Summary and Quick Revision

Card 1 — Physical Equilibrium

Every physical equilibrium is a closed-system balance between two opposing physical changes running at the same rate.

The four types and the constant each one fixes

Process	Equation	What stays constant at a fixed temperature	Name of the constant
Solid-liquid	$\mathrm{H_2O(s)} \rightleftharpoons \mathrm{H_2O(l)}$	The melting point at a fixed pressure; the two phases coexist only at that one temperature	Melting point $273\ \mathrm{K}$ at $1.013\ \mathrm{bar}$
Liquid-vapour	$\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2O(g)}$	The vapour pressure above the liquid	Equilibrium vapour pressure $p_{\mathrm{H_2O}}$
Solid-vapour	$\mathrm{I_2(s)} \rightleftharpoons \mathrm{I_2(g)}$	The vapour pressure of the solid (sublimation)	$p_{\mathrm{I_2}}$
Solid dissolving	$\mathrm{Sugar(s)} \rightleftharpoons \mathrm{Sugar(solution)}$	The concentration of the saturated solution	Solubility
Gas dissolving	$\mathrm{CO_2(g)} \rightleftharpoons \mathrm{CO_2(aq)}$	The ratio $[\mathrm{CO_2(aq)}]/[\mathrm{CO_2(g)}]$	The equilibrium constant of the dissolution, related to Henry's constant $K_H$ by Henry's law

Henry's law in one line

Key Point (Definition): The mass of a gas dissolved in a given mass of solvent at a fixed temperature is proportional to the pressure of that gas above the solvent, $m \propto p$, equivalently $p = K_H x$ where $x$ is the mole fraction of the dissolved gas.

• Raising pressure raises dissolved gas. Opening a soda bottle drops the $\mathrm{CO_2}$ pressure, so gas fizzes out until the new equilibrium is reached.
• Raising temperature lowers gas solubility. Dissolution of a gas is exothermic.
• A boiling liquid is the case where its vapour pressure equals the external pressure, so the boiling point rises when the external pressure rises.

General characteristics of every physical equilibrium

1. Possible only in a closed system at a given temperature. Leave the vessel open and the vapour escapes, and no equilibrium is reached.
2. Both opposing processes continue at the same rate — a dynamic, not a static, condition.
3. All measurable properties (pressure, concentration, colour, density) stop changing.
4. Each equilibrium is characterised by a constant value of one parameter at a given temperature, listed in the table above.
5. The magnitude of that parameter says how far the physical process went before stopping.

Fast facts worth a mark each

Statement	Correct or wrong
At the melting point the ice stops melting	Wrong. Melting and freezing continue at equal rates
Vapour pressure depends on the amount of liquid present	Wrong. It depends only on the liquid and the temperature
Vapour pressure depends on the surface area of the liquid	Wrong. Surface area changes the rate, not the equilibrium pressure
Adding more solid sugar to a saturated solution raises the concentration	Wrong. Concentration is fixed at saturation
Radioactive sugar added to a saturated solution ends up in the solution too	Correct, and this is the proof that dissolution is dynamic
Solubility of a gas falls as temperature rises	Correct

[NEET] The single most-asked line here is the definition of the dynamic condition: the two rates are equal, and nothing stops.

Card 2 — Chemical Equilibrium and Its Dynamic Nature

Reversible and irreversible

Reversible reaction	Irreversible reaction
Products re-form the reactants under the same conditions	Products do not re-form reactants
Written with $\rightleftharpoons$	Written with $\rightarrow$
Reaches equilibrium with all species present	Goes to completion; one reactant is exhausted
$\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$	Burning of a hydrocarbon in excess oxygen

Equilibrium can be reached from either direction. Start with only $\mathrm{N_2}$ and $\mathrm{H_2}$, or only $\mathrm{NH_3}$, and at the same temperature the same equilibrium constant results.

What is constant and what is not

Constant at equilibrium	Not constant, and never claimed to be
Concentration of every species	The two rates before equilibrium is reached
Total pressure and density	The amounts of reactant and product, which are almost never equal
Colour, refractive index, every measurable property	The individual molecules, which keep reacting
The rate of the forward reaction, now equal to the reverse rate	The reaction itself — it does not stop

Key Point (Definition): Chemical equilibrium is the state of a reversible reaction at which the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of all reactants and products remain constant with time while both reactions continue.

