Comprehensive Inorganic Chemistry Exam Guide: Equilibria, Thermodynamics, Kinetics, and Coordination Complexes

Aqueous Equilibria and Precipitation of Barium Fluoride

  • Initial Solution Parameters and Mixing Conditions:

    • Solution A contains barium nitrate: VA=50 mL=0.050 LV_A = 50\,\text{mL} = 0.050\,\text{L} of Ba(NO3)2\text{Ba}(\text{NO}_3)_2 at a concentration of CA=0.1 MC_A = 0.1\,\text{M}.

    • Solution B contains sodium fluoride: VB=70 mL=0.070 LV_B = 70\,\text{mL} = 0.070\,\text{L} at a concentration of 0.08 M0.08\,\text{M}.

    • Total volume upon mixing (assuming volume additivity): Vtotal=VA+VB=50 mL+70 mL=120 mL=0.120 LV_{\text{total}} = V_A + V_B = 50\,\text{mL} + 70\,\text{mL} = 120\,\text{mL} = 0.120\,\text{L}.

  • Initial Molar Quantities Before Precipitation:

    • Barium ions (Ba2+\text{Ba}^{2+}):

    • n(Ba2+)=CA×VA=0.1 mol/L×0.050 L=0.0050 mol=5.0 mmoln(\text{Ba}^{2+}) = C_A \times V_A = 0.1\,\text{mol/L} \times 0.050\,\text{L} = 0.0050\,\text{mol} = 5.0\,\text{mmol}

    • Nitrate ions (NO3−\text{NO}_3^-):

    • Dissociation: Ba(NO3)2(aq)→Ba2+(aq)+2 NO3−(aq)\text{Ba}(\text{NO}_3)_2(aq) \rightarrow \text{Ba}^{2+}(aq) + 2\,\text{NO}_3^-(aq)

    • n(NO3−)=2×n(Ba2+)=2×5.0 mmol=10.0 mmol=0.0100 moln(\text{NO}_3^-) = 2 \times n(\text{Ba}^{2+}) = 2 \times 5.0\,\text{mmol} = 10.0\,\text{mmol} = 0.0100\,\text{mol}

    • Sodium ions (Na+\text{Na}^+):

    • For standard sodium fluoride (NaF\text{NaF}):

    • n(Na+)=0.08 mol/L×0.070 L=0.0056 mol=5.6 mmoln(\text{Na}^+) = 0.08\,\text{mol/L} \times 0.070\,\text{L} = 0.0056\,\text{mol} = 5.6\,\text{mmol}

    • Fluoride ions (F−\text{F}^-):

    • Under standard stoichiometry (NaF→Na++F−\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-):

    • n(F−)=5.6 mmol=0.0056 moln(\text{F}^-) = 5.6\,\text{mmol} = 0.0056\,\text{mol}

    • If formulated strictly as written (NaF2\text{NaF}_2 yielding 2 F−2\,\text{F}^- per formula unit):

    • n(F−)=2×5.6 mmol=11.2 mmol=0.0112 moln(\text{F}^-) = 2 \times 5.6\,\text{mmol} = 11.2\,\text{mmol} = 0.0112\,\text{mol}

  • Concentrations of Spectator Ions at Equilibrium:

    • Neither sodium nor nitrate participates in precipitation equilibria; their concentrations depend solely on dilution.

    • Concentration of nitrate ions:

    • [NO3−]=n(NO3−)Vtotal=0.0100 mol0.120 L=0.0833 M[\text{NO}_3^-] = \frac{n(\text{NO}_3^-)}{V_{\text{total}}} = \frac{0.0100\,\text{mol}}{0.120\,\text{L}} = 0.0833\,\text{M}

    • Concentration of sodium ions:

    • [Na+]=n(Na+)Vtotal=0.0056 mol0.120 L=0.0467 M[\text{Na}^+] = \frac{n(\text{Na}^+)}{V_{\text{total}}} = \frac{0.0056\,\text{mol}}{0.120\,\text{L}} = 0.0467\,\text{M}

  • Precipitation Equilibrium of Barium Fluoride:

    • Precipitation reaction:

    • Ba2+(aq)+2 F−(aq)⇌BaF2(s)\text{Ba}^{2+}(aq) + 2\,\text{F}^-(aq) \rightleftharpoons \text{BaF}_2(s)

    • Solubility product expression:

    • Ksp(BaF2)=[Ba2+][F−]2K_{\text{sp}}(\text{BaF}_2) = [\text{Ba}^{2+}][\text{F}^-]^2

