpoons2NO_{2}(g)}$, $K_{p}=0.15\ \mathrm{bar}$ at $298\ \mathrm{K}$. A mixture has $p_{\mathrm{N_{2}O_{4}}}=0.50\ \mathrm{bar}$ and $p_{\mathrm{NO_{2}}}=0.30\ \mathrm{bar}$. Which way will it go?
8 For $\mathrm{A(g)\rightleftharpoons B(g)}$, $K_{c}=1.0\times10^{4}$. What fraction of the total remains as A at equilibrium?

💡 Show solutions▶

1 $K\ll10^{-3}$: the reactants; practically no products form.

2 $Q_{c}=\dfrac{(1.0)^{2}}{(0.50)^{2}(0.10)}=40\lt280$: forward.

3 $Q_{c}=\dfrac{(5.0)^{2}}{(0.50)^{2}(0.10)}=1.0\times10^{3}\gt280$: backward.

4 $Q_{c}=\dfrac{(0.20)(0.30)}{0.40}=0.15\gt0.13$: backward (some $\mathrm{PCl_{3}}$ and $\mathrm{Cl_{2}}$ combine).

5 Both have the same form; $Q$ uses the concentrations at any moment, $K$ only the equilibrium concentrations. $Q$ changes as the reaction proceeds; $K$ is fixed at a given temperature.

6 $K$ tells how far a reaction can go, not how fast. At room temperature the reaction is extremely slow until a spark or catalyst starts it.

7 $Q_{p}=\dfrac{(0.30)^{2}}{0.50}=0.18\gt0.15$: backward ($\mathrm{NO_{2}}$ combines).

8 $[\mathrm{B}]/[\mathrm{A}]=10^{4}$, so the fraction of A is $\dfrac{1}{1+10^{4}}=1.0\times10^{-4}$ (0.01%): practically complete.

Calculating equilibrium compositions (check every answer by substituting into $K$).

1 $0.40\ \mathrm{mol}$ each of $\mathrm{H_{2}}$ and $\mathrm{I_{2}}$ are heated in a $2.0\ \mathrm{L}$ flask ($K_{c}=49$). Find the equilibrium concentrations.
2 For $\mathrm{CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}(g)+H_{2}(g)}$, $K_{c}=4.0$. A $1.0\ \mathrm{L}$ vessel starts with $0.20\ \mathrm{mol}$ each of CO and $\mathrm{H_{2}O}$ and $0.10\ \mathrm{mol}$ each of $\mathrm{CO_{2}}$ and $\mathrm{H_{2}}$. Find the direction and the equilibrium concentrations.
3 At a certain temperature $K_{c}=0.040\ \mathrm{mol\,L^{-1}}$ for $\mathrm{PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)}$. Starting with $[\mathrm{PCl_{5}}]=0.20\ \mathrm{mol\,L^{-1}}$, find the equilibrium concentrations and the percentage dissociated.
4 Use the result of Example 4 to find the degree of dissociation of $\mathrm{N_{2}O_{4}}$ at $298\ \mathrm{K}$ and a total pressure of $2.00\ \mathrm{bar}$. Compare with the value at $1.00\ \mathrm{bar}$.
5 For $\mathrm{N_{2}(g)+O_{2}(g)\rightleftharpoons2NO(g)}$, $K_{c}=4.0\times10^{-4}$ at about $2000\ \mathrm{K}$. In hot gas with $[\mathrm{N_{2}}]=0.040$ and $[\mathrm{O_{2}}]=0.010\ \mathrm{mol\,L^{-1}}$ initially, estimate $[\mathrm{NO}]$ at equilibrium, stating any approximation.
6 For $\mathrm{CO_{2}(g)+C(s)\rightleftharpoons2CO(g)}$, $K_{p}=2.0\ \mathrm{bar}$ at a high temperature. Initially $p_{\mathrm{CO_{2}}}=0.60\ \mathrm{bar}$, no CO, and graphite is present in excess. Find the equilibrium partial pressures.
7 At a total pressure of $1.0\ \mathrm{bar}$, $\mathrm{PCl_{5}}$ is $40\%$ dissociated. Show that $K_{p}=\dfrac{\alpha^{2}P}{1-\alpha^{2}}$ and evaluate it.
8 For $\mathrm{2NOCl(g)\rightleftharpoons2NO(g)+Cl_{2}(g)}$, $K_{c}=1.6\times10^{-5}\ \mathrm{mol\,L^{-1}}$. Starting with $[\mathrm{NOCl}]=0.50\ \mathrm{mol\,L^{-1}}$, find $[\mathrm{Cl_{2}}]$ and $[\mathrm{NO}]$ at equilibrium using a small-$x$ approximation, and check it.