The evidence that equilibrium is dynamic

Experiment	Observation	What it proves
Haber process run with $\mathrm{N_2}$ and $\mathrm{H_2}$, then repeated starting from $\mathrm{NH_3}$	The same equilibrium mixture at the same temperature	Equilibrium is reached from both directions
$\mathrm{D_2}$ mixed with $\mathrm{NH_3}$ over a catalyst at equilibrium	$\mathrm{NH_2D}$, $\mathrm{NHD_2}$ and $\mathrm{ND_3}$ appear	The $\mathrm{N-H}$ bonds keep breaking and re-forming
$\mathrm{H_2}$ and $\mathrm{D_2}$ mixed at equilibrium	$\mathrm{HD}$ forms with no net change in the amounts of $\mathrm{H_2}$ and $\mathrm{D_2}$	Exchange without net reaction — the definition of dynamic
Radioactive iodine added to a saturated iodine solution	Radioactivity appears in the solid as well	Physical equilibria are dynamic too

The shape of the approach to equilibrium

• The forward rate falls as reactants are used up; the reverse rate rises from zero as products build up.
• The two curves meet, and from that moment the concentrations flatten.
• The time taken to reach equilibrium says nothing about how far the reaction goes. A catalyst changes the time, not the destination.

Traps

Wrong statement	Right statement
At equilibrium the forward and reverse reactions stop	Both continue, at equal rates
At equilibrium the concentrations of reactants and products are equal	They are constant, and generally very unequal
Equilibrium can be reached in an open vessel	Only in a closed system, unless nothing escapes
A fast reaction has a large $K$	Speed and extent are unrelated

[Board] The written answer that scores is "the rates become equal, so the concentrations stay constant while both reactions continue".

Card 3 — The Equilibrium Constant

The expression

For $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$ at a fixed temperature,

$$K_c = \frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}$$

• Products on top, reactants below.
• The powers are the stoichiometric coefficients of the balanced equation, nothing else.
• The concentrations are the equilibrium concentrations in $\mathrm{mol\ L^{-1}}$.
• $K_c$ depends only on temperature. Not on pressure, volume, initial concentrations, catalyst, or the direction from which equilibrium was approached.

Key Point (Definition): Law of mass action — at a given temperature the rate of a reaction is proportional to the product of the molar concentrations of the reactants, each raised to the power of its stoichiometric coefficient. Equating forward and reverse rates at equilibrium gives $K_c = k_f/k_r$.

What the magnitude of $K$ tells you

Value of $K_c$	Position of equilibrium	How to describe it
$K_c > 10^{3}$	Products dominate	Reaction is nearly complete; reactants present only in traces
$10^{-3} < K_c < 10^{3}$	Comparable amounts of both	Appreciable quantities of reactants and products
$K_c < 10^{-3}$	Reactants dominate	Reaction barely proceeds

$K$ says how far, never how fast. A large $K$ with a large activation energy gives a reaction that is complete in principle and imperceptible in practice.

The manipulation rules

Operation on the equation	New equilibrium constant	Worked check with $K = 4$
Reverse the equation	$K' = \dfrac{1}{K}$	$K' = 0.25$
Multiply the whole equation by $n$	$K' = K^{n}$	Doubling gives $K' = 16$
Divide the equation by $n$ (halve it)	$K' = K^{1/n}$	Halving gives $K' = 2$
Add two equations	$K' = K_1 \times K_2$	With $K_2 = 3$, $K' = 12$
Subtract equation 2 from equation 1	$K' = K_1/K_2$	$K' = 4/3$

Combine them in order. Reversing and doubling gives $K' = (1/K)^{2} = 1/K^{2}$, which for $K = 4$ is $0.0625$.

Units

$K_c$ carries units $(\mathrm{mol\ L^{-1}})^{\Delta n}$ unless $\Delta n = 0$, and a $K$ computed from concentrations or pressures may be quoted with those units. The thermodynamic $K$ used in $\Delta G^{\circ} = -RT \ln K$ is dimensionless, because each concentration or pressure is first expressed relative to its standard state, $1\ \mathrm{M}$ or $1\ \mathrm{bar}$.

Traps

Error	Fix
Using initial concentrations in $K_c$	Only equilibrium concentrations belong there
Forgetting the power on a coefficient of $2$ or $3$	Square or cube that concentration
Changing $K$ because the volume changed	$K$ is fixed at fixed temperature; the concentrations adjust
Adding $K_1$ and $K_2$ for added equations	Multiply them
Doubling $K$ when the equation is doubled	Square it

[JEE Main] Multi-step questions almost always test the combination rule: write each given step, adjust each $K$, then multiply.

Card 4 — $K_p$ and $K_c$

For gaseous equilibria, partial pressures may replace concentrations.

$$K_p = \frac{p_{\mathrm{C}}^{c}\,p_{\mathrm{D}}^{d}}{p_{\mathrm{A}}^{a}\,p_{\mathrm{B}}^{b}}, \qquad p_i = \frac{n_i RT}{V} = [\,i\,]RT$$

Substituting gives the relation to memorise:

$$\boxed{K_p = K_c (RT)^{\Delta n}}$$

$$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$$

The rules for $\Delta n$

• Count gases only. Solids, pure liquids and dissolved species contribute nothing.
• Use the coefficients of the balanced equation.
• $R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$ when pressures are in bar; $R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ when they are in atm. $T$ is always in kelvin.

