🔥 AS 2: Further Physical & Inorganic Chemistry and Intro Organic
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CCEA GCE Chemistry (1110) · Unit AS 2: Further Physical and Inorganic Chemistry and an Introduction to Organic Chemistry
Mini-Lesson
AS 2: Further Physical & Inorganic Chemistry and Intro Organic
Unit AS 2 adds the energy and rate ideas to your inorganic knowledge, then opens the door to organic chemistry.
You will calculate enthalpy changes from calorimetry, from bond enthalpies and from Hess cycles; handle rates, the Boltzmann distribution and Kc; work through Groups 2 and 7; and meet your first organic mechanisms.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Energetics
Enthalpy and calorimetry
Exothermic: energy is released, ΔH is negative, and the products are lower in enthalpy than the reactants. Endothermic: ΔH is positive.
q = mcΔTm = mass of the solution (g) · c = 4.18 J g⁻¹ K⁻¹ · ΔT in K or °C
To convert to kJ mol⁻¹: convert q to kJ, then divide by the moles of the limiting reactant. Attach a negative sign if the temperature rose.
Why calorimetry always under-reads: heat is lost to the surroundings and to the apparatus, and the specific heat capacity of the container is ignored. These are systematic errors, so repeating the experiment will not fix them — better insulation will.
Calculate
Your turn — calculation 1
1100 cm³ of solution (mass 100 g) rises in temperature by 5.0 °C during a reaction in which 0.0100 mol of the limiting reactant is used up. Take c = 4.18 J g⁻¹ K⁻¹. Calculate ΔH in kJ mol⁻¹ (include the sign).
Hess's law: the enthalpy change of a reaction is the same whatever route is taken, as long as the start and end points are the same.
ΔHr = Σ ΔHf(products) − Σ ΔHf(reactants)enthalpy of formation of an element in its standard state = zero
Bond breaking is endothermic; bond making is exothermic. So:
ΔH = Σ(bonds broken) − Σ(bonds made)
Bond enthalpies are means — averages taken across many different compounds — and they apply to gases. That is why a value calculated this way rarely matches the experimental one exactly.
Calculate
Your turn — calculation 2
2Calculate ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g) using enthalpies of formation: C₂H₄ = +52.2, H₂ = 0, C₂H₆ = −84.7 kJ mol⁻¹. Include the sign; give your answer to 1 decimal place.
ΔH = (−84.7) − (+52.2) = −136.9 kJ mol⁻¹. Remember ΔHf of H₂, an element in its standard state, is zero.
Calculate
Your turn — calculation 3
3Use mean bond enthalpies to find ΔH for N₂(g) + 3H₂(g) → 2NH₃(g). N≡N = +945, H–H = +436, N–H = +391 kJ mol⁻¹. Include the sign.
kJ mol⁻¹
Hint: Broken: 945 + (3 × 436). Made: 6 × 391 (each NH₃ has three N–H bonds, and there are two NH₃).
Method
Broken = 945 + 1308 = 2253. Made = 6 × 391 = 2346. ΔH = 2253 − 2346 = −93 kJ mol⁻¹. The Haber process is exothermic.
Kinetics
Collision theory, Boltzmann and catalysts
Particles react only if they collide with energy at least equal to the activation energy, Ea, and in the right orientation.
The Maxwell–Boltzmann distribution shows the spread of molecular energies. Only the molecules to the right of Ea can react. The curve starts at the origin (no molecule has zero energy) and never touches the x-axis again.
Higher temperature — the peak moves right and lowers; a much larger fraction of molecules now exceed Ea. This is the dominant reason the rate rises so steeply, far outweighing the modest increase in collision frequency.
Higher concentration / pressure / surface area — more frequent collisions. The distribution curve is unchanged.
Catalyst — an alternative route with lower Ea. The curve is unchanged; Ea simply moves left, so more molecules can react.
Sort it
Magnesium + hydrochloric acid — what happens to the rate?
The acid is already in excess. Tap a change, then tap its effect on the rate of reaction.
📈 Increases the rate
📉 Decreases the rate
➖ No effect on the rate
Quick check
Think it through
?A student says: raising the temperature speeds up a reaction mainly because the particles collide more often. Why is this a poor answer?
Equilibrium
Le Chatelier and Kc
At dynamic equilibrium in a closed system, the forward and reverse reactions continue at equal rates, so concentrations remain constant.
Le Chatelier's principle: the system shifts to oppose any change imposed on it.
