Equilibrium — JEE Mains Chemistry Formulas and Traps

Equilibrium Constant and Its Forms

Writing K from the equation

Law of mass action

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

Reversing, scaling and adding equations

Combining equilibria

K_{\text{reverse}}=\frac1K,\qquad K_{n\times}=K^n,\qquad K_{1+2}=K_1K_2

Kp and Kc through the change in gas moles

Kp and Kc

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

Common traps

A solid left in the expression

In

\mathrm{C(s)+CO_2(g)\rightleftharpoons 2CO(g)}

,

K_p=p_{\mathrm{CO}}^2/p_{\mathrm{CO_2}}

. Carbon does not appear. The same holds for

\mathrm{CaCO_3}

,

\mathrm{CaO}

, metals and liquid water in a reaction mixture.

Equal rates, not equal amounts

A concentration-time plot reaches equilibrium when every curve goes flat. The curves do not have to meet. A plot where the reactant falls to zero shows a reaction that went to completion, not an equilibrium.

Added equations multiply their K

For

\mathrm{X\rightleftharpoons Y}

,

\mathrm{Y\rightleftharpoons Z}

and

\mathrm{Z\rightleftharpoons W}

with

K=1,2,4

, the K for

\mathrm{X\rightleftharpoons W}

is

1\times2\times4=8

, not the sum 7.

Scaling is a power, not a factor

Dividing every coefficient by 3 turns K into

K^{1/3}

, not

K/3

. For

K=2.7\times10^{-5}

that is

3\times10^{-2}

. Cubing it, or taking a square root, are the offered wrong answers.

The sign of Δn

For

\mathrm{CO+\tfrac12O_2\rightleftharpoons CO_2}

,

\Delta n=1-\tfrac32=-\tfrac12

, so

K_p/K_c=1/\sqrt{RT}

. Writing reactants minus products gives

\sqrt{RT}

, which is always offered.

Reading Δn from a ratio

At 400 K,

RT\approx32.8

. If

K_p/K_c\approx33

, then

\Delta n=+1

; if it is about

1/33

,

\Delta n=-1

. Work out the ratio first, then match coefficients.

Equilibrium Composition from ICE Tables

Learn this subtopic in the notes

ICE tables in moles and concentration

Reaction quotient

Q=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\ \text{(any moment)},\qquad Q<K\Rightarrow\text{forward}

ICE tables in partial pressures

Partial pressure

p_i=\frac{n_iRT}{V}=x_i\,P

Common traps

The change row follows the coefficients

In

\mathrm{H_2+I_2\rightleftharpoons 2HI}

, forming 3 mol of HI uses 1.5 mol each of

\mathrm{H_2}

and

\mathrm{I_2}

, not 3 mol. Write x in front of each coefficient before filling the row.

Check the direction before you write the table

If all four species of

\mathrm{A+B\rightleftharpoons C+D}

start at 1 M and

K=100

, then

Q=1<K

, so C and D grow. A table that lets them shrink gives a root that leaves a negative amount.

The inert gas is in the total, not in K

Subtract the inert gas's moles before you find x, and leave it out of K. Using the total pressure as if it came only from the reacting gases gives the wrong x.

Carbon adds nothing to the pressure

In

\mathrm{C(s)+CO_2\rightleftharpoons 2CO}

, x of

\mathrm{CO_2}

gives 2x of CO, so the total rises by x, not by 2x.

Degree of Dissociation and Gibbs Energy

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Degree of dissociation and Kp

A ⇌ B + C

K_p=\frac{\alpha^2P}{1-\alpha^2}\qquad\Longleftrightarrow\qquad \alpha=\sqrt{\frac{K_p}{K_p+P}}

K, ΔG° and temperature

Gibbs energy and K

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

Common traps

Pressure pushes α down, not up

From

\alpha=\sqrt{K_p/(K_p+P)}

: when P is far larger than

K_p

, α is close to 0; when

K_p

is far larger than P, α is close to 1. The options often swap these two.

The fraction is K over the rest

For

\mathrm{A_xB_y}

,

\alpha=\left(\frac{K}{x^xy^yc^{x+y-1}}\right)^{1/(x+y)}

. The inverted fraction, or a power of

x+y

in place of the root, are the usual distractors.

K belongs to the equation as written

Doubling an equation squares K and doubles

\Delta G^\circ

. For HI decomposition,

\mathrm{HI\rightleftharpoons\tfrac12H_2+\tfrac12I_2}

and

\mathrm{2HI\rightleftharpoons H_2+I_2}

give answers that differ by a factor of 2. Use the equation the question writes.

Multiply the slope by 2.303

The slope of a

\log_{10}K

plot is

-\frac{\Delta H^\circ}{2.303R}

. A slope of

-100

gives

\frac{\Delta H^\circ}{R}=230.3

K, not 100.