    • Standard literature value at 25 ∘C25\,^\circ\text{C}: Ksp=1.7×10−6K_{\text{sp}} = 1.7 \times 10^{-6}

    • Initial concentrations prior to reaction:

    • [Ba2+]0=5.0 mmol120 mL=0.0417 M[\text{Ba}^{2+}]_0 = \frac{5.0\,\text{mmol}}{120\,\text{mL}} = 0.0417\,\text{M}

    • Under standard NaF\text{NaF} stoichiometry: [F−]0=5.6 mmol120 mL=0.0467 M[\text{F}^-]_0 = \frac{5.6\,\text{mmol}}{120\,\text{mL}} = 0.0467\,\text{M}

    • Ion product (QspQ_{\text{sp}}):

    • Qsp=[Ba2+]0[F−]02=(0.0417)×(0.0467)2=9.1×10−5Q_{\text{sp}} = [\text{Ba}^{2+}]_0 [\text{F}^-]_0^2 = (0.0417) \times (0.0467)^2 = 9.1 \times 10^{-5}

    • Because Qsp>KspQ_{\text{sp}} > K_{\text{sp}}, precipitation of BaF2(s)\text{BaF}_2(s) occurs spontaneously.

  • Equilibrium Ion Concentrations Calculation (Standard Stoichiometry):

    • Stoichiometry requires 2 mol2\,\text{mol} of F−\text{F}^- per 1 mol1\,\text{mol} of Ba2+\text{Ba}^{2+}.

    • Complete consumption of 5.6 mmol5.6\,\text{mmol} of F−\text{F}^- consumes: 5.6 mmol2=2.8 mmol\frac{5.6\,\text{mmol}}{2} = 2.8\,\text{mmol} of Ba2+\text{Ba}^{2+}.

    • Remaining unprecipitated Ba2+\text{Ba}^{2+} in solution:

    • n(Ba2+)excess=5.0 mmol−2.8 mmol=2.2 mmoln(\text{Ba}^{2+})_{\text{excess}} = 5.0\,\text{mmol} - 2.8\,\text{mmol} = 2.2\,\text{mmol}

    • Concentration of excess barium ions:

    • [Ba2+]excess=2.2 mmol120 mL=0.0183 M[\text{Ba}^{2+}]_{\text{excess}} = \frac{2.2\,\text{mmol}}{120\,\text{mL}} = 0.0183\,\text{M}

    • Equilibrium concentration of dissolved fluoride determined by KspK_{\text{sp}}:

    • [F−]=Ksp[Ba2+]=1.7×10−60.0183=9.64×10−3 M[\text{F}^-] = \sqrt{\frac{K_{\text{sp}}}{[\text{Ba}^{2+}]}} = \sqrt{\frac{1.7 \times 10^{-6}}{0.0183}} = 9.64 \times 10^{-3}\,\text{M}

    • Precise equilibrium concentration of barium ions:

    • [Ba2+]=0.0183 M+[F−]2≈0.0183 M+0.0048 M=0.0231 M[\text{Ba}^{2+}] = 0.0183\,\text{M} + \frac{[\text{F}^-]}{2} \approx 0.0183\,\text{M} + 0.0048\,\text{M} = 0.0231\,\text{M}

Selective Sulfide Precipitation and pH-Dependent Separation

  • System Composition and Speciation:

    • Initial metal ion concentrations:

    • [Mn2+]=0.2 M[\text{Mn}^{2+}] = 0.2\,\text{M}

    • [Fe2+]=0.1 M[\text{Fe}^{2+}] = 0.1\,\text{M}

    • Analytical total concentration of hydrogen sulfide species:

    • Ctotal=[H2S]+[HS−]+[S2−]=0.2 MC_{\text{total}} = [\text{H}_2\text{S}] + [\text{HS}^-] + [\text{S}^{2-}] = 0.2\,\text{M}

    • Initial solution acidity: pH=1\text{pH} = 1, corresponding to [H+]=10−1 M=0.1 M[\text{H}^+] = 10^{-1}\,\text{M} = 0.1\,\text{M}.