💡 Show solutions▶

1 Initial $0.20\ \mathrm{mol\,L^{-1}}$ each; $\dfrac{2x}{0.20-x}=7$ → $x=\dfrac{1.40}{9}=0.156$. $[\mathrm{H_{2}}]=[\mathrm{I_{2}}]=0.044$, $[\mathrm{HI}]=0.311\ \mathrm{mol\,L^{-1}}$.

2 $Q_{c}=\dfrac{(0.10)(0.10)}{(0.20)(0.20)}=0.25\lt4.0$: forward. $\dfrac{(0.10+x)^{2}}{(0.20-x)^{2}}=4.0$ → $0.10+x=2(0.20-x)$ → $x=0.10$. $[\mathrm{CO}]=[\mathrm{H_{2}O}]=0.10$, $[\mathrm{CO_{2}}]=[\mathrm{H_{2}}]=0.20\ \mathrm{mol\,L^{-1}}$.

3 $\dfrac{x^{2}}{0.20-x}=0.040$ → $x^{2}+0.040x-0.0080=0$ → $x=\dfrac{-0.040+\sqrt{0.0016+0.032}}{2}=0.0717$. $[\mathrm{PCl_{3}}]=[\mathrm{Cl_{2}}]=0.072$, $[\mathrm{PCl_{5}}]=0.128\ \mathrm{mol\,L^{-1}}$; $36\%$ dissociated.

4 $\alpha=\sqrt{\dfrac{K_{p}}{K_{p}+4P}}=\sqrt{\dfrac{0.15}{8.15}}=0.136$ (14%), less than 19% at $1.00\ \mathrm{bar}$: higher pressure favours the side with fewer gas molecules.

5 $\dfrac{(2x)^{2}}{(0.040-x)(0.010-x)}=4.0\times10^{-4}$. $K$ is small, so take $x\ll0.010$: $4x^{2}=4.0\times10^{-4}\times4.0\times10^{-4}=1.6\times10^{-7}$, $2x=[\mathrm{NO}]=4.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$. Check: $x=2\times10^{-4}$ is 2% of $0.010$ — acceptable (the exact quadratic gives $3.95\times10^{-4}$).

6 Let $x$ bar of $\mathrm{CO_{2}}$ react: $\dfrac{(2x)^{2}}{0.60-x}=2.0$ → $4x^{2}+2.0x-1.20=0$ → $x=\dfrac{-2.0+\sqrt{4.0+19.2}}{8}=0.352$. $p_{\mathrm{CO}}=0.70\ \mathrm{bar}$, $p_{\mathrm{CO_{2}}}=0.25\ \mathrm{bar}$. Check: $(0.704)^{2}/0.248=2.0$ ✓.

7 Per mole of $\mathrm{PCl_{5}}$: $1-\alpha$, $\alpha$, $\alpha$; total $1+\alpha$. $K_{p}=\dfrac{(\alpha P/(1+\alpha))^{2}}{(1-\alpha)P/(1+\alpha)}=\dfrac{\alpha^{2}P}{(1+\alpha)(1-\alpha)}$. With $\alpha=0.40$, $P=1.0$: $K_{p}=\dfrac{0.16}{0.84}=0.19\ \mathrm{bar}$.

8 $[\mathrm{NO}]=2x$, $[\mathrm{Cl_{2}}]=x$, $[\mathrm{NOCl}]=0.50-2x\approx0.50$: $\dfrac{4x^{3}}{0.25}=1.6\times10^{-5}$ → $x^{3}=1.0\times10^{-6}$ → $x=0.010$. $[\mathrm{Cl_{2}}]=0.010$, $[\mathrm{NO}]=0.020\ \mathrm{mol\,L^{-1}}$. Check: $2x=0.020$ is 4% of $0.50$ — a fair approximation (solving exactly gives $x=0.0097$).

III

$K$, $Q$ and Gibbs energy ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$; $2.303RT=5.706\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$).

1 $\Delta_{r}G^{\circ}=-11.4\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. Find $K$.
2 $\Delta_{r}G^{\circ}=+10.0\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. Find $K$.
3 $K=1.0\times10^{-5}$ at $298\ \mathrm{K}$. Find $\Delta_{r}G^{\circ}$.
4 What is $\Delta_{r}G^{\circ}$ for a reaction with $K=1$? What does this mean?
5 What is the sign of $\Delta_{r}G$ when $Q\gt K$, and which direction is spontaneous?
6 A reaction has $K=10$ at $298\ \mathrm{K}$. Find $\Delta_{r}G$ for a mixture with $Q=100$.
7 Find $\Delta_{r}G^{\circ}$ at $720\ \mathrm{K}$ for $\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$, $K=49$.
8 Explain, using $\Delta_{r}G^{\circ}=-RT\ln K$, why changing concentrations or pressure does not change $K$, but changing temperature does.

