Concours CCINP - independent subject practice: Physical chemistry, amounts, reactions and equilibria

Chemical Reactions and Stoichiometry

This independent foundation chapter develops the mole as the currency of chemistry: balancing equations, identifying the limiting reagent, calculating yields and atom economy, and handling concentrations, dilutions and gas volumes confidently.

Core relationships

n = m/M; n = cV with V in dm3; n = V/24.0 for a gas at room temperature and pressure, using the value given in the data provided.
One mole contains approximately 6.02 × 10^23 specified entities.
The balanced equation supplies the molar ratio; never use the ratio from an unbalanced equation.
For solutions, c in mol dm−3 = mol/dm3; convert cm3 to dm3 by dividing by 1000.

Limiting reagents

Convert every reactant to moles, divide each by its coefficient, and the smallest quotient is limiting.
Product amount is fixed by the limiting reagent; the excess reagent remains unreacted and can be calculated by subtraction.

Yield and atom economy

Percentage yield = actual mass or moles ÷ theoretical × 100; it measures how much of the expected product was obtained.
Atom economy = stoichiometrically weighted formula mass of the desired product ÷ total stoichiometrically weighted formula masses of the reactants × 100; it measures how much reactant mass ends up in the desired product, independent of how well the reaction runs.

Dilution and titration

Dilution: c1V1 = c2V2 because moles of solute are unchanged.
Titration: moles from cV, then use the equation ratio to reach the unknown.

Worked example

4.60 g of sodium (0.200 mol) reacts with 8.00 g of chlorine (0.113 mol) in 2Na + Cl2 → 2NaCl. The sodium needs 0.100 mol of chlorine, which is available, so sodium is limiting. n(NaCl) = 0.200 mol, so m = 0.200 × 58.5 = 11.7 g. The chlorine is in excess by 0.013 mol.

Traps that cost marks

Forgetting to balance before applying a mole ratio.
Quoting product mass from the excess reagent, which always overestimates the yield.
Adding masses or volumes of solutions instead of moles when deciding which is limiting.
Omitting water of crystallisation from a hydrated molar mass.
Treating atom economy and percentage yield as the same quantity.

Thermochemistry and Energetics

This independent foundation chapter develops enthalpy change: sign conventions, calorimetry calculations, Hess's law cycles, standard enthalpy of formation and combustion data, mean bond enthalpies, and the entropy and Gibbs free energy link to whether a reaction is feasible.

Enthalpy basics

ΔH is negative for exothermic reactions and positive for endothermic ones; standard pressure is 100 kPa; a temperature must be specified, commonly 298.15 K for tabulated thermochemical data. Standard state does not itself fix one universal temperature.
ΔHf° for an element in its standard state is zero by definition.
ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants). For combustion data the sign convention reverses: ΔH = ΣΔHc(reactants) − ΣΔHc(products).

Calorimetry

q = mcΔT; for dilute aqueous solutions assume density 1.00 g cm−3 and c = 4.18 J g−1 K−1, so the mass is the total mass of solution.
ΔH per mole = −q/n, converted to kJ mol−1.

Hess's law and bond enthalpies

Any route between the same initial and final states gives the same ΔH; construct cycles, including Born–Haber cycles for ionic compounds.
ΔH = Σ bonds broken − Σ bonds formed. Mean bond enthalpies are averages over many compounds, so answers differ slightly from formation-data values.

Feasibility

ΔG = ΔH − TΔS; at specified constant temperature and pressure, negative actual reaction ΔG indicates a thermodynamically favorable forward change. It does not guarantee a measurable rate; ΔG° and actual ΔG are not interchangeable away from standard-state conditions. Entropy rises with more moles of gas, higher temperature and greater disorder; a reaction with positive ΔH can become feasible at high T if ΔS is positive.

Worked example

50.0 cm3 of 1.00 M HCl is mixed with 50.0 cm3 of 1.00 M NaOH and the temperature rises by 6.8 K. q = 100 g × 4.18 × 6.8 = 2842 J. n = 0.0500 mol of water formed. ΔH = −2.842/0.0500 = −56.8 kJ mol−1; the negative sign follows from the temperature rise.

Traps that cost marks

Omitting the minus sign, or using only one solution's mass in a neutralisation calculation.
Expecting the textbook value exactly; heat losses make the measured value less exothermic.
Getting the bond enthalpy sign the wrong way round.
Assigning a non-zero ΔHf° to an element.
Forgetting that ΔG, not ΔH alone, decides feasibility.

Chemical Kinetics and Equilibrium

This independent foundation chapter develops how reaction rates are measured and explained, how rate laws are deduced from experiment, and how equilibrium constants and Le Chatelier's principle predict the response of a system to changing conditions.

Rates and orders

Rate is the change in concentration per unit time, with units mol dm−3 s−1; for a reactant it is quoted as a positive value from the decrease.
Order is experimental and is not taken from the stoichiometric coefficients. Order with respect to a reactant is found by varying one concentration at a time.
Zero order: rate constant, so a concentration–time plot is linear. First order: constant half-life. For the simple integrated rate law −d[A]/dt = k[A]², 1/[A] against time is linear; an arbitrary overall second-order multireactant law needs its own analysis.

Collision theory and Arrhenius

A reaction occurs when collisions have energy at least equal to Ea and correct orientation. Increasing temperature often raises the fraction of collisions able to cross an activation barrier. Concentration and exposed surface area can affect encounter rates where the mechanism permits; their effects are not universally positive or explained by a higher energetic fraction.
k = Ae−Ea/RT; a plot of ln k against 1/T has gradient −Ea/R. Catalysts lower Ea.

