Course

A2 Chemistry: Energetics and Equilibria (9701)

Name: A2 Chemistry: Energetics and Equilibria (9701)
Price: 1499 BDT

Five constants that are really one, then the two quantities that decide whether a reaction happens at all. Derive every relation rather than recalling it.

22 lessons2h 23mEnglish

What you'll learn

Write a Kc or Kp expression with coefficients as indices, and derive its units by cancelling
Use Δn to predict the units of a constant, whether Kp equals Kc, and whether pressure shifts the position
Derive Kp = Kc(RT)^Δn from the ideal gas equation rather than recalling it
Build partial pressures from mole fractions and check that they sum to the total pressure
Use Q against K to predict the direction of any disturbance, without a memorised rule
State that only temperature changes an equilibrium constant, and say which way for a given enthalpy sign
Calculate K from initial amounts using an initial-change-equilibrium table with the change row in the coefficient ratio
Explain that neutral means equal ion concentrations, so neutral water above 25 °C has a pH below 7
Calculate pH for a strong acid, a weak acid and a strong alkali, choosing the route from what is given
Explain how a buffer resists added acid or alkali with an equation for the ion that reacts
Calculate a buffer pH, and use pH = pKa to choose a weak acid for a target pH
Build a solubility relation from a dissolving equation, including the factor of 4 for a 1:2 salt
Explain the common ion effect as a change of position rather than a change of constant
Define lattice energy with state symbols, and rank salts by charge product before radius
Assign a sign to every step of a Born-Haber cycle, including the endothermic second electron affinity
Calculate a lattice energy as the step that closes a cycle, doubling the steps a 1:2 salt repeats
Calculate an enthalpy of solution from lattice and hydration enthalpies, reversing the lattice term
Predict the sign of an entropy change by counting the moles of gas on each side
Calculate ΔG, converting the entropy from joules to kilojoules before substituting
Find the temperature at which a reaction becomes feasible, and say which sign pairs have no such temperature

Requirements

AS equilibria: dynamic equilibrium, Le Chatelier and the shape of a pH curve
AS stoichiometry: the mole, concentration, and the ideal gas equation
AS energetics: enthalpy changes, Hess cycles and the meaning of an exothermic reaction

Course content

22 lessons · 2h 23m

Equilibrium constants

Δn decides the units, whether Kp equals Kc, and whether pressure shifts anything. Then Q against K, and one table that handles every calculation.

0/4

Kc and where its units come from

A coefficient is an index, not a multiplier, and the units are derived by cancelling rather than recalled.

Preview8m 2.

Kp, and the link to Kc

Partial pressures, why solids are left out, and Kp = Kc(RT)^Δn straight out of the ideal gas equation.

Preview8m

What changes K, and what only shifts the position

Only temperature moves the constant. Q against K gives the direction of every other disturbance without a rule to memorise.

Calculating K from initial amounts

One three-row table, a change row in the coefficient ratio, and a reading habit for choosing between Kc and Kp.

25.1-25.2 · Acids, buffers and solubility

Kw, Ka and Ksp are the same constant in three settings. Then buffers from a rearranged Ka, and why a common ion moves the solubility and not the constant.

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Ka, Kw and pH

Neutral means equal ion concentrations, not pH 7. Then the four routes to a pH, and why the square root is for weak acids only.

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How we know an acid is weak

Strong and weak are measured, not given. A resistance reading separates them by the shape of one curve, and gives Ka for an acid too weak to titrate.

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Buffers

Large stores of a weak acid and its conjugate base, a rearranged Ka for the pH, and pH = pKa when the two are equal.

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Ksp and the common ion effect

A saturated solution is an equilibrium. Build the solubility relation from the equation, because a 1:2 salt gives 4s³.

One solute, two layers

Follow a solute between two immiscible liquids, calculate its partition coefficient and see why several small extractions beat one large extraction.

Equilibria: exam question

One structured Paper 4 question across a Kp expression, what moves a constant, a buffer pH and a solubility product.

23.1-23.2 · Lattice energy and Born-Haber cycles

A lattice energy cannot be measured, so it is the closing step of a cycle. Signs first, then the subtraction, then why a salt dissolves at all.

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Lattice energy

Gaseous ions into one mole of solid, so always exothermic. Then charge and radius, which decide how exothermic.

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Born-Haber cycles

Five measurable steps and one that is not. Assign every sign, then take the route from the enthalpy of formation.

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Enthalpy of solution and hydration

Breaking the lattice costs more than a thousand kilojoules. Hydration pays it back, and the difference is what a thermometer reads.

23.3-23.4 · Entropy and Gibbs free energy

Exothermic is not the criterion. Count the ways a state can be arranged, then combine entropy with enthalpy to decide what actually happens.

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Entropy

The number of ways a state can be arranged, not a vague disorder. Why a gas dwarfs a liquid, and why the units are joules.

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Predicting the entropy change

Count the moles of gas on each side. That one comparison settles the sign of almost every entropy change you are asked about.

Preview7m

Gibbs free energy

One number that settles whether a change can happen. Convert the entropy to kilojoules first, or the answer is out by a thousand.

Temperature and feasibility

Set ΔG to zero and solve for T. Four sign combinations, and only two of them depend on temperature at all.

Energetics: exam question

One structured Paper 4 question across a Born-Haber cycle, an enthalpy of solution, an entropy sign and a feasibility temperature.

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