Chemistry Labs

International Chemistry Olympiad · 2010

Problems

  1. Problem 1In 1894 Lord Rayleigh found that nitrogen prepared chemically (mean mass in a fixed vessel: 2.2990 g) was lighter than \“atmospheric nitrogen\” freed of OX2\ce{O2} (2.3102 g), at 15.0 \°C and 1.013×1051.013 \times 10^{5} Pa; the difference was later attributed to argon. (a) Calculate the vessel volume and the mole fraction of Ar in atmospheric nitrogen. (b) Ramsay isolated helium from cleveite; it emits the D3_3 line at 587.7 nm. Calculate the photon energy and identify the corresponding transition in the He orbital diagram (answer: the highest-energy allowed transition, 3d \→ 2p) and the nuclear process that produces He in uranium minerals (α\alpha-decay) and Ar in rocks (electron capture/positron decay of X40X2240K\ce{^{40}K}). (c) An unknown gas at 15.0 \°C and atmospheric pressure has density 0.850±0.005 kg m−30.850 \pm 0.005\ \text{kg m}^{-3}; sound in it has λ=0.116\lambda = 0.116 m at f=3520f = 3520 Hz, with vs=fλ=γRT/Mv_s = f\lambda = \sqrt{\gamma RT/M}. Find MM and γ\gamma and identify the gas among HCl, HF, Ne, Ar.Solutions: 1
  2. Problem 2(a) In the NaCl crystal, NaX+\ce{Na+} and ClX−\ce{Cl-} each form a face-centred cubic lattice; the ionic radii are r(NaX+)=0.102r(\ce{Na+}) = 0.102 nm and r(ClX−)=0.181r(\ce{Cl-}) = 0.181 nm. Give the numbers of NaX+\ce{Na+} and ClX−\ce{Cl-} per unit cell, their coordination numbers, and calculate the crystal density. (b) From the Born\–Haber data \— ΔfH(NaCl(s))=−411\Delta_fH(\ce{NaCl(s)}) = -411, ΔsubH(Na)=+109\Delta_{sub}H(\ce{Na}) = +109, IE(Na)=+496IE(\ce{Na}) = +496, D(ClX2)=+242D(\ce{Cl2}) = +242, EA(Cl)=−349EA(\ce{Cl}) = -349 kJ mol−1^{-1} \— write the equations for the formation step and the direct dissociation NaCl(s)→NaX+(g)+ClX−(g)\ce{NaCl(s) -> Na+(g) + Cl-(g)}, and calculate the lattice formation enthalpy. (c) The Solvay process achieves 2 NaCl+CaCOX3→NaX2COX3+CaClX2\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2} through the cycle: CaCOX3→ΔA+B\ce{CaCO3 ->[\Delta] A + B}; NaCl+NHX3+B+HX2O→C+D\ce{NaCl + NH3 + B + H2O -> C + D}; 2 C→ΔNaX2COX3+HX2O+B\ce{2C ->[\Delta] Na2CO3 + H2O + B}; A+HX2O→E\ce{A + H2O -> E}; E+2 D→CaClX2+2 HX2O+2 NHX3\ce{E + 2D -> CaCl2 + 2H2O + 2NH3}. Identify compounds A\–E.Solutions: 1