The three cases

Case	Meaning	Example	Relation
$\Delta n > 0$	Gas moles increase	$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$, $\Delta n = 1$	$K_p = K_c(RT)$, so $K_p > K_c$
$\Delta n = 0$	Gas moles unchanged	$\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\mathrm{HI(g)}$, $\Delta n = 0$	$K_p = K_c$, both dimensionless
$\Delta n < 0$	Gas moles decrease	$\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$, $\Delta n = -2$	$K_p = K_c(RT)^{-2}$, so $K_p < K_c$

Worked $\Delta n$ values to have ready

Equilibrium	$\Delta n$	$K_p$
$2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$	$-1$	$K_c(RT)^{-1}$
$\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$	$+1$	$K_c(RT)$
$2\mathrm{NO(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$	$-1$	$K_c(RT)^{-1}$
$\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$	$+1$	$K_c(RT)$
$\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftharpoons 2\mathrm{CO(g)}$	$+1$	$K_c(RT)$
$\mathrm{H_2(g)} + \mathrm{Cl_2(g)} \rightleftharpoons 2\mathrm{HCl(g)}$	$0$	$K_c$
$\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)}$	$0$	$K_c$

Partial pressure shortcuts

• $p_i = x_i \times P_{\text{total}}$, where $x_i$ is the mole fraction.
• Mole fraction $x_i = n_i / n_{\text{total}}$, and mole fractions of all species sum to $1$.
• Only $K_c$, $K_p$ and the dimensionless thermodynamic $K$ appear in this chapter; partial pressures are the bridge between the first two.

Traps

Error	Value it produces	Fix
Reactants minus products	Sign of the exponent flips	Products minus reactants, always
Counting a solid or a pure liquid	Wrong $\Delta n$	Gases only
Ignoring coefficients	For ammonia, $\Delta n = -3$ instead of $-2$	Use the balanced coefficients
Using $T$ in degrees Celsius	$(RT)$ far too small	Kelvin only
Using $R = 8.314$ with L and bar	Factor of $100$ error	$R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$

[NEET] Reading $\Delta n$ off the equation and substituting is a twenty-second question. It appears almost every year in some form.

Card 5 — Heterogeneous Equilibria

A heterogeneous equilibrium has reactants and products in more than one phase.

The omission rule and its reason

Key Point: Pure solids and pure liquids do not appear in the equilibrium constant expression, because their concentration — mass per unit volume, fixed by density and molar mass — cannot change while any of the substance is present. Formally their activity is $1$.

• Density and molar mass are constants of the substance, so $[\text{pure solid}] = \rho/M$ is a constant. Constants are absorbed into $K$.
• Amount of solid does not matter. Doubling the mass of $\mathrm{CaCO_3}$ changes nothing.
• Presence of the solid does matter. Remove all of it and there is no equilibrium left to speak of.
• A solvent present in large excess, such as water in a dilute aqueous equilibrium, is treated the same way and left out.

Equations beside their expressions

Equilibrium	$K_p$	$K_c$
$\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$	$p_{\mathrm{CO_2}}$	$[\mathrm{CO_2}]$
$\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftharpoons 2\mathrm{CO(g)}$	$\dfrac{p_{\mathrm{CO}}^{2}}{p_{\mathrm{CO_2}}}$	$\dfrac{[\mathrm{CO}]^{2}}{[\mathrm{CO_2}]}$
$3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} \rightleftharpoons \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}$	$\dfrac{p_{\mathrm{H_2}}^{4}}{p_{\mathrm{H_2O}}^{4}}$	$\dfrac{[\mathrm{H_2}]^{4}}{[\mathrm{H_2O}]^{4}}$
$\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2O(g)}$	$p_{\mathrm{H_2O}}$	$[\mathrm{H_2O(g)}]$
$\mathrm{NH_4HS(s)} \rightleftharpoons \mathrm{NH_3(g)} + \mathrm{H_2S(g)}$	$p_{\mathrm{NH_3}}\,p_{\mathrm{H_2S}}$	$[\mathrm{NH_3}][\mathrm{H_2S}]$
$\mathrm{CaO(s)} + \mathrm{CO_2(g)} \rightleftharpoons \mathrm{CaCO_3(s)}$	$\dfrac{1}{p_{\mathrm{CO_2}}}$	$\dfrac{1}{[\mathrm{CO_2}]}$
$\mathrm{Ag_2O(s)} + 2\mathrm{HNO_3(aq)} \rightleftharpoons 2\mathrm{AgNO_3(aq)} + \mathrm{H_2O(l)}$	not used	$\dfrac{[\mathrm{AgNO_3}]^{2}}{[\mathrm{HNO_3}]^{2}}$

The single-gas cases

When one gas alone appears, $K_p$ equals its partial pressure and therefore equals the total pressure of the closed vessel, once any air is excluded. This gives the fastest questions in the chapter: for $\mathrm{CaCO_3}$ decomposing at a stated temperature, $K_p = p_{\mathrm{CO_2}} = P_{\text{total}}$.