Le Chatelier's Principle

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Finding the new equilibrium

Same K before and after

K=\frac{[\mathrm{B}]_{\text{old}}}{[\mathrm{A}]_{\text{old}}}=\frac{[\mathrm{B}]_{\text{new}}}{[\mathrm{A}]_{\text{new}}}\quad(\mathrm{A\rightleftharpoons B},\ \text{same }T)

Which way the equilibrium shifts

Change	Which way it shifts	Effect on K
Add a reactant gas or solute	Forward, using up some of it	None
Add a pure solid or liquid	No shift	None Adding $\mathrm{Fe_2O_3}$ to the blast-furnace equilibrium changes nothing.
Raise the pressure by compressing	Toward fewer gas moles	None
Compress when $\Delta n=0$ , as in $\mathrm{H_2+I_2\rightleftharpoons 2HI}$	No shift	None
Add an inert gas at constant volume	No shift	None
Add an inert gas at constant pressure	Toward more gas moles	None The volume grows, so every partial pressure falls, like a pressure drop.
Heat an exothermic forward reaction	Backward	K falls
Heat an endothermic forward reaction	Forward	K rises
Add a catalyst	No shift; equilibrium comes sooner	None
Raise the pressure on ice and water at 0 °C	Toward water, which takes less volume	Melting point falls

Only temperature changes K. Every other change moves the mixture while K stays fixed.

Common traps

Pressure moves the mixture, not K

Compressing

\mathrm{CO+3H_2\rightleftharpoons CH_4+H_2O}

raises every concentration and shifts it forward, but K is unchanged. "K increases because products increase" is the offered wrong statement.

Read how the inert gas is added

At constant volume the partial pressures do not change, so nothing moves. At constant pressure the volume grows and the reaction shifts toward more gas moles. If the question does not say, the standard answer assumes constant volume.

Start from the disturbed mixture

The Initial row of the new table holds the amounts just after the addition, not the original starting amounts and not the old equilibrium without the addition.

The change is only partly undone

An added species ends above its old equilibrium value. In the example, B was 0.8 M, rose to 1.0 M on adding, and settles at 0.933 M. An answer that returns B to 0.8 M or below is wrong.

pH of Acids and Bases

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Strong acids, bases and their mixtures

pH of a strong acid-base mixture

\mathrm{pH}=-\log\frac{n_{\mathrm{H^+}}-n_{\mathrm{OH^-}}}{V_{\text{total}}}

Weak acids and bases from Ka and Kb

Weak acid

[\mathrm{H^+}]=\sqrt{K_aC},\qquad \mathrm{pH}=\tfrac12\left(\mathrm{p}K_a-\log C\right)

Common traps

Sulphuric acid counts twice

0.01 M

\mathrm{H_2SO_4}

gives 0.02 M

\mathrm{H^+}

, and 0.01 M

\mathrm{Ca(OH)_2}

gives 0.02 M

\mathrm{OH^-}

. Forgetting the 2 is the most common wrong option in mixture questions.

Dilution does not cross 7

Diluting HCl to

10^{-8}

M does not give pH 8. An acid stays acidic; water's

10^{-7}

M of

\mathrm{H^+}

must be added in.

Hot water is neutral at pH below 7

Heating raises

K_w

, so both

[\mathrm{H^+}]

and

[\mathrm{OH^-}]

rise together. The pH falls, but the water is still neutral. "

\mathrm{H^+}

rises and

\mathrm{OH^-}

falls" is wrong.

Take the square root

[\mathrm{H^+}]

is

\sqrt{K_aC}

, not

K_aC

, and pH is half of

\mathrm{p}K_a-\log C

. Forgetting the half gives a pH twice too large.

In strong acid, divide by the strong acid's $\mathrm{H^+}$

A weak acid dissolved in 0.1 M HCl barely ionises. With

[\mathrm{H_2A}]=[\mathrm{H^+}]=0.1

M,

[\mathrm{HA^-}]=K_{a1}

. The option 0.1 M treats the weak acid as fully ionised.

Buffer Solutions

Learn this subtopic in the notes

The Henderson equation

Henderson equation

\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}

Buffers made by part-neutralisation

Weak acid part-neutralised by strong base

\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_b}{n_a-n_b}

Common traps

The ratio flips for a basic buffer

pOH uses salt over base:

\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\text{salt}]}{[\text{base}]}

. Written for pH it becomes

\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{base}]}{[\text{salt}]}

. Mixing the two forms gives a pH on the wrong side of

\mathrm{p}K_a

.

Salt over acid, not acid over salt

For a weak acid ionised to a fraction x, the salt form is x and the acid form is

1-x

, so

\mathrm{pH}-\mathrm{p}K_a=\log\frac{x}{1-x}

. The inverted fraction is always offered.

Base left is y minus x, not y

Mixing x mL of HCl with y mL of a weak base of the same molarity leaves salt x and base

y-x

. Putting

x/y

into the Henderson equation instead of

x/(y-x)

gives a wrong pair of volumes that is always among the options.

