24 g of N2 and 8 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?
M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
2.StoichiometryHard
In the reaction N2 + 3H2 → 2NH3, 24 g of N2 reacts completely (the other reactants are in excess) and 18.09 g of NH3 is collected. Find the percent yield Y.
M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 18.09 g
Y = %
3.Molarity & dilutionMedium
You need 500 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?
c1 = 1 mol/Lc2 = 0.3 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
150 mL of a 0.5 mol/L solution is mixed with 400 mL of a 3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 0.5 mol/LV1 = 150 mLc2 = 3 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 20 L contains 140 g of CO₂ at 95 °C. Treat it as an ideal gas and find the pressure P.
A 200 g block of zinc (c = 0.388 J/(g·°C)) at 90 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 21 °C in an insulated cup. Find the final temperature Tf.
0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 4
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.25 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.25 mol/LKc = 0.5
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
c = 0.25 mol/LKa = 1.8 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L hydrofluoric acid and 0.5 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.5Ka = 0.00068
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.25 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 15 min.
k = 0.25 L/(mol·min)CA0 = 1 mol/Lt = 15 min
CA = mol/L
14.Reaction kineticsHard
The rate constant of a reaction is 0.002 min⁻¹ at 30 °C and 0.0056 min⁻¹ at 45 °C. Find the activation energy Ea.
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.
Moles of each reactant
n(N2) = 24/28.02 = 0.8565 mol
n(H2) = 8/2.016 = 3.968 mol
Limiting reactant
0.8565/1 = 0.8565, 3.968/3 = 1.323 → N2
Mass of product
n(NH3) = 0.8565·2 = 1.713 mol
m = 1.713·17.03 = 29.17 g
Answer
m = 29.17 g
2.Stoichiometry
Answer: Y = 62.01 %
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.
Theoretical yield
n(N2) = 24/28.02 = 0.8565 mol
m = 0.8565·2/1·17.03 = 29.17 g
Percent yield
Y = 18.09/29.17·100 = 62.01%
Answer
Y = 62.01%
3.Molarity & dilution
Answer: Vwater = 350 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.
Volume of stock solution
V1 = c2·V2/c1 = 0.3·500/1 = 150 mL
Water to add
Vwater = 500 − 150 = 350 mL
Answer
Vwater = 350 mL
4.Molarity & dilution
Answer: c = 2.318 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.
Total amount
n = 0.5·150 + 3·400 = 1275 mmol
Concentration of the mixture
c = 1275/550 = 2.318 mol/L
Answer
c = 2.318 mol/L
5.Ideal gas law
Answer: P = 486.8 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 140/44.01 = 3.181 mol
T = 95 + 273.15 = 368.1 K
Pressure
P = nRT/V = 3.181·8.314·368.1/0.02 = 486.8 kPa
Answer
P = 486.8 kPa
6.Ideal gas law
Answer: V2 = 14.63 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 308.1 K, T2 = 413.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 20·(300/550)·(413.1/308.1) = 14.63 L
Answer
V2 = 14.63 L
7.Calorimetry & enthalpy
Answer: Tf = 26.86 °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.
Heat balance
200·0.388·(90 − Tf) = 200·4.18·(Tf − 21)
Solve for the final temperature
Tf = (77.6·90 + 836·21)/(77.6 + 836) = 26.86 °C
Answer
Tf = 26.86 °C
8.Calorimetry & enthalpy
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).
9.Chemical equilibrium
Answer: [HI] = 0.5 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.
Equilibrium expression
4 = (2x)²/(0.5 − x)²
Take the square root
2 = 2x/(0.5 − x) → x = 0.5·2/(2 + 2) = 0.25 mol/L
Equilibrium concentration
[HI] = 2x = 0.5 mol/L
Answer
[HI] = 0.5 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.183 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.
Equilibrium expression
0.5 = x²/(0.25 − x) → x² + 0.5·x − 0.125 = 0
Solve the quadratic
x = (−0.5 + √(0.5² + 4·0.125))/2 = 0.183 mol/L
Equilibrium concentration
[Cl₂] = x = 0.183 mol/L
Answer
[Cl2] = 0.183 mol/L
11.pH & buffers
Answer: pH = 2.675
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]).
Equilibrium
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Ka = x²/(0.25 − x) → x = 0.002112 mol/L
pH
pH = −log(0.002112) = 2.675
Answer
pH = 2.675
12.pH & buffers
Answer: pH = 3.565
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]).
pKa
pKa = 3.167
Henderson–Hasselbalch
pH = 3.167 + log(0.5/0.2) = 3.565
Answer
pH = 3.565
13.Reaction kinetics
Answer: CA = 0.2105 mol/L
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).
Integrated second-order law
1/CA = 1/CA0 + kt = 1 + 0.25·15 = 4.75 L/mol
CA = 1/4.75 = 0.2105 mol/L
Answer
CA = 0.2105 mol/L
14.Reaction kinetics
Answer: Ea = 55.04 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).
Absolute temperatures
T1 = 303.1 K, T2 = 318.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.03/(-1.555e-4) = 55041 J/mol