For $\mathrm{NH_4HS(s)}$, the two gases are produced in a $1:1$ ratio, so each has partial pressure $P/2$ and

$$K_p = \left(\frac{P}{2}\right)^{2} = \frac{P^{2}}{4}$$

The same reasoning gives $K_p = P^{2}/4$ for $\mathrm{NH_4Cl(s)} \rightleftharpoons \mathrm{NH_3(g)} + \mathrm{HCl(g)}$ and $K_p = 4P^{3}/27$ for $\mathrm{NH_4CO_2NH_2(s)} \rightleftharpoons 2\mathrm{NH_3(g)} + \mathrm{CO_2(g)}$.

Traps

Error	Fix
Writing $[\mathrm{CaCO_3}]$ in the denominator	Omit every pure solid
Including liquid water in an aqueous equilibrium	Omit it; it is the solvent in excess
Omitting $\mathrm{H_2O(g)}$ because it is water	Steam is a gas and is included
Saying more solid shifts the equilibrium	Amount of solid has no effect
Forgetting the power $4$ on steam and hydrogen	Coefficients still become powers

[Board] State the reason with the answer: their concentrations are constant, so they are absorbed into $K$.

Card 6 — $Q$, Direction of Reaction, and the ICE Method

The reaction quotient

$Q_c$ is the same expression as $K_c$, evaluated with the concentrations present at any instant, not necessarily at equilibrium.

Comparison	Meaning	Which way the reaction moves
$Q < K$	Too few products	Forward, left to right, until $Q$ rises to $K$
$Q > K$	Too many products	Backward, right to left, until $Q$ falls to $K$
$Q = K$	The mixture already satisfies the expression	No net change; the system is at equilibrium

The same test works in pressures with $Q_p$ against $K_p$.

What $K$ predicts and what it does not

$K$ predicts	$K$ does not predict
The extent of reaction — how far it goes	The rate, or the time taken
The direction of net change from any given mixture, through $Q$	The mechanism
The equilibrium concentrations from the initial ones	Whether the reaction happens at a measurable speed

The ICE recipe

1. Write the balanced equation. Every power and every coefficient in the rest of the working comes from it.
2. Rule three rows: Initial, Change, Equilibrium. Work in $\mathrm{mol\ L^{-1}}$ — divide moles by the volume of the vessel first.
3. Enter the initial concentrations. Anything not present starts at zero.
4. Enter the change as $-ax$, $-bx$, $+cx$, $+dx$, with the coefficients as multipliers. One unknown $x$ only.
5. Add the columns to get the equilibrium row.
6. Substitute into $K_c$ and solve for $x$.
7. Approximate if $K$ is small: assume $x$ is negligible beside the initial concentration. Valid when $x$ comes out below about $5\%$ of that concentration; check it afterwards and solve the quadratic if it fails.
8. Reject any root that makes a concentration negative or larger than the total available.
9. Convert $x$ back into the quantity that was asked — concentration, moles, partial pressure or degree of dissociation.

Degree of dissociation

For $\mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C}$ starting from $c$ with degree of dissociation $\alpha$, the equilibrium amounts are $c(1-\alpha)$, $c\alpha$, $c\alpha$, giving

$$K_c = \frac{c\alpha^{2}}{1-\alpha}$$

In terms of a total pressure $P$, the same reaction — one mole giving two different products — has

$$K_p = \frac{\alpha^{2}P}{1-\alpha^{2}} \qquad \text{for } \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C}$$

A reactant giving two moles of the same product carries an extra factor of 4, from squaring the coefficient 2:

$$K_p = \frac{4\alpha^{2}P}{1-\alpha^{2}} \qquad \text{for } \mathrm{A} \rightleftharpoons 2\mathrm{B}$$

So $\mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2}$ takes the first form and $\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$ the second. Attaching the 4 to the first pattern is the commonest slip in the topic.

Traps

Error	Fix
Using moles instead of concentrations	Divide by the volume before starting
Forgetting the coefficient in the change row	$-2x$ for a coefficient of $2$
Making the approximation when $K$ is large	Solve the quadratic
Keeping a negative root	Discard it
Answering with $x$ when the question asked for a concentration	Read the last line of the question again

[JEE Main] The most common single mistake in a full ICE problem is dropping a stoichiometric multiplier in the change row.