Raise the pressure → shifts to the side with fewer moles of gas.
Raise the temperature → shifts in the endothermic direction.
A catalyst changes neither the position of equilibrium nor Kc — it only shortens the time to reach equilibrium.
Kc = [products] ÷ [reactants], each raised to its coefficientKc is changed only by temperature
Calculate
Your turn — calculation 4
4For N₂O₄(g) ⇌ 2NO₂(g), a 2.00 dm³ vessel at equilibrium contains 0.0400 mol N₂O₄ and 0.0800 mol NO₂. Calculate Kc in mol dm⁻³, to 3 significant figures.
mol dm⁻³
Hint: First find concentrations: divide each amount by 2.00 dm³. Then Kc = [NO₂]² ÷ [N₂O₄].
Group 2 — reactivity increases down the group: the atomic radius and shielding both increase, so the two outer electrons are lost more easily.
Ca + 2H₂O → Ca(OH)₂ + H₂ (balanced: 1 Ca, 2 O, 4 H each side ✓)
Hydroxides get more soluble down the group; sulfates get less soluble down the group — which is why BaSO₄ is the white precipitate in the sulfate test.
Group 7 — the halogens get less reactive down the group as oxidising agents. A halogen higher in the group will displace a halide lower down:
Cl₂ + 2KBr → 2KCl + Br₂chlorine is the stronger oxidising agent, so it takes electrons from Br⁻
Halide tests with silver nitrate: Cl⁻ gives a white precipitate (dissolves in dilute NH₃); Br⁻ gives a cream precipitate (dissolves only in concentrated NH₃); I⁻ gives a yellow precipitate (insoluble even in concentrated NH₃).
Quick check
Think it through
?Bromine water is added to separate solutions of potassium chloride and potassium iodide. What is observed?
Introduction to organic
Alkanes, alkenes and their mechanisms
Alkanes (CnH2n+2) are saturated and unreactive. With a halogen and UV light they undergo free-radical substitution:
Termination — two radicals combine, e.g. •CH₃ + Cl• → CH₃Cl
Alkenes (CnH2n) have a C=C with an exposed π bond, so they are attacked by electrophiles — electrophilic addition.
Markovnikov, properly explained: propene + HBr gives mainly 2-bromopropane, because the reaction goes through the more stable secondary carbocation. Alkyl groups release electron density and spread the positive charge, so tertiary > secondary > primary in stability.
Test for a C=C: bromine water is decolourised from orange to colourless.
Quick check
Think it through
?Propene reacts with HBr. Which is the major product, and why?
Introduction to organic
Haloalkanes and alcohols
The C–X bond in a haloalkane is polar (Cδ+), so a nucleophile (an electron-pair donor) attacks the carbon — nucleophilic substitution.
+ aqueous KOH, warm → alcohol
+ hot ethanolic KOH → alkene (elimination — OH⁻ acts as a base, not a nucleophile)
+ ethanolic KCN, reflux → nitrile (adds a carbon to the chain)
Alcohols hydrogen-bond, so they have much higher boiling points than alkanes of similar Mr.
Primary + acidified dichromate: distil at once → aldehyde; reflux with excess → carboxylic acid.
Tertiary — not oxidised; the dichromate stays orange.
The single most-tested distinction in this unit: the same reagent (KOH) gives a completely different product depending on the solvent. Aqueous → nucleophilic substitution → alcohol. Hot and ethanolic → elimination → alkene.
Match it
Match the reagents to the product
Tap a set of reagents on the left, then the organic product on the right.
Reagents and conditions
Product
Calculate
Your turn — calculation 5
5Calculate the mass of carbon dioxide produced when 2.20 g of propane is completely burned: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. (M: C₃H₈ = 44.0, CO₂ = 44.0)
g
Hint: n(C₃H₈) = 2.20 ÷ 44.0 = 0.0500 mol. Ratio 1 : 3, so n(CO₂) = 0.150 mol. mass = n × M.
Method
n(C₃H₈) = 0.0500 mol → n(CO₂) = 3 × 0.0500 = 0.150 mol → mass = 0.150 × 44.0 = 6.60 g. (Check the equation balances: C 3 = 3 ✓, H 8 = 8 ✓, O 10 = 6 + 4 = 10 ✓.)
Quick check
Think it through
?Which alcohol is NOT oxidised by warm acidified potassium dichromate(VI)?
Quick check
Think it through
?A catalyst is added to an equilibrium mixture. Which statement is correct?