Chemistry Labs

International Chemistry Olympiad · 2011

Problems

  1. Problem 1Nitric oxide reacts with hydrogen at 820 °C: 2 NO(g)+HX2(g)→NX2O(g)+HX2O(g)\ce{2NO(g) + H2(g) -> N2O(g) + H2O(g)}. Initial rates of NX2O\ce{N2O} formation were measured at various initial partial pressures (all pressures in torr, times in seconds; do not use concentrations): Exp. 1: pNO=120.0p_{\ce{NO}} = 120.0, pHX2=60.0p_{\ce{H2}} = 60.0 → rate =8.66×10−2= 8.66 \times 10^{-2} torr s−1^{-1}; Exp. 2: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=60.0p_{\ce{H2}} = 60.0 → 2.17×10−22.17 \times 10^{-2}; Exp. 3: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=180.0p_{\ce{H2}} = 180.0 → 6.62×10−26.62 \times 10^{-2} torr s−1^{-1}. (a) Find the rate law and the rate constant. (b) Find the initial rate of disappearance of NO when pNO=200p_{\ce{NO}} = 200 torr and pHX2=100p_{\ce{H2}} = 100 torr. (c) Find the time to halve pHX2p_{\ce{H2}} when pNO=800p_{\ce{NO}} = 800 torr and pHX2=1.0p_{\ce{H2}} = 1.0 torr. (d) The proposed mechanism is 2 NO⇌NX2OX2\ce{2NO <=> N2O2} (rate constants k1,k−1k_1, k_{-1}) followed by NX2OX2+HX2→kX2NX2O+HX2O\ce{N2O2 + H2 ->[k_2] N2O + H2O}. Derive the rate law using the steady-state approximation for NX2OX2\ce{N2O2}, state the condition under which it reduces to the experimental law, and express kk in terms of k1k_1, k−1k_{-1}, k2k_2.Solutions: 1
  2. Problem 2Anhydrous ammonia is a clean, energy-dense fuel. In a fixed-volume container, gaseous NHX3\ce{NH3} burns according to 4 NHX3(g)+3 OX2(g)→2 NX2(g)+6 HX2O(l)\ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(l)}; initial and final states are at 298 K and after combustion of 14.40 g of OX2\ce{O2} some NHX3\ce{NH3} remains (ΔfH∘(NHX3(g))=−46.11\Delta_fH^{\circ}(\ce{NH3(g)}) = -46.11 kJ mol−1^{-1}, ΔfH∘(HX2O(l))=−285.83\Delta_fH^{\circ}(\ce{H2O(l)}) = -285.83 kJ mol−1^{-1}). (a) Calculate the heat released. (b) To determine dissolved NHX3\ce{NH3}, a 10.00 cm3^3 sample of the resulting aqueous solution was added to 15.0 cm3^3 of HX2SOX4\ce{H2SO4} (c=0.0100c = 0.0100 mol dm−3^{-3}) and back-titrated with NaOH (c=0.0200c = 0.0200 mol dm−3^{-3}), equivalence at 10.64 cm3^3 (Kb(NHX3)=1.8×10−5K_b(\ce{NH3}) = 1.8 \times 10^{-5}; Ka(HSOX4X−)=1.1×10−2K_a(\ce{HSO4-}) = 1.1 \times 10^{-2}). Calculate the pH of the solution in the container. (c) At the equivalence point NHX4X+\ce{NH4+} and SOX4X2−\ce{SO4^{2-}} are present; write the relevant equilibria and predict whether the equivalence-point pH is above, below, or equal to 7.Solutions: 1