Card 7 — Equilibrium and Gibbs Energy

The two equations

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

At equilibrium $\Delta G = 0$ and $Q = K$, which gives

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

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

Use $R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$ here, and answer in $\mathrm{J\ mol^{-1}}$ or $\mathrm{kJ\ mol^{-1}}$. At $298\ \mathrm{K}$, $2.303\,RT = 5705\ \mathrm{J\ mol^{-1}} = 5.705\ \mathrm{kJ\ mol^{-1}}$, so

$$\Delta G^{\circ} = -5.705 \log K \ \ \mathrm{kJ\ mol^{-1}} \quad \text{at } 298\ \mathrm{K}$$

The sign table

$\Delta G^{\circ}$	$\log K$	$K$	Position of equilibrium	Reaction as written
Large and negative	Large positive	$K \gg 1$	Far to the right	Spontaneous; nearly complete
Slightly negative	Small positive	$K > 1$	Products favoured	Proceeds appreciably
Zero	$0$	$K = 1$	Balanced	At standard conditions the mixture is already at equilibrium
Slightly positive	Small negative	$K < 1$	Reactants favoured	Proceeds only slightly
Large and positive	Large negative	$K \ll 1$	Far to the left	Non-spontaneous as written; the reverse is spontaneous

$\Delta G$ against $\Delta G^{\circ}$

$\Delta G$	$\Delta G^{\circ}$
Refers to the actual mixture at the actual composition	Refers to all species in their standard states
Changes continuously as the reaction proceeds	A fixed number for a given reaction at a given temperature
Zero at equilibrium, always	Zero only when $K = 1$
Sign gives the direction of spontaneous change now	Sign gives whether $K$ exceeds $1$

A negative $\Delta G^{\circ}$ does not mean the reaction goes to completion, and a positive $\Delta G^{\circ}$ does not mean nothing happens. Both statements are about the size of $K$, not about totality.

Useful conversions

$K$	$\log K$	$\Delta G^{\circ}$ at $298\ \mathrm{K}$
$10^{6}$	$6$	$-34.2\ \mathrm{kJ\ mol^{-1}}$
$10^{3}$	$3$	$-17.1\ \mathrm{kJ\ mol^{-1}}$
$10$	$1$	$-5.7\ \mathrm{kJ\ mol^{-1}}$
$1$	$0$	$0$
$10^{-3}$	$-3$	$+17.1\ \mathrm{kJ\ mol^{-1}}$
$10^{-6}$	$-6$	$+34.2\ \mathrm{kJ\ mol^{-1}}$

Traps

Error	Fix
Dropping the minus sign in $-RT\ln K$	A negative $\Delta G^{\circ}$ must give $K > 1$; check the sign against the table
Mixing $\ln$ and $\log$	$\ln K = 2.303 \log K$
Using $R = 0.0821$ in this equation	Use $8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$
Reporting $\Delta G^{\circ}$ in $\mathrm{kJ}$ after using $R$ in $\mathrm{J}$	Divide by $1000$ once, deliberately
Setting $\Delta G^{\circ} = 0$ at equilibrium	$\Delta G = 0$ at equilibrium; $\Delta G^{\circ}$ is a constant

[JEE/NEET] The favourite one-liner: $\Delta G^{\circ} = 0$ implies $K = 1$, not $K = 0$.

Card 8 — Le Chatelier's Principle

Key Point (Definition): When a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract the effect of the imposed change.

The full grid

Reference reaction: $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$, $\Delta H = -92.38\ \mathrm{kJ\ mol^{-1}}$, $\Delta n = -2$.

Change imposed	Shift	Effect on $K$	Reason
Add a reactant ($\mathrm{N_2}$ or $\mathrm{H_2}$)	Forward	None	$Q$ falls below $K$
Remove a reactant	Backward	None	$Q$ rises above $K$
Add a product ($\mathrm{NH_3}$)	Backward	None	$Q$ rises above $K$
Remove a product as it forms	Forward	None	$Q$ falls below $K$; the basis of driving a reaction to completion
Increase pressure by decreasing volume	Towards fewer gas moles — forward here	None	The system reduces the number of gas molecules
Decrease pressure by increasing volume	Towards more gas moles — backward here	None	Same rule, other way
Pressure change when $\Delta n = 0$	No shift	None	Both sides have the same gas count
Increase temperature, exothermic reaction	Backward	$K$ decreases	Heat is a product; the system absorbs the added heat
Increase temperature, endothermic reaction	Forward	$K$ increases	Heat is a reactant
Decrease temperature, exothermic reaction	Forward	$K$ increases	Yield rises but the rate falls
Add an inert gas at constant volume	No shift	None	Partial pressures of the reacting gases are unchanged
Add an inert gas at constant pressure	Towards more gas moles — backward here	None	The volume must expand, so the reacting partial pressures fall; behaves like dilution
Dilute an aqueous equilibrium with water	Towards more dissolved particles	None	Same logic as constant-pressure inert gas
Add a catalyst	No shift	None	Both rates are raised equally; only the time to equilibrium falls
Add more pure solid or pure liquid	No shift	None	Its concentration is constant

Temperature is the only change that alters $K$. Everything else moves the composition to a new position with the same $K$.