💡 Show solutions▶

1 $\log K=\dfrac{11.4}{5.706}=2.00$, $K=1.0\times10^{2}$.

2 $\log K=-\dfrac{10.0}{5.706}=-1.753$, $K=1.8\times10^{-2}$.

3 $\Delta_{r}G^{\circ}=-5.706\log(1.0\times10^{-5})=+28.5\ \mathrm{kJ\,mol^{-1}}$.

4 $\Delta_{r}G^{\circ}=-RT\ln1=0$: under standard conditions the system is already at equilibrium; neither side is favoured.

5 $\Delta_{r}G=RT\ln(Q/K)\gt0$: the forward reaction is non-spontaneous, the backward reaction is spontaneous.

6 $\Delta_{r}G=RT\ln\dfrac{Q}{K}=2.478\ln10=+5.71\ \mathrm{kJ\,mol^{-1}}$: net backward reaction.

7 $\Delta_{r}G^{\circ}=-(8.314)(720)\ln49=-(5986)(3.892)=-2.33\times10^{4}\ \mathrm{J\,mol^{-1}}=-23.3\ \mathrm{kJ\,mol^{-1}}$.

8 $\Delta_{r}G^{\circ}$ refers to fixed standard states, so it depends only on the reaction and the temperature; hence $K$ depends only on $T$. Changing amounts or pressure changes $Q$, not $K$; changing $T$ changes both $\Delta_{r}G^{\circ}$ and the $RT$ factor, so $K$ changes.

Le Chatelier's principle.

1 State Le Chatelier's principle.
2 Predict the effect of increasing the pressure (by compression) on: (i) $\mathrm{N_{2}O_{4}(g)\rightleftharpoons2NO_{2}(g)}$; (ii) $\mathrm{H_{2}(g)+Cl_{2}(g)\rightleftharpoons2HCl(g)}$; (iii) $\mathrm{CO(g)+3H_{2}(g)\rightleftharpoons CH_{4}(g)+H_{2}O(g)}$; (iv) $\mathrm{CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)}$.
3 For an endothermic reaction, what happens to $K$ when the temperature is raised?
4 A red solution contains the equilibrium $\mathrm{Fe^{3+}(aq)+SCN^{-}(aq)}$ $\mathrm{\rightleftharpoons[Fe(SCN)]^{2+}(aq)}$. What happens to the colour when (i) KSCN crystals are added; (ii) oxalic acid is added? Explain.
5 A blue-violet solution of cobalt(II) chloride in hydrochloric acid turns pink in iced water. What does this tell you about the sign of $\Delta_{r}H$ for $\mathrm{[Co(H_{2}O)_{6}]^{2+}+4Cl^{-}}$ $\mathrm{\rightleftharpoons[CoCl_{4}]^{2-}+6H_{2}O}$?
6 What effect does a catalyst have on (i) $K$; (ii) the equilibrium composition; (iii) the time needed to reach equilibrium?
7 In the Haber process, why is ammonia removed continuously, and why is the reactor run at about $700$–$775\ \mathrm{K}$ rather than at room temperature, although $K$ is larger at room temperature?
8 Applied. In the lungs haemoglobin picks up oxygen: $\mathrm{Hb(aq)+O_{2}(aq)\rightleftharpoons HbO_{2}(aq)}$. Use Le Chatelier's principle to explain why climbers feel unwell at high altitude, and why after some weeks there the body makes more haemoglobin.

💡 Show solutions▶

1 When a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that tends to counteract the change.

2 (i) Backward (2 mol gas → 1 mol); (ii) no effect ($\Delta n_{g}=0$); (iii) forward (4 mol gas → 2 mol); (iv) backward (1 mol gas → 0).

3 $K$ increases: raising $T$ favours the endothermic (forward) direction.

4 (i) Deeper red: a reactant is added, $Q\lt K$, more complex forms. (ii) Paler: oxalate binds $\mathrm{Fe^{3+}}$, removing a reactant, so the equilibrium shifts backward.

5 Cooling favours the pink aqua ion (backward reaction), so the backward reaction is exothermic and the forward reaction endothermic: $\Delta_{r}H\gt0$.