  • Hydrogen Sulfide Acid Dissociation Equilibria:

    • First dissociation step:

    • H2S(aq)⇌H+(aq)+HS−(aq)\text{H}_2\text{S}(aq) \rightleftharpoons \text{H}^+(aq) + \text{HS}^-(aq)

    • Ka1=[H+][HS−][H2S]≈1.0×10−7K_{a1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \approx 1.0 \times 10^{-7}

    • Second dissociation step:

    • HS−(aq)⇌H+(aq)+S2−(aq)\text{HS}^-(aq) \rightleftharpoons \text{H}^+(aq) + \text{S}^{2-}(aq)

    • Ka2=[H+][S2−][HS−]≈1.0×10−14K_{a2} = \frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]} \approx 1.0 \times 10^{-14}

    • Overall dissociation constant:

    • Ka=Ka1×Ka2=[H+]2[S2−][H2S]≈1.0×10−21K_a = K_{a1} \times K_{a2} = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]} \approx 1.0 \times 10^{-21}

    • Sulfide fraction (alpha value α2\alpha_2):

    • [S2−]=Ka1Ka2[H+]2+Ka1[H+]+Ka1Ka2×Ctotal[\text{S}^{2-}] = \frac{K_{a1} K_{a2}}{[\text{H}^+]^2 + K_{a1}[\text{H}^+] + K_{a1} K_{a2}} \times C_{\text{total}}

    • In acidic solutions where [H+]2≫Ka1[H+][\text{H}^+]^2 \gg K_{a1}[\text{H}^+], the equation simplifies to:

    • [S2−]≈Ka1Ka2Ctotal[H+]2[\text{S}^{2-}] \approx \frac{K_{a1} K_{a2} C_{\text{total}}}{[\text{H}^+]^2}

  • Precipitation Thresholds of Metal Sulfides:

    • Solubility product constants:

    • For iron(II) sulfide (FeS\text{FeS}): Ksp(FeS)=[Fe2+][S2−]≈6.0×10−19K_{\text{sp}}(\text{FeS}) = [\text{Fe}^{2+}][\text{S}^{2-}] \approx 6.0 \times 10^{-19}

    • For manganese(II) sulfide (MnS\text{MnS}): Ksp(MnS)=[Mn2+][S2−]≈3.0×10−13K_{\text{sp}}(\text{MnS}) = [\text{Mn}^{2+}][\text{S}^{2-}] \approx 3.0 \times 10^{-13}

    • Critical sulfide concentration to initiate FeS\text{FeS} precipitation:

    • [S2−]crit, FeS=Ksp(FeS)[Fe2+]=6.0×10−190.1 M=6.0×10−18 M[\text{S}^{2-}]_{\text{crit, FeS}} = \frac{K_{\text{sp}}(\text{FeS})}{[\text{Fe}^{2+}]} = \frac{6.0 \times 10^{-19}}{0.1\,\text{M}} = 6.0 \times 10^{-18}\,\text{M}

    • Critical sulfide concentration to initiate MnS\text{MnS} precipitation:

    • [S2−]crit, MnS=Ksp(MnS)[Mn2+]=3.0×10−130.2 M=1.5×10−12 M[\text{S}^{2-}]_{\text{crit, MnS}} = \frac{K_{\text{sp}}(\text{MnS})}{[\text{Mn}^{2+}]} = \frac{3.0 \times 10^{-13}}{0.2\,\text{M}} = 1.5 \times 10^{-12}\,\text{M}

    • Sequence of precipitation:

    • Because [S2−]crit, FeS≪[S2−]crit, MnS[\text{S}^{2-}]_{\text{crit, FeS}} \ll [\text{S}^{2-}]_{\text{crit, MnS}}, iron(II) sulfide (FeS\text{FeS}) precipitates first.

  • Determination of pH at Onset of Precipitation:

    • At initial pH=1\text{pH} = 1 ([H+]=0.1 M[\text{H}^+] = 0.1\,\text{M}):

    • [S2−]init=(1.0×10−21)×0.2(0.1)2=2.0×10−20 M[\text{S}^{2-}]_{\text{init}} = \frac{(1.0 \times 10^{-21}) \times 0.2}{(0.1)^2} = 2.0 \times 10^{-20}\,\text{M}

    • Since 2.0×10−20 M<6.0×10−18 M2.0 \times 10^{-20}\,\text{M} < 6.0 \times 10^{-18}\,\text{M}, no precipitation occurs at pH=1\text{pH} = 1.