Reading the shift from the equation

1. Write $\Delta n$ for gases. Pressure questions are decided by its sign alone.
2. Write $\Delta H$ as heat on the correct side. Temperature questions are then a concentration question.
3. For a concentration change, ask which way $Q$ moved relative to $K$.

Industrial applications

Process	Equation	Conditions and why
Haber synthesis of ammonia	$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$, exothermic, $\Delta n = -2$	High pressure $200\ \mathrm{atm}$ favours the forward shift; temperature about $773\ \mathrm{K}$, near $500$ degrees Celsius, is a compromise between yield and rate; iron catalyst; ammonia removed by liquefaction
Contact process	$2\mathrm{SO_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{SO_3}$, exothermic, $\Delta n = -1$	Near-atmospheric pressure, about $2\ \mathrm{atm}$ — with $K \approx 10^{26}$ the conversion is already near-complete, so high pressure is not worth paying for; temperature near $720\ \mathrm{K}$ as a compromise; $\mathrm{V_2O_5}$ catalyst; excess air
Manufacture of $\mathrm{CaO}$	$\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$, endothermic	High temperature and continuous removal of $\mathrm{CO_2}$

[NEET] Every Le Chatelier question reduces to two numbers written on the equation: the sign of $\Delta n$ and the side heat sits on.

Card 9 — Acids and Bases

The three definitions

Definition	Acid	Base	Scope	Fails on
Arrhenius	Gives $\mathrm{H^+}$ in water	Gives $\mathrm{OH^-}$ in water	Aqueous solutions only	$\mathrm{NH_3}$, which has no $\mathrm{OH}$; non-aqueous systems
Bronsted-Lowry	Proton donor	Proton acceptor	Any proton-transfer system, any solvent	$\mathrm{BF_3}$, $\mathrm{AlCl_3}$ — no proton is involved
Lewis	Electron-pair acceptor	Electron-pair donor	The widest in reach: it covers species carrying no proton at all	$\mathrm{NH_4^+}$, a Bronsted acid with no vacant orbital and so not a Lewis acid

Every Bronsted base is a Lewis base, and every Arrhenius acid is a Bronsted acid, but the acid classes do not nest. The Lewis definition is wider only in the sense that it reaches species with no proton at all ($\mathrm{BF_3}$, $\mathrm{AlCl_3}$, $\mathrm{Co^{3+}}$); $\mathrm{NH_4^+}$ is a Bronsted acid and not a Lewis acid.

Conjugate pairs

Key Point: A conjugate acid-base pair is two species differing by exactly one proton. The acid is the member with the extra proton.

Acid	Its conjugate base	Base	Its conjugate acid
$\mathrm{HCl}$	$\mathrm{Cl^-}$	$\mathrm{NH_3}$	$\mathrm{NH_4^+}$
$\mathrm{H_2SO_4}$	$\mathrm{HSO_4^-}$	$\mathrm{OH^-}$	$\mathrm{H_2O}$
$\mathrm{H_3O^+}$	$\mathrm{H_2O}$	$\mathrm{H_2O}$	$\mathrm{H_3O^+}$
$\mathrm{H_2CO_3}$	$\mathrm{HCO_3^-}$	$\mathrm{CO_3^{2-}}$	$\mathrm{HCO_3^-}$
$\mathrm{HCO_3^-}$	$\mathrm{CO_3^{2-}}$	$\mathrm{HPO_4^{2-}}$	$\mathrm{H_2PO_4^-}$
$\mathrm{NH_4^+}$	$\mathrm{NH_3}$	$\mathrm{F^-}$	$\mathrm{HF}$
$\mathrm{CH_3COOH}$	$\mathrm{CH_3COO^-}$	$\mathrm{S^{2-}}$	$\mathrm{HS^-}$

The stronger the acid, the weaker its conjugate base. $\mathrm{Cl^-}$ from the strong acid $\mathrm{HCl}$ is so weak a base that it does not hydrolyse at all.

Amphiprotic species

Species that can both donate and accept a proton: $\mathrm{H_2O}$, $\mathrm{HCO_3^-}$, $\mathrm{HSO_4^-}$, $\mathrm{HS^-}$, $\mathrm{H_2PO_4^-}$, $\mathrm{HPO_4^{2-}}$, and amino acids.

$$\mathrm{HCO_3^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CO_3^{2-}} \quad \text{(acting as an acid)}$$ $$\mathrm{HCO_3^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2CO_3} + \mathrm{OH^-} \quad \text{(acting as a base)}$$

