Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
12 g of N2 and 10 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?
In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 4 g of C3H8 reacts completely (the other reactants are in excess) and 6.15 g of H2O is collected. Find the percent yield Y.
You need 200 mL of 0.3 mol/L solution and have a 1 mol/L stock solution. How much water Vwater do you add to the stock solution?
250 mL of a 1.5 mol/L solution is mixed with 500 mL of a 3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 5 L contains 180 g of N₂ at 70 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CH₄ at 500 kPa and 180 °C, treating it as an ideal gas.
A 250 g block of zinc (c = 0.388 J/(g·°C)) at 150 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find ΔH° for the reaction 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l).
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 36. Find the equilibrium concentration [H2].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.25 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
A 1.00 L buffer contains 0.2 mol hypochlorous acid and 0.25 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.01 mol of NaOH is added. Find the new pH pH.
In a first-order reaction the concentration of A falls from 0.5 mol/L to 0.2 mol/L in 20 min. Find the rate constant k.
A first-order reaction has k = 0.002 min⁻¹ at 30 °C and an activation energy of 50 kJ/mol. The batch reactor is now run at 55 °C for 45 min. Find the conversion X reached.
Answer: m = 14.59 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 = 94.07 %
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 = 140 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 = 2.5 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 = 3667 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 2.129 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 32.12 °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° = −3119 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: [H2] = 0.0625 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] = 0.04689 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 = 2.405
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 = 7.659
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: k = 0.04581 min⁻¹
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).
Answer: X = 33.5 %
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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