Chemistry Labs

Problem 2

Anhydrous 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.
Step 1 of 4: Heat released at constant volume
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ΔU=ΔH−ΔngRT=−382.64−(−1.25)(8.314)(298)×10−3=−379.5 kJ per mol NHX3;qV=14.4032.0×43×(−379.5)=−227.7 kJ\Delta U = \Delta H - \Delta n_g RT = -382.64 - (-1.25)(8.314)(298) \times 10^{-3} = -379.5\ \text{kJ per mol } \ce{NH3};\quad q_V = \tfrac{14.40}{32.0} \times \tfrac{4}{3} \times (-379.5) = -227.7\ \text{kJ}
Analysis

Per mole of NHX3\ce{NH3}: ΔH=32(−285.83)−(−46.11)=−382.64\Delta H = \tfrac{3}{2}(-285.83) - (-46.11) = -382.64 kJ and Δng=−1.25\Delta n_g = -1.25 mol, so ΔU=−379.5\Delta U = -379.5 kJ. With n(OX2)=0.450n(\ce{O2}) = 0.450 mol limiting, n(NHX3)reacted=0.600n(\ce{NH3})_{reacted} = 0.600 mol and qV=0.600×(−379.5)=−227.7q_V = 0.600 \times (-379.5) = -227.7 kJ.

Common pitfall. At constant volume the measured heat is ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta n_g RT, not ΔH\Delta H.