6 (i) No change; (ii) no change; (iii) shorter — the forward and backward rates are both increased.

7 Removing the product keeps $Q\lt K$, so more ammonia keeps forming. At room temperature $K$ is large but the rate is negligibly small; at $700$–$775\ \mathrm{K}$ with an iron catalyst equilibrium is reached quickly, at the cost of a smaller (but acceptable) yield per pass.

8 At altitude the partial pressure of oxygen is low, so $[\mathrm{O_{2}}]$ in the blood falls and the equilibrium shifts backward: less $\mathrm{HbO_{2}}$ reaches the tissues (hypoxia — headache, fatigue). Making more Hb raises the concentration of a reactant and shifts the equilibrium forward again, restoring the amount of $\mathrm{HbO_{2}}$.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET For $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, $\Delta_{r}H\lt0$, the equilibrium yield of ammonia is increased by (A) raising the temperature and lowering the pressure (B) lowering the temperature and raising the pressure (C) raising both the temperature and the pressure (D) adding a catalyst

💡 Show solution▶

Idea Exothermic: low $T$ raises $K$. Fewer gas moles on the right: high pressure shifts forward. A catalyst does not change the yield.

Answer (B).

E2. JEE Main Numerical answer. For $\mathrm{A(g)+B(g)\rightleftharpoons C(g)+D(g)}$, $K_{c}=9.0$. $1.0\ \mathrm{mol}$ each of A and B are placed in a $1.0\ \mathrm{L}$ vessel. The equilibrium concentration of C, in $\mathrm{mol\,L^{-1}}$, multiplied by 100 is ____.

💡 Show solution▶

ICE $\dfrac{x^{2}}{(1.0-x)^{2}}=9.0$ → $\dfrac{x}{1.0-x}=3.0$ → $x=0.75$.

Answer $[\mathrm{C}]=0.75\ \mathrm{mol\,L^{-1}}$ → 75.

E3. Board · 3 marks (a) Define the reaction quotient $Q_{c}$ and state how comparing it with $K_{c}$ predicts the direction of a reaction. (2 marks) (b) For $\mathrm{2HI(g)\rightleftharpoons H_{2}(g)+I_{2}(g)}$, $K_{c}=0.020$ at a certain temperature. A mixture has $[\mathrm{HI}]=0.50$, $[\mathrm{H_{2}}]=[\mathrm{I_{2}}]=0.10\ \mathrm{mol\,L^{-1}}$. Which way does it react? (1 mark)

💡 Show solution▶

(a) $Q_{c}$ has the same form as $K_{c}$ but uses the concentrations at any instant. If $Q_{c}\lt K_{c}$ the reaction goes forward; if $Q_{c}\gt K_{c}$, backward; if $Q_{c}=K_{c}$ it is at equilibrium.

(b) $Q_{c}=\dfrac{(0.10)(0.10)}{(0.50)^{2}}=0.040\gt0.020$: backward — $\mathrm{H_{2}}$ and $\mathrm{I_{2}}$ combine to form more HI.

E4. JEE Main The standard Gibbs energy change of a reaction at $300\ \mathrm{K}$ is $-11.49\ \mathrm{kJ\,mol^{-1}}$. Taking $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, its equilibrium constant is closest to (A) $10$ (B) $100$ (C) $1000$ (D) $0.01$

💡 Show solution▶

Idea $\log K=\dfrac{-\Delta_{r}G^{\circ}}{2.303RT}=\dfrac{11\,490}{2.303\times8.314\times300}=\dfrac{11\,490}{5744}=2.00$.

Answer $K=100$: (B).

E5. JEE Advanced One or more options correct. (A) A catalyst does not change the equilibrium constant of a reaction. (B) Adding helium at constant volume to the equilibrium $\mathrm{PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)}$ increases the dissociation of $\mathrm{PCl_{5}}$. (C) For an exothermic reaction, $K$ decreases as the temperature rises. (D) Changing the pressure has no effect on the equilibrium composition of $\mathrm{H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)}$.

💡 Show solution▶

(A) True: it speeds up both directions equally.

(B) False: at constant volume the partial pressures of $\mathrm{PCl_{5}}$, $\mathrm{PCl_{3}}$ and $\mathrm{Cl_{2}}$ are unchanged, so $Q=K$ still.

(D) True: $\Delta n_{g}=0$, so compression multiplies numerator and denominator of $Q$ by the same factor.

Answer (A), (C), (D).