    • Threshold hydrogen ion concentration for FeS\text{FeS} precipitation:

    • [H+]2=Ka1Ka2Ctotal[S2−]crit, FeS=(1.0×10−21)×0.26.0×10−18=3.33×10−5[\text{H}^+]^2 = \frac{K_{a1} K_{a2} C_{\text{total}}}{[\text{S}^{2-}]_{\text{crit, FeS}}} = \frac{(1.0 \times 10^{-21}) \times 0.2}{6.0 \times 10^{-18}} = 3.33 \times 10^{-5}

    • [H+]=3.33×10−5=5.77×10−3 M[\text{H}^+] = \sqrt{3.33 \times 10^{-5}} = 5.77 \times 10^{-3}\,\text{M}

    • Calculation of critical pH:

    • pH=−log⁡(5.77×10−3)≈2.24\text{pH} = -\log(5.77 \times 10^{-3}) \approx 2.24

    • Iron(II) sulfide (FeS\text{FeS}) begins precipitating when base titration elevates the pH to approximately 2.242.24.

Thermodynamics and Bomb Calorimetry of Ammonium Nitrate Decomposition

  • Reaction Stoichiometry and Gas Generation:

    • Decomposition equation:

    • NH4NO3(s)→N2O(g)+2 H2O(g)\text{NH}_4\text{NO}_3(s) \rightarrow \text{N}_2\text{O}(g) + 2\,\text{H}_2\text{O}(g)

    • Change in stoichiometric moles of gas (Δngas\Delta n_{\text{gas}}):

    • Δngas=nproducts, gas−nreactants, gas=(1+2)−0=3 mol\Delta n_{\text{gas}} = n_{\text{products, gas}} - n_{\text{reactants, gas}} = (1 + 2) - 0 = 3\,\text{mol}

  • Standard Enthalpy and Internal Energy Changes:

    • Standard enthalpies of formation (ΔHf∘\Delta H_f^\circ at 298.15 K298.15\,\text{K}):

    • ΔHf∘[NH4NO3(s)]=−365.6 kJ/mol\Delta H_f^\circ[\text{NH}_4\text{NO}_3(s)] = -365.6\,\text{kJ/mol}

    • ΔHf∘[N2O(g)]=+82.05 kJ/mol\Delta H_f^\circ[\text{N}_2\text{O}(g)] = +82.05\,\text{kJ/mol}

    • ΔHf∘[H2O(g)]=−241.82 kJ/mol\Delta H_f^\circ[\text{H}_2\text{O}(g)] = -241.82\,\text{kJ/mol}

    • Enthalpy change of reaction (ΔHrxn∘\Delta H_{\text{rxn}}^\circ):

    • ΔHrxn∘=ΔHf∘[N2O(g)]+2 ΔHf∘[H2O(g)]−ΔHf∘[NH4NO3(s)]\Delta H_{\text{rxn}}^\circ = \Delta H_f^\circ[\text{N}_2\text{O}(g)] + 2\,\Delta H_f^\circ[\text{H}_2\text{O}(g)] - \Delta H_f^\circ[\text{NH}_4\text{NO}_3(s)]

    • ΔHrxn∘=82.05+2(−241.82)−(−365.6)=−35.99 kJ/mol\Delta H_{\text{rxn}}^\circ = 82.05 + 2(-241.82) - (-365.6) = -35.99\,\text{kJ/mol}

    • Internal energy change (ΔUrxn\Delta U_{\text{rxn}}):

    • Derived from thermodynamic definition: ΔH=ΔU+Δ(PV)=ΔU+ΔngasRT\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_{\text{gas}} R T

    • ΔUrxn=ΔHrxn−ΔngasRT\Delta U_{\text{rxn}} = \Delta H_{\text{rxn}} - \Delta n_{\text{gas}} R T

    • At reference temperature T=483.15 KT = 483.15\,\text{K} (210 ∘C210\,^\circ\text{C}):

    • ΔngasRT=3 mol×8.3145 J/(mol K)×483.15 K=12051 J/mol=12.05 kJ/mol\Delta n_{\text{gas}} R T = 3\,\text{mol} \times 8.3145\,\text{J/(mol K)} \times 483.15\,\text{K} = 12051\,\text{J/mol} = 12.05\,\text{kJ/mol}

    • ΔUrxn=−35.99 kJ/mol−12.05 kJ/mol=−48.04 kJ/mol\Delta U_{\text{rxn}} = -35.99\,\text{kJ/mol} - 12.05\,\text{kJ/mol} = -48.04\,\text{kJ/mol}

  • Constant-Volume Bomb Calorimeter Calculations:

    • Given operational parameters:

    • Quantity decomposed: n=1.68 moln = 1.68\,\text{mol}

    • Initial temperature: T1=210 ∘CT_1 = 210\,^\circ\text{C}

    • Total calorimeter heat capacity constant: Ccal=5.625 kJ/KC_{\text{cal}} = 5.625\,\text{kJ/K}