Excess strong acid is not a buffer

\mathrm{NH_4OH}

with an equal or larger amount of HCl leaves no free base. For a pH of

\mathrm{p}K_a(\mathrm{NH_4^+})=9.25

the mixture must hold equal amounts of

\mathrm{NH_3}

and

\mathrm{NH_4^+}

, so the base must be exactly twice the acid.

Salt Hydrolysis and Indicators

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pH of a salt solution

Salt of a weak acid and a strong base

\mathrm{pH}=7+\tfrac12\,\mathrm{p}K_a+\tfrac12\log C

Acid-base indicators and titrations

Indicator	Nature	Colour in acid, then in base	Works around	Used for
Phenolphthalein	Weak acid	Colourless (un-ionised HIn), then pink (ionised $\mathrm{In^-}$ )	pH 8.3 to 10	Weak acid vs strong base; strong acid vs strong base It does ionise in base; the pink colour is the ionised form.
Methyl orange	Weak base	Red quinonoid form, then yellow benzenoid form	pH 3.1 to 4.4	Strong acid vs weak base; strong acid vs strong base The quinonoid (acid) form is the more deeply coloured one.
Eriochrome black T	Metal-ion indicator	Wine red with $\mathrm{Ca^{2+}}$ or $\mathrm{Mg^{2+}}$ , then blue when free	pH about 10	EDTA titrations of $\mathrm{Ca^{2+}}$ and $\mathrm{Mg^{2+}}$
Diphenylamine	Redox indicator	Colourless when reduced, then violet when oxidised	A change in electrode potential, not pH	Redox titrations, such as $\mathrm{Fe^{2+}}$ with dichromate

Acid-base indicators respond to pH; redox indicators respond to electrode potential.

Common traps

Use the anion's concentration

Calcium lactate gives two lactate ions per formula, so a 0.005 M solution has 0.01 M lactate. Put 0.01 into the formula, not 0.005.

Exact neutralisation leaves a salt, not a buffer

Equal moles of a weak base and HCl leave only the salt, in the total volume. Use the hydrolysis formula, not the Henderson equation, and divide by the combined volume.

Phenolphthalein is an acid, and it ionises in base

Phenolphthalein is colourless in acid and pink in base because its ionised form is pink. "It does not dissociate in basic medium" is false.

Redox and acid-base indicators are not swapped

Acid-base indicators respond to pH; redox indicators respond to electrode potential. A statement pair that swaps the two makes both statements false.

Solubility Product

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Ksp and molar solubility

Solubility product

K_{sp}=x^x\,y^y\,s^{x+y}\qquad(\mathrm{A_xB_y})

Common ion and solubility

Solubility with a common ion

s=\frac{K_{sp}}{C^{\,n}}\qquad(n=\text{count of the common ion in the formula})

Will it precipitate, and in what order

Precipitation condition

Q>K_{sp}\ \Rightarrow\ \text{precipitate}

Common traps

Each ion is its count times s

In

\mathrm{Ag_2CrO_4}

,

[\mathrm{Ag^+}]=2s

, so

K_{sp}=(2s)^2s=4s^3

. Writing

s^2\cdot s

drops the factor 4.

Divide first, then take the root

From

K_{sp}=27s^4

,

s=\left(\frac{K_{sp}}{27}\right)^{1/4}

. Taking the root of

K_{sp}

alone, or a square root out of habit, gives the offered wrong answers.

Square the common ion for AB₂

For

\mathrm{Zn(OH)_2}

in NaOH,

s=\frac{K_{sp}}{[\mathrm{OH^-}]^2}

. Dividing by

[\mathrm{OH^-}]

once gives an answer too large by a factor of

1/[\mathrm{OH^-}]

.

Convert to grams only at the end

Find s in mol/L first, then multiply by the molar mass if the answer is asked in g/L. Using a mass concentration inside

K_{sp}

is wrong.

Mixing halves both concentrations

When equal volumes of two solutions are mixed, each ion is at half its original concentration. Using the original values makes Q four times too large for an AB salt, and eight times for

\mathrm{AY_2}

.

The smaller Ksp is not always first

For salts of different types, compare the reagent concentration each one needs, not the Ksp values. A hydroxide with a larger Ksp can still start first.

Equilibrium formulas

Equilibrium Constant and Its Forms

Writing K from the equation

Reversing, scaling and adding equations

Kp and Kc through the change in gas moles

Common traps

Equilibrium Composition from ICE Tables

ICE tables in moles and concentration

ICE tables in partial pressures

Common traps

Degree of Dissociation and Gibbs Energy

Degree of dissociation and Kp

K, ΔG° and temperature

Common traps

Le Chatelier's Principle

Finding the new equilibrium

Which way the equilibrium shifts

Common traps

pH of Acids and Bases

Strong acids, bases and their mixtures

Weak acids and bases from Ka and Kb

Common traps

Buffer Solutions

The Henderson equation

Buffers made by part-neutralisation

Common traps

Salt Hydrolysis and Indicators

pH of a salt solution

Acid-base indicators and titrations

Common traps

Solubility Product

Ksp and molar solubility

Common ion and solubility

Will it precipitate, and in what order

Common traps

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