Equilibrium

Equilibrium is dynamic: forward and reverse rates are equal and concentrations remain constant, but they are not equal to each other.
Kc = [products]^coefficients/[reactants]^coefficients, omitting pure solids and liquids.
K changes only with temperature. A catalyst supplies an alternative kinetic pathway and speeds attainment of equilibrium without changing the thermodynamic equilibrium constant or equilibrium composition.
Le Chatelier: raising concentration of a reactant drives the forward reaction; increasing pressure favours the side with fewer moles of gas; increasing temperature favours the endothermic direction.

Worked example

Doubling [A] while [B] is fixed quadruples the rate, so the reaction is second order in A; doubling [B] doubles the rate, so it is first order in B. Rate = k[A]²[B]. When rate = 2.0 × 10−3 mol dm−3 s−1 at [A] = [B] = 0.10 mol dm−3, k = 2.0 × 10−3/(0.0100 × 0.10) = 2.0 dm6 mol−2 s−1.

Traps that cost marks

Deriving the rate law from the balanced equation rather than from data.
Giving k units that do not match the overall order.
Claiming a catalyst increases yield or shifts equilibrium.
Adding an inert gas at constant volume and expecting a pressure shift; partial pressures of the reactants are unchanged.
Forgetting that K is unaffected by pressure and concentration changes.

Acids, Bases, and Salt Solutions

This independent foundation chapter develops Brønsted–Lowry theory, the distinction between strong and weak acids, pH and pKa calculations, buffer behaviour, titration curves and the hydrolysis that decides whether a salt solution is acidic, neutral or alkaline.

Theory

An acid donates a proton, a base accepts one; a conjugate pair differs by a single H+.
The first dissociation of H2SO4 and the usual aqueous dissociation of acids such as HCl, HNO3 and HClO4 are treated as strong in the relevant dilute-solution model; the second dissociation of H2SO4 requires a separate equilibrium calculation when significant; weak acids such as ethanoic acid (Ka ≈ 1.8 × 10−5) dissociate only partially.
Kw = [H+][OH−] = 1.0 × 10−14 at 298 K, so pH + pOH = 14 at that temperature.

Calculations

For a dilute fully dissociated monoprotic acid dominating over water autoionisation, [H+] ≈ c and pH ≈ −log c. Within that model, tenfold dilution raises pH by about one.
Weak acid: Ka = [H+]²/(c − [H+]), simplified to [H+]²/c when dissociation is under about 5%.
Buffer: pH = pKa + log([A−]/[HA]); at half neutralisation pH = pKa.
pKa = −log Ka; a smaller pKa means a stronger acid.

Salts and indicators

Strong acid + strong base gives a neutral salt; weak acid + strong base gives an alkaline solution because the anion hydrolyses; strong acid + weak base gives an acidic solution.
An indicator’s transition should lie within a sufficiently steep portion of the actual titration curve so its endpoint approximates equivalence to the required precision. Phenolphthalein and methyl orange are conventional examples, but suitability depends on the specific curve; merely naming acid/base classes is not enough.

Worked example

0.100 M ethanoic acid with Ka = 1.8 × 10−5: x²/0.100 = 1.8 × 10−5, so x = [H+] = 1.34 × 10−3 M and pH = −log(1.34 × 10−3) = 2.87. Only about 1.3% is dissociated, confirming weak behaviour and justifying the approximation.

Traps that cost marks

Using pH + pOH = 14 at temperatures other than 298 K.
Assuming the equivalence point of a weak acid–strong base titration is pH 7; it is above 7.
Inverting the buffer ratio, or forgetting that dilution hardly changes the pH of a buffer.
Ignoring the contribution of water autoionisation for very dilute or very weak acids.
Confusing strength (degree of dissociation) with concentration (amount per dm3).

Electrochemistry and Redox Reactions

This independent foundation chapter develops oxidation numbers, half-equation balancing in acidic and alkaline media, galvanic cell predictions from standard electrode potentials, and the quantitative link between charge, moles of electrons and mass deposited in electrolysis.

Oxidation numbers

A free element has oxidation number zero; a monatomic ion equals its charge; oxygen is usually −2 but −1 in peroxides; hydrogen is +1 except in hydrides; the sum equals the overall charge.
Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease.
Disproportionation is simultaneous oxidation and reduction of the same element, as in Cl2 with cold dilute NaOH.

Balancing redox equations

Split into half-equations, balance atoms other than O and H, balance O with H2O, balance H with H+, and balance charge with electrons.
In alkaline solution, add OH− to both sides to neutralise H+, then cancel water.
Multiply each half-equation so the electron counts match before adding.

Cells and potentials

E°cell = E°cathode − E°anode; a positive standard cell potential corresponds to negative standard reaction Gibbs energy as written; actual composition and conditions can alter the actual cell potential.
E° is intensive: do not multiply it when you double a half-equation. ΔG° = −nFE° with F = 96 485 C mol−1.
Oxidation occurs at the anode. In a galvanic cell the anode is the negative electrode; in an electrolytic cell it is the positive one.

Worked example

Silver is deposited from AgNO3 solution using a current of 2.00 A for 30.0 min. Q = It = 2.00 × 1800 = 3600 C. n(e−) = 3600/96 485 = 0.0373 mol. Since Ag+ + e− → Ag, n(Ag) = 0.0373 mol and m = 0.0373 × 107.9 = 4.03 g.

Traps that cost marks

Reversing the sign conventions for galvanic and electrolytic cells.
Multiplying E° values when balancing a cell reaction.
Forgetting to balance charge with electrons before adding half-equations.
Assigning oxygen −2 in peroxides or hydrogen +1 in metal hydrides.
Mixing seconds and minutes in Q = It, or using the wrong number of electrons per ion.