Chemistry Labs

Problem 1

A compound Q (molar mass 122.0 g mol−1^{-1}) consists of carbon, hydrogen and oxygen. PART A. The standard enthalpies of formation of COX2(g)\ce{CO2(g)} and HX2O(l)\ce{H2O(l)} at 25.00 ∘C25.00\ ^\circ\mathrm{C} are −393.51-393.51 and −285.83-285.83 kJ mol−1^{-1}; R=8.314R = 8.314 J K−1^{-1} mol−1^{-1} (atomic masses: H = 1.0, C = 12.0, O = 16.0). A 0.6000 g sample of solid Q is combusted in excess oxygen in a bomb calorimeter initially containing 710.0 g of water at 25.000 ∘C25.000\ ^\circ\mathrm{C}. After reaction the temperature is 27.250 ∘C27.250\ ^\circ\mathrm{C} and 1.5144 g of COX2(g)\ce{CO2(g)} and 0.2656 g of HX2O(l)\ce{H2O(l)} are produced. (1.1) Determine the molecular formula of Q and write a balanced equation, with states, for its combustion. (1.2) Given the specific heat of water 4.184 J g−1^{-1} K−1^{-1} and the internal energy change of the reaction ΔU∘=−3079\Delta U^\circ = -3079 kJ mol−1^{-1}, calculate the heat capacity of the calorimeter (excluding the water). (1.3) Calculate the standard enthalpy of formation ΔHf∘\Delta H_f^\circ of Q. PART B. The distribution of Q between benzene and water at 6 ∘C6\ ^\circ\mathrm{C} gave the equilibrium concentrations (cB,cW)(c_B, c_W) in mol dm−3^{-3}: (0.0118, 0.00281), (0.0478, 0.00566), (0.0981, 0.00812), (0.156, 0.0102). Assume only one species of Q exists in benzene and that Q is a monomer in water. (1.4) Show by calculation whether Q is a monomer or a dimer in benzene. (1.5) For an ideal dilute solution, ΔTf=R(Tf0)2Xs/ΔHf\Delta T_f = R(T_f^0)^2 X_s/\Delta H_f; the molar mass of benzene is 78.0 g mol−1^{-1}, pure benzene freezes at 5.40 ∘C5.40\ ^\circ\mathrm{C} at 1 atm and ΔHf=9.89\Delta H_f = 9.89 kJ mol−1^{-1}. Calculate the freezing point of a solution of 0.244 g of Q in 5.85 g of benzene.
Step 4 of 5: Monomer or dimer in benzene?
Intuition

Test both hypotheses against the data: the partition law that stays constant reveals the aggregation state.

cBcW2=1.49, 1.49, 1.49, 1.50×103 dm3mol−1 (constant)  ⇒  2 Q⇌QX2 in benzene\frac{c_B}{c_W^2} = 1.49,\ 1.49,\ 1.49,\ 1.50\times10^{3}\ \text{dm}^3\text{mol}^{-1}\ (\text{constant}) \;\Rightarrow\; \ce{2Q <=> Q2}\ \text{in benzene}
Analysis

If Q dimerizes in benzene, cB∝cW2c_B \propto c_W^2, so cB/cW2c_B/c_W^2 should be constant while cB/cWc_B/c_W varies. The ratios cB/cWc_B/c_W = 4.20, 8.44, 12.1, 15.3 drift strongly, whereas cB/cW2c_B/c_W^2 stays at ≈1.49×103\approx 1.49\times10^{3} — Q is a dimer in benzene.