Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
30 g of H2 and 24 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?
In the reaction N2 + 3H2 → 2NH3, 30 g of N2 reacts completely (the other reactants are in excess) and 25.53 g of NH3 is collected. Find the percent yield Y.
You need 500 mL of 0.3 mol/L solution and have a 0.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
100 mL of a 0.5 mol/L solution is mixed with 50 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
What volume V does 60 g of N₂ occupy at 600 kPa and 70 °C? Treat it as an ideal gas.
Find the density ρ of Ar at 950 kPa and 190 °C, treating it as an ideal gas.
A 250 g block of iron (c = 0.449 J/(g·°C)) at 150 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 50 g of C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 64. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.01 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A 1.00 L buffer contains 0.2 mol acetic acid and 0.15 mol of its conjugate base (Ka = 1.8 × 10⁻⁵). 0.025 mol of NaOH is added. Find the new pH pH.
A second-order reaction (−rA = k·CA²) has k = 0.4 L/(mol·min) and an initial concentration of 0.8 mol/L. Find its first half-life t½.
The rate constant of a reaction is 0.01 min⁻¹ at 40 °C and 0.171 min⁻¹ at 65 °C. Find the activation energy Ea.
Answer: m = 27.03 g
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: Y = 70.01 %
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: Vwater = 200 mL
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: c = 1.167 mol/L
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: V = 10.19 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 9.856 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 34.35 °C
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: q = 2475 kJ
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: [HI] = 0.32 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: [PCl5] = 0.3772 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: pH = 3.382
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Answer: pH = 4.745
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Answer: t½ = 3.125 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Each later half-life is twice as long, because C₀ has halved.
Answer: Ea = 99.98 kJ/mol
Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).
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