    • In a rigid bomb calorimeter, volume is constant (ΔV=0\Delta V = 0), meaning work w=0w = 0 and heat released equals internal energy change:

    • qv=ΔUtotal=n×ΔUrxnq_v = \Delta U_{\text{total}} = n \times \Delta U_{\text{rxn}}

    • qv=1.68 mol×(−48.04 kJ/mol)=−80.71 kJq_v = 1.68\,\text{mol} \times (-48.04\,\text{kJ/mol}) = -80.71\,\text{kJ}

    • Heat absorbed by the calorimeter assembly:

    • qcal=−qv=+80.71 kJq_{\text{cal}} = -q_v = +80.71\,\text{kJ}

    • Temperature increase (ΔT\Delta T):

    • ΔT=qcalCcal=80.71 kJ5.625 kJ/K=14.35 K\Delta T = \frac{q_{\text{cal}}}{C_{\text{cal}}} = \frac{80.71\,\text{kJ}}{5.625\,\text{kJ/K}} = 14.35\,\text{K}

    • Final calorimeter temperature (T2T_2):

    • T2=T1+ΔT=210 ∘C+14.35 ∘C=224.35 ∘CT_2 = T_1 + \Delta T = 210\,^\circ\text{C} + 14.35\,^\circ\text{C} = 224.35\,^\circ\text{C}

Thermochemistry and Enthalpy of Combustion of Adamantane

  • Structure and Chemical Formula:

    • Adamantane is a highly symmetrical, cage-like polycyclic alkane consisting of four interconnected cyclohexane rings in chair conformations.

    • Chemical formula: C10H16\text{C}_{10}\text{H}_{16}

Chemical structure of adamantane

  • Thermochemical Parameters and Atomization Enthalpies:

    • Atomization enthalpy of adamantane: ΔHatom∘[C10H16]=2157 kJ/mol\Delta H_{\text{atom}}^\circ[\text{C}_{10}\text{H}_{16}] = 2157\,\text{kJ/mol}

    • Atomization enthalpy of solid carbon (graphite):

    • C(graphite)→C(g)\text{C}(\text{graphite}) \rightarrow \text{C}(g)

    • ΔHatom∘[C(graphite)]=700 kJ/mol\Delta H_{\text{atom}}^\circ[\text{C}(\text{graphite})] = 700\,\text{kJ/mol}

    • Atomization enthalpy of diatomic hydrogen gas:

    • 12 H2(g)→H(g)\frac{1}{2}\,\text{H}_2(g) \rightarrow \text{H}(g)

    • ΔHatom∘[H]=12D(H−H)≈218 kJ/mol\Delta H_{\text{atom}}^\circ[\text{H}] = \frac{1}{2} D(\text{H}-\text{H}) \approx 218\,\text{kJ/mol}

  • Derivation of Standard Enthalpy of Formation (ΔHf∘\Delta H_f^\circ):

    • Formation reaction from elements in standard states:

    • 10 C(graphite)+8 H2(g)→C10H16(s)10\,\text{C}(\text{graphite}) + 8\,\text{H}_2(g) \rightarrow \text{C}_{10}\text{H}_{16}(s)

    • Enthalpy balance based on gaseous atomization intermediate state:

    • ΔHf∘[C10H16]=10 ΔHatom∘[C(graphite)]+16 ΔHatom∘[H]−ΔHatom∘[C10H16]\Delta H_f^\circ[\text{C}_{10}\text{H}_{16}] = 10\,\Delta H_{\text{atom}}^\circ[\text{C}(\text{graphite})] + 16\,\Delta H_{\text{atom}}^\circ[\text{H}] - \Delta H_{\text{atom}}^\circ[\text{C}_{10}\text{H}_{16}]

    • ΔHf∘[C10H16]=10(700 kJ/mol)+16(218 kJ/mol)−2157 kJ/mol\Delta H_f^\circ[\text{C}_{10}\text{H}_{16}] = 10(700\,\text{kJ/mol}) + 16(218\,\text{kJ/mol}) - 2157\,\text{kJ/mol}

    • ΔHf∘[C10H16]=7000+3488−2157=8331 kJ/mol\Delta H_f^\circ[\text{C}_{10}\text{H}_{16}] = 7000 + 3488 - 2157 = 8331\,\text{kJ/mol}

  • Combustion Reaction and Combustion Enthalpy:

    • Balanced combustion equation:

