Hydrogen and s-Block

Question 1

(a) (5 marks) "20-volume" means $1$ volume of the solution liberates $20$ volumes of $O_2$ at STP on complete decomposition. (1) Decomposition: $2H_2O_2 \rightarrow 2H_2O + O_2$. (1) $1\ \text{L} = 1000\ \text{cm}^3$ solution gives $20\times1000 = 20000\ \text{cm}^3\ O_2$. Moles $O_2 = 20000/22400 = 0.8929\ \text{mol}$. (1) Moles $H_2O_2 = 2\times0.8929 = 1.786\ \text{mol L}^{-1}$. (1) Concentration $= 1.786 \times 34.0 = 60.7\ \text{g L}^{-1}$. (1)

(b) (5 marks) Diluted conc $= 1.786/10 = 0.1786\ \text{mol L}^{-1}$. (1) Ionic equation: $2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$. (2) Moles $H_2O_2$ in aliquot $= 0.1786 \times 0.0250 = 4.464\times10^{-3}$ mol. (1) Moles $KMnO_4 = \frac{2}{5}\times4.464\times10^{-3} = 1.786\times10^{-3}$ mol. Volume $= 1.786\times10^{-3}/0.0200 = 0.0893\ \text{L} = 89.3\ \text{cm}^3$. (1)

(c) (3 marks) Open-book / non-planar structure: O–O single bond, each O bonded to one H. (1) O–O–H angle $\approx 94.8^\circ$ (gas), dihedral $\approx 111.5^\circ$ (gas phase). (1) Non-planar because lone-pair–lone-pair repulsions on the two oxygens twist the two O–H bonds out of a common plane. (1)

(d) (5 marks) Half-life $t_{1/2} = \ln 2/k = 0.6931/1.06\times10^{-3} = 653.9\ \text{s} \approx 654\ \text{s}$. (2) For $10\%$: $t = \frac{1}{k}\ln(1/0.10) = \frac{\ln 10}{1.06\times10^{-3}} = \frac{2.3026}{1.06\times10^{-3}} = 2172\ \text{s}$. (3) ($\approx 36.2$ min.)

Question 2

(a) (5 marks) (1 each equation, 1 for KHCO₃ reason)

1. $NH_3 + H_2O + CO_2 \rightarrow NH_4HCO_3$
2. $NH_4HCO_3 + NaCl \rightarrow NaHCO_3\downarrow + NH_4Cl$
3. $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$
4. $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$ (ammonia recovery). $KHCO_3$ is too soluble to precipitate out from solution, so the analogous step fails. (1)

(b) (5 marks) $M(NaHCO_3) = 84.0\ \text{g mol}^{-1}$; moles $= 1000/84.0 = 11.905$ mol. (1) $Na_2CO_3$ moles $= 11.905/2 = 5.952$ mol; $M(Na_2CO_3)=106.0$. (1) Mass $Na_2CO_3 = 5.952\times106.0 = 630.9\ \text{g} \approx 631\ \text{g}$. (2) $CO_2$ moles $= 5.952$; $V = 5.952\times22.4 = 133.3\ \text{L}$ at STP. (1)

(c) (4 marks) At equilibrium $\Delta G^\circ = 0 \Rightarrow T = \Delta H^\circ/\Delta S^\circ$. (1) $T = \frac{178000}{161} = 1105.6\ \text{K}$. (1) $= 1105.6 - 273 = 832.6^\circ\text{C} \approx 833^\circ\text{C}$. (1) Thermal stability of carbonates increases down the group (larger, less polarising cation stabilises the large carbonate ion): $BeCO_3 < MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$. (1)

(d) (3 marks) (1.5 each, any two)

• Both form covalent, polymeric/bridged chlorides ($BeCl_2$, $Al_2Cl_6$); soluble in organic solvents.
• Both oxides/hydroxides are amphoteric ($BeO$, $Al_2O_3$; $Be(OH)_2$, $Al(OH)_3$).
• Both form complex ions ($BeF_4^{2-}$, $AlF_6^{3-}$); both passivated by conc. $HNO_3$; carbides give methane.

Question 3

(a) (4 marks) $M(H_2O)=18.015$, $M(D_2O)=20.028$. (1) % difference $= \frac{20.028-18.015}{18.015}\times100 = 11.17\% \approx 11.2\%$. (2) $D_2O$ slows (moderates) fast neutrons via elastic collisions with low neutron capture, sustaining chain reaction; it is not consumed chemically, hence a moderator not a reactant. (1)

(b) (6 marks) Total hardness ion conc $= (1.62 + 0.90)\times10^{-3} = 2.52\times10^{-3}\ \text{mol L}^{-1}$. (1) Each mole hardness ion $\equiv$ 1 mole $CaCO_3$ ($M=100.0$). (1) Mass equiv $CaCO_3 = 2.52\times10^{-3}\times100.0 = 0.252\ \text{g L}^{-1} = 252\ \text{mg L}^{-1}$. (2) $= 252$ ppm (since $1\ \text{L}\equiv1\ \text{kg}$). (1) $Ca(HCO_3)_2$ → temporary (162 ppm); $MgSO_4$ → permanent (90 ppm). (1)

(c) (5 marks) Conversion constant: multiply molar concentration by $100{,}000\ \text{mg mol}^{-1}$ ... i.e. $M(CaCO_3)=100\ \text{g/mol} = 100000\ \text{mg/mol}$, since 1 mol divalent ion ≡ 1 mol $CaCO_3$. (1)

def hardness_ppm(conc_list):
    # conc_list: list of (molarity_mol_per_L, is_temporary)
    M_CaCO3 = 100000.0  # mg per mol
    total = 0.0
    temp = 0.0
    for c, is_temp in conc_list:
        ppm = c * M_CaCO3   # 1 L = 1 kg
        total += ppm
        if is_temp:
            temp += ppm
    frac = temp / total if total else 0.0
    return total, frac

(3 marks logic, 1 mark correct constant with justification above) Justification: 1 mol of any divalent hardness ion is chemically equated to 1 mol $CaCO_3$; $g/L \to ppm$ uses $1L\equiv1kg$.

[
  {"claim":"20-vol H2O2 is 1.786 mol/L","code":"nO2=20000/22400; c=2*nO2; result=abs(c-1.786)<0.01"},
  {"claim":"KMnO4 titre volume is 89.3 cm3","code":"cdil=0.1786; nH2O2=cdil*0.025; nMnO4=Rational(2,5)*nH2O2; V=nMnO4/0.02*1000; result=abs(float(V)-89.3)<0.5"},
  {"claim":"H2O2 half-life ~654 s","code":"import sympy as sp; t=sp.log(2)/1.06e-3; result=abs(float(t)-654)<2"},
  {"claim":"Na2CO3 mass from 1kg NaHCO3 is ~631 g","code":"n=1000/84.0; m=(n/2)*106.0; result=abs(m-631)<2"},
  {"claim":"CaCO3 decomposition T is ~1106 K","code":"T=178000/161; result=abs(T-1105.6)<1"},
  {"claim":"Total hardness is 252 ppm","code":"h=(1.62e-3+0.90e-3)*100000; result=abs(h-252)<1"}
]