Lewis acids and bases

Lewis acids	Why	Lewis bases	Why
$\mathrm{BF_3}$, $\mathrm{AlCl_3}$, $\mathrm{BCl_3}$	Incomplete octet, six electrons	$\mathrm{NH_3}$, $\mathrm{H_2O}$, $\mathrm{ROH}$, $\mathrm{R_2O}$	Lone pair on N or O
$\mathrm{H^+}$, $\mathrm{Ag^+}$, $\mathrm{Cu^{2+}}$, $\mathrm{Fe^{3+}}$	Cations accepting lone pairs	$\mathrm{OH^-}$, $\mathrm{Cl^-}$, $\mathrm{CN^-}$, $\mathrm{F^-}$	Anions with lone pairs
$\mathrm{SiF_4}$, $\mathrm{PF_5}$, $\mathrm{SF_4}$	Expandable octet	$\mathrm{CO}$, alkenes	Lone pair or $\pi$ electrons
$\mathrm{CO_2}$, $\mathrm{SO_2}$	Multiply bonded central atom accepts a pair	$\mathrm{NH_2^-}$, $\mathrm{H^-}$	Strongly electron-rich anions

Traps

Error	Fix
Calling $\mathrm{H_2SO_4}$ and $\mathrm{SO_4^{2-}}$ a conjugate pair	They differ by two protons; the pair is $\mathrm{H_2SO_4}$ and $\mathrm{HSO_4^-}$
Calling $\mathrm{NH_3}$ an Arrhenius base	It has no $\mathrm{OH}$; it is Bronsted and Lewis
Calling $\mathrm{BF_3}$ a Bronsted acid	No proton; Lewis only
Saying a strong acid has a strong conjugate base	The stronger the acid, the weaker its conjugate base

[Board] Write both equations for an amphiprotic species. One equation alone does not prove the term.

Card 10 — Water, pH and pOH

The ionic product of water

$$2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}, \qquad K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$$

• $K_w = 1.0 \times 10^{-14}$ at $298\ \mathrm{K}$, so in pure water $[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\ \mathrm{M}$.
• Autoionisation is endothermic, so $K_w$ rises with temperature.
• The degree of ionisation of water at $298\ \mathrm{K}$ is about $1.8 \times 10^{-9}$, or roughly two molecules in a billion.
• $K_w$ holds in every aqueous solution, acidic or basic, not only in pure water.

Temperature	$K_w$	Neutral $[\mathrm{H_3O^+}]$	Neutral pH
$283\ \mathrm{K}$	$0.29 \times 10^{-14}$	$5.4 \times 10^{-8}$	$7.27$
$298\ \mathrm{K}$	$1.0 \times 10^{-14}$	$1.0 \times 10^{-7}$	$7.00$
$310\ \mathrm{K}$	$2.7 \times 10^{-14}$	$1.64 \times 10^{-7}$	$6.78$
$373\ \mathrm{K}$	$51 \times 10^{-14}$	$7.1 \times 10^{-7}$	$6.15$

Key Point: Neutrality means $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$, not pH $= 7$. Water at $310\ \mathrm{K}$ has pH $6.78$ and is still perfectly neutral.

Definitions

$$\mathrm{pH} = -\log[\mathrm{H_3O^+}], \qquad \mathrm{pOH} = -\log[\mathrm{OH^-}], \qquad \mathrm{p}K_w = -\log K_w$$

$$\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w = 14 \quad \text{at } 298\ \mathrm{K \ only}$$

Solution at $298\ \mathrm{K}$	$[\mathrm{H_3O^+}]$	pH
Acidic	$> 10^{-7}\ \mathrm{M}$	$< 7$
Neutral	$= 10^{-7}\ \mathrm{M}$	$= 7$
Basic	$< 10^{-7}\ \mathrm{M}$	$> 7$

A change of one pH unit is a tenfold change in $[\mathrm{H_3O^+}]$. pH values below $0$ and above $14$ are perfectly possible in concentrated solutions.

Strong acids and bases

Strong electrolytes are completely ionised, so the ion concentration comes straight from the stoichiometry.

Solution	$[\mathrm{H_3O^+}]$ or $[\mathrm{OH^-}]$	pH at $298\ \mathrm{K}$
$0.1\ \mathrm{M}\ \mathrm{HCl}$	$10^{-1}\ \mathrm{M}$	$1$
$10^{-3}\ \mathrm{M}\ \mathrm{HNO_3}$	$10^{-3}\ \mathrm{M}$	$3$
$0.05\ \mathrm{M}\ \mathrm{H_2SO_4}$ (both protons)	$0.1\ \mathrm{M}$	$1$
$10^{-3}\ \mathrm{M}\ \mathrm{NaOH}$	$[\mathrm{OH^-}] = 10^{-3}$, pOH $= 3$	$11$
$0.01\ \mathrm{M}\ \mathrm{Ca(OH)_2}$	$[\mathrm{OH^-}] = 0.02$, pOH $= 1.70$	$12.30$
$10^{-8}\ \mathrm{M}\ \mathrm{HCl}$	Water contributes; solve $x^{2} + 10^{-8}x - 10^{-14} = 0$	$6.98$, never $8$

The last row is the standard trap. Below about $10^{-6}\ \mathrm{M}$ the ions from water can no longer be ignored, and an acid can never give pH above $7$.

