Question 1 (a)

(ii) Determine the value of AHS In for LiCl in kJ/molrxn. 1 mol LiCl 10.0 gLiC1 x = 0.236 mol LiC1 42.39 g LiC1 -9.47 k.J 00.1 kJ/molrxn 0.236 mol LiC1 1 point is earned for the number of moles of LiCl. 1 point is earned for the correct Al-Isoln and the correct Sign.

Question 1 (c)

(c) Using principles of atomic structure, explain why the Na ion is larger than the Li+ ion. a ion are ina hi her rinci al ener level the valence 1 point is earned for a correct explanation elecfrons in the Li ion. Electrons inhi er based on occupied principal energy levels. principal energy levels are, on average farther from the nucleus.

Question 1 (d)

(d) Which salt, LiCl or NaCl, has the greater lattice enthalpy? Justify your answer. LiCl. Because the Li ion is smaller than the Na ion, ions in LiCl are stronger than in NaCl. This results in a greater lattice enthalpy. 1 point is earned for the correct choice and justification.

Question 1 (f)

(f) The lattice enthalpy of LiCl is positive, indicating that it takes energy to break the ions apart. in LiCl. However, the dissolution of LiCl in water is an exothermic process. Identify all particle-particle interactions that contribute significantly to the exothermic dissolution process being exothermic. For each interaction, include the particles that interact and the specific type of intermolecular force between those particles. There are interactions between Li+ ions and polar water molecules and between Cl- ions and polar water molecules. These ar ion-dipole interactions. 1 point is earned for identifying the particles that interact. 1 point is earned for correctly identifying the type of interaction.

Positive ends of polar molecules are oriented toward negatively charged anion Negative ends of polar molecules are oriented toward positively charged cation

Solute—solu te (positive AH) Solvent—solvent (positive AH) Solute + solvent Solution Solven t—solute (negative AH) SO In Solute—solu te (positive AH) Solvent—solvent (positive AH) Solute + solvent (a) Negative AH so In Solven t—solute (negative AH) Solution - - (b) Positive AH soln

Question 2 (e)

The HCOg- ion has three carbon-to-oxygen bonds. Two of the carbon-to-oxygen bonds have the same length and the third carbon-to-oxygen bond is longer than the other two. The hydrogen atom is bonded to one of the oxygen atoms. In the box below, draw a Lewis electron-dot diagram (or diagrams) for the HC03- ion that is (are) consistent with the given information.

Question 2 (f)

(f) A student prepares a solution containing equimolar amounts of HC2H302 and NaC2H302. The pH of the solution is measured to be 4.7. The student adds two drops of 3.0 M HN03(aq) and stirs the sample, observing that the pH remains at 4.7. Write a balanced, net-ionic equation for the reaction between HN03(aq) and the chemical species in the sample that is responsible for the pH remaining at 4.7. C2H302- + H30+ HC2H302 + H20 OR C2H302- + H+ HC2H302 I point is earned for a correct equation.

Presence of HF counteracts addtion of base; pH increase is small After addition Initial buffered solution Presence of F- counteracts addition of add; pH decrease is small After addition of H+ Add H+ Add OH H20 + F HF -F OH A Figure 17.2 Buffer action. The pH of an HF/F- buffered solution changes by only a small amount in response to addition of an acid or base.

Question 3 (d)

(d) Explain why 12 is a solid at room temperature whereas Br2 is a liquid. Your explanation should clearly reference the types and relative strengths of the intermolecular forces present in each substance. Both Br2 and 12 molecules are nonpolar molecules, therefore the only possible intermolecular forces are London dispersion forces. The London dispersion forces are stronger in 12 because it is larger in size with electron cloud. The stronger London dispersion forces in 12 result in a higher melting point,which makes 12 a solid at room temperature. 1 point is earned for identifying the forces in each substance as London dispersion forces. 1 point is earned for explaining why the forces are stronger in 12 than in Br2 .

Question 5 (b)

![Zeroth Order First Order Second Order Differential rate law Concentration vs. time Integrated rate law Straight-line plot to determine rate constant Relative rate vs. concentration Half-life Units of k, rate constant Rate = o o o o ALA] At Time Slope = —k Time Rate — o o ALA] = MA] At Time o o —kt or Slope = —k Time [A], M Rate, M/S o o o o 2 3 M Rate, M/s 2k 2 3 Rate = 2 3 A[A] = At Time Slope = k Time Rate, M/S 4 9 2 3 0.693 l/s ](./media/image22.png)

Order Rate Law Rate = Rate = kl Concentration - Time Equation = kot 2.303 Half Life 2k o o .aga Graphical

Zero order 1st order 2nd Order (b) Plots from integrated rate equations: Zero order Conc. 1st order Conc. 2nd order (Conc.)2 3rd order 3rd order (Conc.)

Question 6 (a)

(a) Considering the value of K for the reaction, determine the concentration of Ba(EDTA)2-(aq) in the 100.0 mL of solution. Justify your answer. Based on the K value, the reaction goes essentially to Ba2+(aq) is the limiting reactant. The concentration of Ba2+ when the solutions are first mixed but before any reaction takes place is 0.20 M/2 = 0.10 M. Thus the equilibrium concentration of Ba(EDTA)2-(aq) is 0.10M. 1 point is earned for indicating that the equilibrium concentration of Ba(EDTA)2-(aq) is the same as the original concentration of Ba2+ when the solutions are mixed. 1 point is earned for the concentration with appropriate calculations.

Question 7 (a)

  • Since your buret is graduated to 0.1 mL, you will read your buret to 0.01 ml. The second decimal place is an estimate, but should be recorded.

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