E6. Board · Case study Ammonia is made by the Haber process, $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, $\Delta_{r}H^{\circ}=-92.4\ \mathrm{kJ\,mol^{-1}}$, at about $200\ \mathrm{atm}$ and $773\ \mathrm{K}$ over an iron catalyst. (i) Explain the effect of pressure and of temperature on the equilibrium yield. (2 marks) (ii) Why is a catalyst used, although it does not change the yield? (1 mark) (iii) At $773\ \mathrm{K}$, $K_{c}=0.061\ \mathrm{L^{2}\,mol^{-2}}$. A sample of gas has $[\mathrm{N_{2}}]=2.0$, $[\mathrm{H_{2}}]=1.0$ and $[\mathrm{NH_{3}}]=0.30\ \mathrm{mol\,L^{-1}}$. Will more ammonia form? (1 mark)

💡 Show solution▶

(i) 4 mol of gas give 2 mol, so higher pressure shifts the equilibrium forward and raises the yield. The forward reaction is exothermic, so a higher temperature lowers $K$ and the yield; a lower temperature raises it.

(ii) At temperatures low enough to give a good yield the reaction is too slow; the catalyst lets equilibrium be reached quickly at a workable temperature.

(iii) $Q_{c}=\dfrac{(0.30)^{2}}{(2.0)(1.0)^{3}}=0.045\lt0.061$: yes, the reaction proceeds forward.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much ammonia can a Haber reactor produce at different temperatures and pressures, and why do real plants accept a yield of only about 15–20% per pass?

Why it matters. Ammonia plants use about 1–2% of the world's energy, mostly to make hydrogen and to compress the gases. Engineers in fertiliser plants from India to the Netherlands balance three things: equilibrium yield (thermodynamics), rate (kinetics) and the cost of high pressure. Research on new catalysts that work at lower temperatures — and on "green ammonia" made with hydrogen from renewable electricity — aims to change that balance.

What to do.

For a feed of $1$ mol $\mathrm{N_{2}}$ and $3$ mol $\mathrm{H_{2}}$, let $z$ mol of $\mathrm{N_{2}}$ react. Write the mole fractions of the three gases in terms of $z$ and show that $K_{p}=\dfrac{(x_{\mathrm{NH_{3}}})^{2}}{x_{\mathrm{N_{2}}}(x_{\mathrm{H_{2}}})^{3}P^{2}}$ at total pressure $P$.
Use $K_{p}=1.45\times10^{-5}\ \mathrm{bar^{-2}}$ at $773\ \mathrm{K}$ and $K_{p}\approx1.2\times10^{-4}\ \mathrm{bar^{-2}}$ at $673\ \mathrm{K}$. With a spreadsheet (trial values of $z$) find the equilibrium mole percentage of ammonia at $100$, $200$ and $300\ \mathrm{bar}$ for both temperatures.
Plot mole % $\mathrm{NH_{3}}$ against pressure for the two temperatures. Which variable has the larger effect?
Use Interactive 2 to explain, in terms of $Q$ and $K$, why removing ammonia and recycling unreacted gas lets a plant convert almost all of its feed in the end.

Tools: Rules 6.8 and 6.9, Principle 6.10, Interactive 2, a spreadsheet.

Hand in: Your table and graph, and a half-page recommendation of operating conditions that weighs yield against rate and cost.

Going further. A catalyst fast enough at $673\ \mathrm{K}$ would allow a lower pressure for the same yield. Estimate, with your method, the pressure needed at $673\ \mathrm{K}$ to match the yield at $773\ \mathrm{K}$ and $200\ \mathrm{bar}$.

Instructor note: with these $K_{p}$ values, mole % $\mathrm{NH_{3}}$ ≈ 10.0, 17.0, 22.4 at 100, 200, 300 bar (773 K) and 21.9, 32.6, 39.5 (673 K) — measured values are a little higher at high pressure because the gases are not ideal. Cooling by 100 K roughly doubles the yield at each pressure, and so does tripling the pressure: both variables matter. Removing $\mathrm{NH_{3}}$ keeps $Q\lt K$; recycling gives an overall conversion of about 97%. Matching 17% at 673 K needs only about 70 bar (students should find 60–80 bar).

🧪 Self-check quiz

0/5 correct

Q1. For a reaction mixture, $Q_{c}=2.0\times10^{-3}$ and $K_{c}=1.5\times10^{-2}$. The mixture will

Q2. Adding a catalyst to a system at equilibrium

Q3. For an exothermic reaction, raising the temperature

Q4. A reaction with $\Delta_{r}G^{\circ}\lt0$ has

Q5. Increasing the pressure by compression has no effect on the equilibrium

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