    • C10H16(s)+14 O2(g)→10 CO2(g)+8 H2O(l)\text{C}_{10}\text{H}_{16}(s) + 14\,\text{O}_2(g) \rightarrow 10\,\text{CO}_2(g) + 8\,\text{H}_2\text{O}(l)

    • Standard formation enthalpies of combustion products:

    • ΔHf∘[CO2(g)]=−393.5 kJ/mol\Delta H_f^\circ[\text{CO}_2(g)] = -393.5\,\text{kJ/mol}

    • ΔHf∘[H2O(l)]=−285.8 kJ/mol\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -285.8\,\text{kJ/mol}

    • ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\text{O}_2(g)] = 0\,\text{kJ/mol}

    • Standard combustion enthalpy expression:

    • ΔHcomb∘=10 ΔHf∘[CO2(g)]+8 ΔHf∘[H2O(l)]−ΔHf∘[C10H16(s)]\Delta H_{\text{comb}}^\circ = 10\,\Delta H_f^\circ[\text{CO}_2(g)] + 8\,\Delta H_f^\circ[\text{H}_2\text{O}(l)] - \Delta H_f^\circ[\text{C}_{10}\text{H}_{16}(s)]

    • ΔHcomb∘=10(−393.5)+8(−285.8)−8331\Delta H_{\text{comb}}^\circ = 10(-393.5) + 8(-285.8) - 8331

    • ΔHcomb∘=−3935−2286.4−8331=−14552.4 kJ/mol\Delta H_{\text{comb}}^\circ = -3935 - 2286.4 - 8331 = -14552.4\,\text{kJ/mol}

Reaction Kinetics and Mechanism Evaluation

  • Elementary Reaction Steps of Proposed Mechanism:

    • Step 1: A+B→C\text{A} + \text{B} \rightarrow \text{C}

    • Step 2: C→D+B\text{C} \rightarrow \text{D} + \text{B}

    • Step 3: C+D→E+B\text{C} + \text{D} \rightarrow \text{E} + \text{B}

  • Overall Reaction Stoichiometry:

    • Linear combination of elementary steps to eliminate intermediates:

    • Intermediate D\text{D} produced in Step 2 is consumed in Step 3 in a 1:11:1 ratio.

    • Intermediate C\text{C} is consumed in both Step 2 (1 mol1\,\text{mol}) and Step 3 (1 mol1\,\text{mol}), requiring 2 mol2\,\text{mol} of Step 1 to produce sufficient C\text{C}.

    • Summation: 2×(Step 1)+1×(Step 2)+1×(Step 3)2 \times (\text{Step 1}) + 1 \times (\text{Step 2}) + 1 \times (\text{Step 3}):

    • Left-hand side: 2 A+2 B+C+C+D2\,\text{A} + 2\,\text{B} + \text{C} + \text{C} + \text{D}

    • Right-hand side: 2 C+D+B+E+B2\,\text{C} + \text{D} + \text{B} + \text{E} + \text{B}

    • Net canceled equation:

    • 2 A→E2\,\text{A} \rightarrow \text{E}

  • Classification of Chemical Species:

    • Reactant: A\text{A}

    • Product: E\text{E}

    • Catalyst: B\text{B}

    • Consumed in Step 1 (2 B2\,\text{B} consumed) and subsequently regenerated in Steps 2 and 3 (1 B+1 B=2 B1\,\text{B} + 1\,\text{B} = 2\,\text{B} regenerated).

    • Intermediates: C\text{C} and D\text{D}

    • Formed during early steps and completely consumed prior to product completion.

  • Theoretical Rate Law Derivation (Step 3 Rate-Determining):

    • Given that Step 3 is rate-determining, the rate of product formation is:

    • r=k3[C][D]r = k_3 [\text{C}][\text{D}]

    • Assuming rapid pre-equilibria for Steps 1 and 2:

    • For Step 1: A+B⇌C\text{A} + \text{B} \rightleftharpoons \text{C}

      • K1=[C][A][B]  ⟹  [C]=K1[A][B]K_1 = \frac{[\text{C}]}{[\text{A}][\text{B}]} \implies [\text{C}] = K_1 [\text{A}][\text{B}]

    • For Step 2: C⇌D+B\text{C} \rightleftharpoons \text{D} + \text{B}

      • K2=[D][B][C]  ⟹  [D]=K2[C][B]=K1K2[A]K_2 = \frac{[\text{D}][\text{B}]}{[\text{C}]} \implies [\text{D}] = K_2 \frac{[\text{C}]}{[\text{B}]} = K_1 K_2 [\text{A}]

    • Substituting equilibrium expressions into the rate law:

