test2s-1 - Chemistry 2000 Spring 2001 Section B Test 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 2000 Spring 2001 Section B Test 2 Solutions 1. (a) The liquid can only exist above the triple point. The triple point pressure is (from the graph) approximately 0.37 atm. (b) The normal boiling point is the equilibrium temperature at 1 atm. Again, we can just read this off the graph: 108 C. ¡ (c) The melting point increases with pressure indicating that the solid occupies a smaller volume than the liquid, i.e. that the solid is more dense. £ aCO aH2 aH2 O ¥ £ ¤ Q ¤ 2. The reaction ratio is ¢ (The graphite is solid so its activity is 1.) In order for the reaction to proceed from left to right, we must have Q K . Suppose that we take PCO PH2 0 and any PH2 O 0. Then Q 0, which is certainly smaller than K . ¢ § ¢ ¦ ¢ 3. (a) KI is not volatile because it is an ionic compound. (b) Tartaric acid can form a number of hydrogen bonds to the solvent, so it is not volatile. (c) Bromine is nonpolar so it should be volatile. ¥  12 © ¥¢ ¥ 10  ¢ 3 47  7 46 8 ¥ ¥ ¢ ¥ 10 10 ¥  ¨ ¢ pKa ¢ ¢ © ¢  10 46   5. ¢ Ka 14  pH Kw 1 47 10 aOH 0 0032 log10 aH 11 3 ¢¨ aH 0 0032. ¥ 4. Potassium hydroxide dissociates completely in water so aOH The Ka is quite small, so it is likely that a very small number of molecules of the acid will dissociate. Thus, aHClO 0 05. On the other hand, the acid is fairly concentrated, so we can probably safely ignore the water equilibrium. This implies that aH aClO . Therefore ¥   ¥ ¥ £ ¤© £ ¢ £ ¥ ¤© ¢¤© ¥ £  ¥ ¥ ¢ ¢ ¥ ¢ ¥ ¢ 1 4 38 ¥ ¤¨ aH 2 0 05 1 73 10 9 4 16 10 5 log10 aH £ 8  aH 2 aH pH aHClO aH 2 aHClO ¤© 10 aClO © © ¤©    3 47 aH ¨  Ka £ ¥  6. We need the mole fraction of water. To calculate this, we need to know the number of moles of solutes and solvent. ¢ ¥ ¥ ¥ ¥£ ¢ ¢ ¥ ¥ ¥ ¢ ¥ ¢¤ ¢ ¥ £ ¢   ¨ ¢ ¢ ¥  ¥ ¢ ¥ £ ¥  ¥ ¥ ¤ © ¥ 300 g 16 7 mol 18 015 g mol 16 7 mol 0 936 16 7 0 76 0 38 mol XH2O PH2 O 0 936 7373 Pa 6902 Pa ¥  ¢¤ ¢ ¥   XH2 O ¥    0 38 mol nH2 O  PH2 O ¥ ¢  4  4 and nSO2  nNH ¢ 2 SO4 ¥ n NH4 132 13 g mol  2MN 8MH MS 4MO 18 015 g mol 50 g 0 38 mol 132 13 g mol 2 0 38 mol 0 76 mol ¥ M NH4 2 SO4 MH2 O ¢ 7. The reaction is H2 g I2 g 2HI g , i.e. we get two HI’s for every H2 or I2 used. This gives us the stoichiometric relationships ¢$ ¢ ¢ ¢ PI2 initial " " # # #   and PI2 1 PHI 2 1 PHI 2 2 PI2 initial # PH2 initial PH2 PH2 2 PH2 initial PI2 ¥$ !  PHI Since P and a only differ by a factor of 1 atm, these relationships apply equally well in terms of activities. ¥ ¥ ¤ ¥ ¥ ¤ £¥ ¥ £¢ 2 2 792 1 355 0 105 ¢ ¥ ¤ a2 HI aH2 aI2 aHI 0 2 79 ¥ The equilibrium constant is therefore £ ¤ £¢ K aI2 1 50 0 105 54 7 ¥¥ aH2 2 75 1 355 initial final ...
View Full Document

This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

Ask a homework question - tutors are online