Theoretical Chemistry: ΔH and mol

The molecular mass of CO₂ is 12+16×2=44, so 0.50 mol has a mass of 44×0.50=22 g. The answer 44 g is a trap for anyone who confuses this with 1 mol. Apply “mass ＝ molar mass × amount of substance” mechanically.

At standard state 1 mol of any gas occupies about 22.4 L, so 11.2÷22.4=0.50 mol. Dividing the other way round, 22.4÷11.2=2.0, is the classic slip. Remember the direction of the division: “volume ÷ 22.4”.

Molar concentration × volume (L) ＝ amount of substance, so 0.20×0.500=0.10 mol. The typical mistake is to forget to convert 500 mL into 0.5 L and answer 0.20 mol. The knack is to do the mL→L conversion first of all.

1 L ＝ 1000 cm³ of the solution has a mass of 1800 g; 98% of this is H₂SO₄, that is 1764 g, and dividing by the molecular mass 98 gives 18 mol/L. Forgetting to multiply by the density leads to wrong answers such as 1.8. Solve it in three steps: “take 1 L → mass of the solution → mass of the solute → mol”.

From PV=nRT, n=PV/RT=(8.3×10⁴×3.0)/(8.3×10³×300)=0.10 mol. If you substitute 27 as it is, the order of magnitude goes wrong. The iron rule is to convert to the absolute temperature T=27+273=300 K.

One water molecule contains 2 H atoms, so 6.0×10²³×2=1.2×10²⁴ atoms. The value 6.0×10²³ is a trap: that is the number of molecules. Never forget “number of molecules × number of atoms in one molecule”.

Take the weighted average using the mole fractions: 28×0.80＋32×0.20＝22.4＋6.4＝28.8. The simple average 30.0 is the biggest trap. Remembering that air has an average molecular mass of about 29 gives you a handy check.

In a body-centred cubic lattice the atom at the centre touches the 8 atoms at the corners of the cube, so the coordination number is 8. The value 12 is the coordination number of the face-centred cubic and hexagonal close-packed structures, which is easy to confuse. Memorise them as a pair: “body-centred ＝ 8, face-centred and hexagonal ＝ 12”.

8 corners×1/8＋6 face centres×1/2＝1＋3＝4 atoms. If you count the corner or face atoms as whole atoms you get 8 or 6. Always use the shares belonging to the unit cell (corner 1/8, face 1/2, edge 1/4).

In graphite one of the four valence electrons of each carbon atom can move freely through the planar structure, so it conducts electricity. Diamond is made of the same element, but all of its electrons are used in bonding, so it is an insulator. Remember that “graphite alone is the exception among covalent crystals”.

The H bonded to O, which has a large electronegativity, forms a hydrogen bond with the O of a neighbouring molecule, so a large amount of energy is needed to separate them. The strength of a covalent bond is a different matter from the boiling point (which is about pulling molecules apart), so do not be caught out. Remember: “H attached to F, O or N means hydrogen bonding”.

The bond formed when the lone pair of N is donated one-sidedly to H⁺ is a coordinate bond. It is also frequently asked that, once formed, it cannot be distinguished at all from the other N–H covalent bonds. Fix it as “a covalent bond in which one side supplies the electron pair ＝ coordinate bond”.

In a face-centred cubic lattice the atoms touch along the diagonal of a face, and the length of that diagonal, √2a, is equal to four radii. √3a is the relation for the body-centred cubic lattice (contact along the body diagonal of the cube) and is the most frequent trap. Learn it from a diagram: “face-centred uses the face diagonal, body-centred the body diagonal”.

In an exothermic reaction the system releases energy and the enthalpy of the system decreases, so ΔH is negative. In the thermochemical equations of the old curriculum heat release was written as ＋Q, so the sign was the opposite — this is the biggest point of confusion. Remember: “it goes out, so it is minus”.

Since 891 kJ is released per 1 mol, 0.50 mol releases 891×0.50≒446 kJ. Note that even though ΔH is negative, the “quantity of heat released” is answered as a positive value. Handle it with “quantity of heat ＝ |ΔH| × mol”.

The target reaction is (equation 1)−(equation 2), so ΔH＝−394−(−283)＝−111 kJ. Adding the two and getting −677 is the typical mistake. Keep in mind the operation of Hess's law: “subtract the equations so that CO₂ cancels out”.