[JEE Main] Remember the two-source calculation: $[\mathrm{H_3O^+}] = 10^{-8} + x$ with $K_w = (10^{-8}+x)x$, giving $[\mathrm{OH^-}] = 9.5 \times 10^{-8}$, pOH $= 7.02$, pH $= 6.98$.

Card 11 — Weak Acids and Bases

The constants

$$\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}, \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

$$\mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-}, \qquad K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$$

$$\mathrm{p}K_a = -\log K_a, \qquad \mathrm{p}K_b = -\log K_b$$

Larger $K_a$ means a stronger acid; larger $\mathrm{p}K_a$ means a weaker acid. The two run in opposite directions, and confusing them is the most expensive one-line error in this half of the chapter.

Ostwald's dilution law

For a weak electrolyte of initial concentration $c$ and degree of ionisation $\alpha$,

$$K_a = \frac{c\alpha^{2}}{1-\alpha}$$

When $\alpha \ll 1$, that is when $c/K_a > 400$ or ionisation is below about $5\%$,

$$\alpha = \sqrt{\frac{K_a}{c}}, \qquad [\mathrm{H_3O^+}] = c\alpha = \sqrt{K_a c}$$

$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{p}K_a - \log c\right)$$

and for a weak base

$$[\mathrm{OH^-}] = \sqrt{K_b c}, \qquad \mathrm{pOH} = \frac{1}{2}\left(\mathrm{p}K_b - \log c\right)$$

$\alpha$ rises on dilution — it varies as $1/\sqrt{c}$ — while $[\mathrm{H_3O^+}]$ falls. Dilute a weak acid tenfold and $\alpha$ rises by $\sqrt{10}$ but the pH rises by $0.5$.

The conjugate relation

$$K_a \times K_b = K_w = 10^{-14} \quad \text{at } 298\ \mathrm{K}, \qquad \mathrm{p}K_a + \mathrm{p}K_b = 14$$

This holds only for a conjugate pair — $\mathrm{CH_3COOH}$ with $\mathrm{CH_3COO^-}$, $\mathrm{NH_4^+}$ with $\mathrm{NH_3}$. It is not a relation between an unrelated acid and base.

Values worth carrying into the exam

Weak acid	$K_a$ at $298\ \mathrm{K}$	$\mathrm{p}K_a$
Hydrofluoric acid, $\mathrm{HF}$	$3.5 \times 10^{-4}$	$3.46$
Nitrous acid, $\mathrm{HNO_2}$	$4.5 \times 10^{-4}$	$3.35$
Formic acid, $\mathrm{HCOOH}$	$1.8 \times 10^{-4}$	$3.74$
Benzoic acid, $\mathrm{C_6H_5COOH}$	$6.5 \times 10^{-5}$	$4.19$
Acetic acid, $\mathrm{CH_3COOH}$	$1.74 \times 10^{-5}$	$4.76$
Niacin, $\mathrm{C_5H_4NCOOH}$	$1.5 \times 10^{-5}$	$4.82$
Hypochlorous acid, $\mathrm{HClO}$	$3.0 \times 10^{-8}$	$7.52$
Hydrocyanic acid, $\mathrm{HCN}$	$4.9 \times 10^{-10}$	$9.31$
Phenol, $\mathrm{C_6H_5OH}$	$1.3 \times 10^{-10}$	$9.89$

Weak base	$K_b$ at $298\ \mathrm{K}$	$\mathrm{p}K_b$
Dimethylamine, $\mathrm{(CH_3)_2NH}$	$5.4 \times 10^{-4}$	$3.27$
Triethylamine, $\mathrm{(C_2H_5)_3N}$	$6.45 \times 10^{-5}$	$4.19$
Ammonia, $\mathrm{NH_3}$	$1.77 \times 10^{-5}$	$4.75$
Pyridine, $\mathrm{C_5H_5N}$	$1.77 \times 10^{-9}$	$8.75$
Aniline, $\mathrm{C_6H_5NH_2}$	$4.27 \times 10^{-10}$	$9.37$

Polyprotic acids and acid strength

• $K_{a_1} > K_{a_2} > K_{a_3}$ always, by three to five orders of magnitude, because pulling a proton off an increasingly negative ion is harder. Use $K_{a_1}$ alone for the pH.
• Down a group, bond strength decides: $\mathrm{HF} \ll \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$, acid strength increasing.
• Across a period, polarity decides: $\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$, acid strength increasing.

[NEET] $\alpha = \sqrt{K_a/c}$ with a clean square root is one of the two or three calculations this chapter contributes to the paper.