    • r=k3(K1[A][B])(K1K2[A])=k3K12K2[A]2[B]r = k_3 (K_1 [\text{A}][\text{B}]) (K_1 K_2 [\text{A}]) = k_3 K_1^2 K_2 [\text{A}]^2 [\text{B}]

    • Theoretical rate law:

    • r=ktheor[A]2[B]r = k_{\text{theor}} [\text{A}]^2 [\text{B}]

  • Empirical Rate Law Determination from Initial Rates Data:

    • Experimental data provided:

    • Run 1: CA=0.1 MC_A = 0.1\,\text{M}, CB=0.1 MC_B = 0.1\,\text{M}, Ri,0=7.5×10−3 M/sR_{i,0} = 7.5 \times 10^{-3}\,\text{M/s}

    • Run 2: CA=0.2 MC_A = 0.2\,\text{M}, CB=0.1 MC_B = 0.1\,\text{M}, Ri,0=3.0×10−2 M/sR_{i,0} = 3.0 \times 10^{-2}\,\text{M/s}

    • Run 3: CA=0.2 MC_A = 0.2\,\text{M}, CB=0.3 MC_B = 0.3\,\text{M}, Ri,0=9.0×10−2 M/sR_{i,0} = 9.0 \times 10^{-2}\,\text{M/s}

    • Determination of reaction order with respect to A\text{A} (xx):

    • Comparing Run 1 and Run 2 where CBC_B is constant at 0.1 M0.1\,\text{M}:

    • R2R1=3.0×10−27.5×10−3=4.0\frac{R_2}{R_1} = \frac{3.0 \times 10^{-2}}{7.5 \times 10^{-3}} = 4.0

    • CA,2CA,1=0.20.1=2.0\frac{C_{A,2}}{C_{A,1}} = \frac{0.2}{0.1} = 2.0

    • (2.0)x=4.0=(2.0)2  ⟹  x=2(2.0)^x = 4.0 = (2.0)^2 \implies x = 2

    • Determination of reaction order with respect to B\text{B} (yy):

    • Comparing Run 2 and Run 3 where CAC_A is constant at 0.2 M0.2\,\text{M}:

    • R3R2=9.0×10−23.0×10−2=3.0\frac{R_3}{R_2} = \frac{9.0 \times 10^{-2}}{3.0 \times 10^{-2}} = 3.0

    • CB,3CB,2=0.30.1=3.0\frac{C_{B,3}}{C_{B,2}} = \frac{0.3}{0.1} = 3.0

    • (3.0)y=3.0=(3.0)1  ⟹  y=1(3.0)^y = 3.0 = (3.0)^1 \implies y = 1

    • Empirical rate expression:

    • R=k[A]2[B]R = k [\text{A}]^2 [\text{B}]

    • Value of empirical rate constant (kk):

    • k=7.5×10−3 M/s(0.1 M)2(0.1 M)=7.5 M−2 s−1k = \frac{7.5 \times 10^{-3}\,\text{M/s}}{(0.1\,\text{M})^2 (0.1\,\text{M})} = 7.5\,\text{M}^{-2}\,\text{s}^{-1}

  • Mechanism Consistency Analysis:

    • The empirical rate law (R=k[A]2[B]R = k [\text{A}]^2 [\text{B}]) is identical in form to the theoretical rate law derived under the assumption that Step 3 is rate-determining preceded by rapid pre-equilibria.

    • Therefore, the proposed mechanism is fully consistent with the experimental kinetic observations.

Coordination Chemistry and Electronic Structure of Iron(III) Complexes

  • Electronic Configuration of Iron(III):

    • Neutral iron ground state: Fe:[Ar] 3d6 4s2\text{Fe}: [\text{Ar}]\,3d^6\,4s^2

    • Trivalent iron cation ground state: Fe3+:[Ar] 3d5\text{Fe}^{3+}: [\text{Ar}]\,3d^5

    • Octahedral coordination number: 66

  • Ligand Field Strengths and Spin States:

    • Hexacyanidoferrate(III), [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-}:

    • Cyanide (CN−\text{CN}^-) is a strong-field ligand (carbon-donor, strong π\pi-acceptor) located high in the spectrochemical series.

    • Octahedral splitting energy exceeds electron pairing energy: Δo>P\Delta_o > P.

    • Low-spin complex results, causing maximum orbital pairing in the lower energy manifold.

    • Hexachloridoferrate(III), [FeCl6]3−[\text{FeCl}_6]^{3-}:

    • Chloride (Cl−\text{Cl}^-) is a weak-field ligand (halide donor, π\pi-donor) located low in the spectrochemical series.