A catalyst provides a different path with a lower activation energy, increasing the proportion of molecules that can get over the pass. The standard trap is ΔH (the difference between the starting point and the finishing point), which a catalyst does not change. Picture it as “a catalyst lowers the pass but does not change the depth of the valley”.

As the temperature rises the distribution of molecular kinetic energies spreads towards higher energies, and the proportion of molecules able to cross the pass increases sharply. The activation energy itself is not changed by temperature (it is a catalyst that changes it). Keep them apart: “temperature → the proportion increases; catalyst → the pass is lowered”.

The dissolution of ammonium nitrate takes heat from the surroundings — an endothermic change that is used in instant cold packs. Combustion, neutralisation and the oxidation of metals are all typical exothermic reactions. Memorise them as a set: “cold pack ＝ dissolving NH₄NO₃ ＝ ΔH＞0”.

When the pressure is raised, the equilibrium shifts in the direction in which the total number of gas molecules decreases (4 molecules on the left → 2 molecules on the right), so as to lower the pressure again. Note that changing the pressure does not change the equilibrium constant. Judge it with “Le Chatelier ＝ the direction that cancels the change”.

On heating, the equilibrium shifts in the endothermic direction (the reverse reaction, to the left), absorbing heat so as to cancel the rise in temperature. It is not “to the right because a higher temperature makes it go faster”. The knack is to separate the rate of reaction from the shift of the equilibrium: they are different questions.

A catalyst speeds up the forward and the reverse reaction by the same factor, so neither the position of the equilibrium nor the equilibrium constant changes. Thinking that “a catalyst increases the yield of NH₃” is the typical error. Remember that the only advantage of a catalyst is that equilibrium is reached sooner.

The equilibrium constant is “products ／ reactants”, with each concentration raised to the power of its coefficient. Multiplying by the coefficients, or forgetting the exponents, are the standard mistakes. Build the expression while chanting “the coefficients ride on the shoulders” (they become exponents).

Since 1.6 mol of HI was formed, 0.80 mol each of H₂ and I₂ reacted, leaving 0.20 mol of each. K＝(1.6)²／(0.20×0.20)＝2.56／0.04＝64 (in this reaction the numbers of molecules on the two sides are equal, so the volume cancels out). Making a table of the amount reacted, x, is the royal road to equilibrium calculations; taking HI＝2x to be x gives answers such as 16.

[H⁺]＝concentration×degree of ionisation＝0.10×0.010＝1.0×10⁻³ mol/L, so pH＝3. The biggest trap is to ignore the degree of ionisation and treat it as a strong acid, giving pH＝1. Use them separately: “for a weak acid cα, for a strong acid the concentration itself”.

If pH＝12, then [H⁺]＝10⁻¹² mol/L, and from the ionic product [OH⁻]＝10⁻¹⁴÷10⁻¹²＝10⁻² mol/L. Answering 10⁻¹², the value of [H⁺] itself, is the trap. If you use “pH＋pOH＝14” you get pOH＝2 immediately.

Taking H as ＋1 and O as −2, (＋1)×2＋S＋(−2)×4＝0 gives S＝＋6. It is easily confused with ＋4 in SO₂ or −2 in H₂S. The basic method is to work backwards from “the sum of the oxidation numbers in a compound ＝ 0”.

Taking O as −2, Mn＋(−2)×4＝−1 gives Mn＝＋7. With an ion, many students forget to make the sum equal to the charge of the ion and drift towards ＋8. Learn it with the reason attached: KMnO₄ is such a powerful oxidising agent because Mn is at its highest oxidation number, ＋7.

Only metals whose ionisation tendency is greater than that of H₂ react with dilute acids and give off hydrogen. Zn lies to the left of H and Cu to the right, so Cu does not dissolve. Always fix the position of H in the ionisation series (K＞Ca＞Na＞Mg＞Al＞Zn＞Fe＞Ni＞Sn＞Pb＞(H₂)＞Cu＞Hg＞Ag＞Pt＞Au).

Zn, which has the greater ionisation tendency, dissolves as Zn²⁺ (oxidation) and releases electrons, so it is the negative electrode. The correspondence “the electrode from which the electrons flow out ＝ the negative electrode ＝ oxidation” is easily muddled. Remember: “the metal with the greater ionisation tendency is the negative electrode”.

During discharge lead(II) sulfate PbSO₄ is formed and sticks to the surface of both electrodes, so both of them gain mass. The trap is to think “it dissolves, so it loses mass”: in fact Pb²⁺ does not go out into the solution but turns into PbSO₄ on the spot. Memorise as a set: “discharge ＝ PbSO₄ on both electrodes, so both gain mass, and the dilute sulfuric acid becomes more dilute”.

