Chemistry Labs

Vietnam National Chemistry Olympiad · 2025

Problems

  1. Problem 1The gas-phase isomerisation between cis-but-2-ene and trans-but-2-ene is reversible and unimolecular, with equilibrium constant KK and forward/reverse rate constants k1k_1 and k−1k_{-1}: \ce{cis-but-2-ene <=>[k_1][][k_{-1}] trans-but-2-ene}. Initially the system contains only cis-but-2-ene at concentration Cs0C_s^0. (a) Write the differential rate equation for CstC_s^t. (b) Derive the expression for the concentration ratio Cst/Cs0C_s^t/C_s^0 as a function of tt, KK and k−1k_{-1}. (c) When K=1K = 1, identify which of the following linear plots is valid: (H-1) ln⁡(Cst−0.5Cs0)\ln(C_s^t - 0.5C_s^0) vs tt; (H-2) ln⁡(Cst)\ln(C_s^t) vs tt; (H-3) 1/ln⁡(Cst−0.5Cs0)1/\ln(C_s^t - 0.5C_s^0) vs tt; (H-4) 1/(Cst−0.5Cs0)1/(C_s^t - 0.5C_s^0) vs tt. (d) At T=690 KT = 690\ \text{K}, K=1.14K = 1.14 and k1=1.6×10−6 s−1k_1 = 1.6 \times 10^{-6}\ \text{s}^{-1}. Calculate the time (in hours) required for cis-but-2-ene to convert to 30 % of its maximum equilibrium conversion. (e) Calculate the percentage of trans-but-2-ene in the mixture after 10 hours.Solutions: 1
  2. Problem 2Cathodic protection is widely used to prevent metal corrosion by attaching a more active sacrificial metal to the structure. (a) In an experiment on protecting steel in seawater, a 25.0 g zinc block was attached to a steel apparatus. After some time, the block was reweighed at 28.0 g; assume that the only oxidation product is Zn(OH)X2\ce{Zn(OH)2} adhering to the block (M(Zn)=65.38M(\ce{Zn}) = 65.38, M(Zn(OH)X2)=99.40M(\ce{Zn(OH)2}) = 99.40 g mol−1^{-1}). Calculate the percentage of zinc that has oxidised. (b) Find the maximum service time (in hours) of this 25.0 g zinc block if the average protective current generated is 25 mA. (c) A steel ship hull with an immersed area of 1000 m2^2 requires an average protective current density of 2.5 mA m−2^{-2}. If zinc sacrificial anodes are used and 10 % of the zinc is lost to secondary processes, calculate the mass of zinc required per year (365 days).Solutions: 1