Ans_Ex2 - Nielsen CHM1041 Answer to Exercise Set 2 1....

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Nielsen Bhaskar/PC/1713S/J 3/05 1 CHM1041 Answer to Exercise Set 2 1. Potassium iodide is more soluble in methyl alcohol, CH 3 OH, than in benzene, C 6 H 6 , whereas naphthalene, C 10 H 8 , is more soluble in benzene. Suggest explanations for both cases. The solvents: a. Methyl alcohol, H 3 COH is a polar covalent molecule capable of London interaction and hydrogen bonding (extreme form of dipole-dipole) b. Benzene, C 6 H 6 , is a nonpolar molecular capable of only London forces. Solid potassium iodide consists of K +1 ions and I -1 ions arranged in a strong crystal lattice. Methyl alcohol is able to dissolve KI for basically two reasons: the permanent dipoles of the alcohol molecules align with the full unit charged K +1 and I -1 ions, collectively solvating and stabilizing the individual ions in solution; the relatively large dielectric constant of methyl alcohol ( =33 ) allows for an “easier” separation of the K +1 and I -1 ions. Nonpolar benzene is unable to significantly solvate the ions and has a very low dielectric constant ( = 2.3 ). KI is essentially insoluble in benzene. Naphthalene is nonpolar hydrocarbon and is capable of interacting and dissolving with nonpolar benzene molecules via London forces. Polar methyl alcohol molecules would prefer to interact and associate with themselves rather than with the nonpoar naphthalene molecules. Methyl alcohol molecules attract one another by London forces and permanent dipole-dipole alignment (actually H-bonding), contrasted with only London attraction between methyl alcohol and naphthalene molecules. Naphthalene would be insoluble in methyl alcohol. 2. When a solid substance dissolves, two processes are of prime importance: the breaking way of particles from the solid and the joining of solute and solvent particles. In terms of these two processes, explain how the dissolving of sodium hydroxide can be exothermic, whereas for ammonium chloride the process is endothermic. ( ) () 2 HO +1 +1 soln sa q 4 a q NaOH Na + OH + heat H - ⎯⎯⎯→ + ( ) 2 +1 -1 4 soln s4 a q a q NH Cl + heat NH + Cl H + + The total enthalpy change for the dissolution process, H soln , is the sum of the heat required to break apart the crystal (– cryst ) and the heat released by the salvation process solv hyd Ho rH + + soln cryst solv H= - H+H ++
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Nielsen Bhaskar/PC/1713S/J 3/05 2 Be careful with the above defined quantities and signs. Recall, H cryst is the heat released as gaseous ions form one mole of solid crystal. As defined, H cryst is negative. During dissolution, essentially the reverse process is occurring and heat would be required; the sign of H cryst is reversed, i.e. , - H cryst . H solv is the heat released when the ions of one mole of an ionic compound are solvated or, specifically for water, hydrated. H solv would be negative.
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This note was uploaded on 12/12/2011 for the course CHM 2041 taught by Professor Staff during the Fall '09 term at Santa Fe College.

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Ans_Ex2 - Nielsen CHM1041 Answer to Exercise Set 2 1....

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