The ionisation tendency of Na⁺ is too great for it to be reduced, so water is reduced instead and H₂ is given off. Thinking “it is the cathode, so Na is deposited” is the typical error. Remember: “in an aqueous solution of a cation with a large ionisation tendency, hydrogen is given off”.

SO₄²⁻ is hard to oxidise, so water is oxidised instead and O₂ is given off. The deposition of copper is the change at the cathode, so do not confuse them. The rule for deciding is “at the anode: Cl₂ if Cl⁻ is present, otherwise O₂ (when the electrodes are Pt or C)”.

The quantity of electricity is 0.50×1930＝965 C, so the electrons amount to 965÷96500＝0.010 mol. From Cu²⁺＋2e⁻→Cu the copper is half of the electrons, 0.0050 mol, and its mass is 0.0050×64＝0.32 g. Forgetting the coefficient “2e⁻ for 1 Cu” and answering 0.64 g is the most frequent mistake.

MnO₄⁻ amounts to 0.020×0.010＝2.0×10⁻⁴ mol. Making the electrons given and received equal gives MnO₄⁻ : H₂O₂＝2 : 5, so 2.0×10⁻⁴×5/2＝5.0×10⁻⁴ mol. Assuming that the mole ratio is 1:1 is the trap; the iron rule is to fix the coefficients from “electrons accepted ＝ electrons released”.

NH₃ is lighter than air and dissolves extremely well in water, so it is collected by upward displacement of air. Collection over water is only for gases that dissolve very little in water, so it cannot be used for NH₃. Decide with the three categories: “light → upward displacement, heavy → downward displacement, slightly soluble → over water”.

Na gives a vivid yellow, which is why it has long been used for lighting in road tunnels. Red belongs to Li or Sr, blue-green to Cu and reddish purple to K, and these are easy to confuse. Learn the whole set together (Li red, Na yellow, K reddish purple, Cu blue-green, Ca orange-red, Sr red, Ba yellow-green).

Cl₂ is yellow-green with a pungent smell and is used to disinfect tap water. NH₃ and H₂S are colourless (with a pungent smell and a rotten-egg smell), and NO₂ is reddish brown, so they can be told apart by colour. It is a real strength to remember that “the only coloured gases are roughly Cl₂ (yellow-green), NO₂ (reddish brown) and O₃ (pale blue)”.

Cu²⁺ is the representative ion that forms a black precipitate of CuS even in acid. Zn²⁺ precipitates ZnS (white) only under neutral to basic conditions, and this difference in conditions is a frequent trap. Organise it as “precipitates even in acid ＝ metals with a small ionisation tendency (Cu, Ag, Pb and so on)”.

Oxidising power (the power to take electrons away) is stronger the higher up the periodic table, so F₂ is the strongest. This is why passing Cl₂ into aqueous KBr liberates Br₂ (the reverse does not happen). Learn it together with the experiment: “a halogen higher up drives out the halide ion of one lower down”.

A salt of a weak base reacts with a strong base, and NH₃, the weak base, is driven out and given off (the standard laboratory preparation). Choosing chlorine or HCl just because the name contains “chloride” is the trap. “The weak one is driven out by the strong one” is the principle behind these gas preparations.

Ag⁺ forms a white precipitate of AgCl with Cl⁻, and because it is photosensitive it is decomposed by light and darkens (the principle of photography). Cu²⁺ and Fe³⁺ do not precipitate with hydrochloric acid. Memorise as a set: “the ions that precipitate with Cl⁻ are Ag⁺ and Pb²⁺”.

Fe, Ni and Al become passive in concentrated nitric acid and the reaction stops. Thinking “nitric acid is a strong oxidising agent, so iron should dissolve too” is the trap; copper and silver, on the contrary, do dissolve in concentrated nitric acid. Learn the three passive metals as one set: Fe, Ni, Al.

For NH₃, a basic gas, use soda lime, a basic drying agent. Concentrated sulfuric acid and tetraphosphorus decaoxide are acidic and would react with NH₃, and even the neutral calcium chloride cannot be used because it exceptionally forms an addition compound with NH₃. Remember: “the drying agent and the gas must be a combination that does not quarrel as acid and base, and CaCl₂ × NH₃ is the exception”.