Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
10 g of N2 and 40 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?
In the reaction 2Al + 3Cl2 → 2AlCl3, 12 g of Al reacts completely (the other reactants are in excess) and 46.26 g of AlCl3 is collected. Find the percent yield Y.
You need 1000 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?
200 mL of a 0.4 mol/L solution is mixed with 250 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 30 L contains 110 g of Ar at 100 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of O₂ goes from 250 kPa, 10 L, 35 °C to 500 kPa and 190 °C. Find the new volume V2.
A 75 g block of zinc (c = 0.388 J/(g·°C)) at 170 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find ΔH° for the reaction C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l).
1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 16. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.2 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.2 mol/L solution of hydrofluoric acid (Ka = 0.00068).
A 1.00 L buffer contains 0.1 mol lactic acid and 0.1 mol of its conjugate base (Ka = 0.00014). 0.01 mol of NaOH is added. Find the new pH pH.
A second-order reaction (−rA = k·CA²) has k = 0.1 L/(mol·min) and an initial concentration of 0.5 mol/L. Find its first half-life t½.
A first-order reaction has k = 0.002 min⁻¹ at 25 °C and an activation energy of 50 kJ/mol. The batch reactor is now run at 55 °C for 10 min. Find the conversion X reached.
Answer: m = 12.16 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 = 78.01 %
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 = 400 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 = 0.3444 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: P = 284.7 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 7.515 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 20.17 °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: ΔH° = −1367 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.
Negative: the reaction gives off heat (exothermic).
Answer: [HI] = 1.333 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] = 0.6417 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 = 1.946
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 = 3.941
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 = 20 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: X = 11.88 %
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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