    • Octahedral splitting energy is smaller than electron pairing energy: Δo<P\Delta_o < P.

    • High-spin complex results, conforming strictly to Hund's rule of maximum multiplicity.

  • Orbital Hybridization (Valence Bond Theory):

    • [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-}:

    • Strong ligand field forces the five 3d3d electrons to pair within three dd-orbitals (dxy,dyz,dxzd_{xy}, d_{yz}, d_{xz}.

    • Leaves two inner 3d3d orbitals (dx2−y2,dz2d_{x^2-y^2}, d_{z^2}) empty.

    • Hybridization: d2sp3d^2sp^3 (inner orbital complex), utilizing two 3d3d, one 4s4s, and three 4p4p orbitals to accommodate six ligand lone pairs.

    • [FeCl6]3−[\text{FeCl}_6]^{3-}:

    • Weak ligand field leaves all five 3d3d orbitals singly occupied.

    • Inner 3d3d orbitals are unavailable for bonding.

    • Hybridization: sp3d2sp^3d^2 (outer orbital complex), utilizing one 4s4s, three 4p4p, and two outer 4d4d orbitals.

  • Crystal Field Orbital Splitting and Orbital Diagrams:

    • In an octahedral crystal field, five degenerate dd-orbitals split into two sets:

    • Triply degenerate, lower energy set: t2gt_{2g} (dxy,dyz,dxzd_{xy}, d_{yz}, d_{xz}, stabilized by −0.4 Δo-0.4\,\Delta_o)

    • Doubly degenerate, higher energy set: ege_g (dx2−y2,dz2d_{x^2-y^2}, d_{z^2}, destabilized by +0.6 Δo+0.6\,\Delta_o)

    • Electron configuration for low-spin [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-}:

    • Configuration: (t2g)5(eg)0(t_{2g})^5 (e_g)^0

    • Orbital occupancy: (dxy)2(dyz)2(dxz)1(d_{xy})^2 (d_{yz})^2 (d_{xz})^1

    • Unpaired electrons: n=1n = 1

    • Crystal Field Stabilization Energy: CFSE=5×(−0.4 Δo)+2P=−2.0 Δo+2P\text{CFSE} = 5 \times (-0.4\,\Delta_o) + 2P = -2.0\,\Delta_o + 2P

    • Electron configuration for high-spin [FeCl6]3−[\text{FeCl}_6]^{3-}:

    • Configuration: (t2g)3(eg)2(t_{2g})^3 (e_g)^2

    • Orbital occupancy: (dxy)1(dyz)1(dxz)1(dx2−y2)1(dz2)1(d_{xy})^1 (d_{yz})^1 (d_{xz})^1 (d_{x^2-y^2})^1 (d_{z^2})^1

    • Unpaired electrons: n=5n = 5

    • Crystal Field Stabilization Energy: CFSE=3×(−0.4 Δo)+2×(+0.6 Δo)=0\text{CFSE} = 3 \times (-0.4\,\Delta_o) + 2 \times (+0.6\,\Delta_o) = 0

  • Magnetic Properties and Paramagnetism Comparison:

    • Paramagnetism arises from the presence of permanent magnetic dipole moments generated by unpaired electron spins.

    • Spin-only magnetic moment formula:

    • μs=n(n+2) μB\mu_s = \sqrt{n(n + 2)}\,\mu_B

    • Where nn is the number of unpaired electrons and μB\mu_B is the Bohr Magneton (μB=eℏ2me≈9.274×10−24 J/T\mu_B = \frac{e\hbar}{2m_e} \approx 9.274 \times 10^{-24}\,\text{J/T}.

    • Magnetic moment of [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-}:

    • n=1n = 1

    • μs=1(1+2)=3 μB≈1.73 μB\mu_s = \sqrt{1(1 + 2)} = \sqrt{3}\,\mu_B \approx 1.73\,\mu_B

    • Magnetic moment of [FeCl6]3−[\text{FeCl}_6]^{3-}:

    • n=5n = 5

    • μs=5(5+2)=35 μB≈5.92 μB\mu_s = \sqrt{5(5 + 2)} = \sqrt{35}\,\mu_B \approx 5.92\,\mu_B

    • [FeCl6]3−[\text{FeCl}_6]^{3-} is significantly more paramagnetic than [Fe(CN)6]3−[\text{Fe}(\text{CN})_6]^{3-} due to possessing five unpaired electrons compared to only one.