Question 3

Which of the following correctly identifies which has the higher first-ionization energy, Cl or Ar, and supplies the best justification? (A) (B) (C) (D) Cl, because of its higher electronegativity Cl, because of its higher electron affinity Ar, because of its completely filled valence shell Ar, because of its higher effective nuclear char e

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest. • From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus. Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy. • From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy nuclear char — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons. The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

Question 4

  • Sealed rigid vessel = constant volume

    There are two cases on which equilibrium depends. These are: Addition of an inert gas at constant volume: When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium. Addition of an inert gas When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Question 6

Real Gases and Limitations of the Kinetic Theory (a) I ligh pressure (b) I -ow temperature The ideal gas model breaks down at high pressures and low temperatures. • high pressure: volume of particles no longer negligible • low temperature: particles move slowly enough to interact

Question 12

Ionization Energies for Period 2 Main Grou Elements Element Lithium (Li) Beryllium (Be) Boron (B) Carbon (C) Ni trogen (N) Oxygen (O) Fluorine (F) Neon (Ne) Electron Configuration \[He\] 2sl \[He\] 2s \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P Ionization Energy 520 kJ/m01 899 kJ/m01 801 kJ/m01 1086 kJ/m01 1400 kJ/m01 1314 kJ/m01 1680 kJ/m01 2081 kJ/m01 @ 000 2s @ 000 2s

Why the drop between groups 5 and 6 (N-O and P-S) ? Once again, you might expect the ionisation energy of the group 6 element to be higher than that of group 5 because of the extra proton. What is offsetting it this time? N 1s22s22px12py12pz1 1st I.E. 1400 kJ mol-I 0 1s22s22px22py12pz1 1st I.E. 1310 kJ mol-I The screening is identical (from the 1s2 and, to some extent, from the 2s2 electrons), and the electron is being removed from an identical orbital. The difference is that in the oxygen case the electron being removed is one of the 2px2 pair. The electrons in the same orbital means that the electron is easier to be. The drop in ionisation energy at sulphur is accounted for in the same way.

Question 27

10 Cuvette 1 I —1 cm Microplate (96- or 384-well) 1 10 I is variable —5 < mm < —10

4

Question 30

  • Common Hydgron bonds: H-O, H-N

    Hydrogen bond Water molecule Hydrogen bond ...H-N-H Ammonia molecule

    fluorine oxygen oxygen nitrogen nitrogen sulfur Donors (in an acid) (in water, or alcohols) Donor strength medium medium weak weak Acceptoe -o-p,-o-s-, H20, H-o\_c, C=N-C 7t -electrons Acceptor strength very strong Strong strong medium—strong medium medium medium—weak weak

Question 38

Because of the incomplete dissociation of the acid, the reaction is in equilibrium, with an acid dissociation constant, Ka, which is specific to that acid. point are the same. Therefore, at the half-equivalence point, the pH is equal to the pKa. Experiment 6 Titration Il www.uccs.edu/Documents/chemistry/nsf/106%20 Expt6V-Titration2. pdf

25.00 0.100 M equivalence point 1/2 equivalence point 0.100 M NaOH added

Question 40

Which of the following best explains why more energy is required for the process occurring at 110 K than for the process occurring at 90 K ? (A) Intermolecular attractions are completely overcome d va rization. (B) Intermolecular attractions in the solid phase are weaker than in the liquid phase. (C) Electron clouds of methane molecules are less polarizable at lower temperatures. (D) Vaporization involves a large increase in temperature.

Question 49

Trend \#2 — For molecules with a given functional group, boiling point increases with molecular weight. Look at the dramatic increases in boi ing points as you increase molecular weight in all of these series: In a given series, boiling point increases with molecular weight. Why? Boiling point -42 oc ALCOHOLS HO Boiling point CIDS Boiling point Boiling oint 97 ac 141 oc -24 oc HO 117 oc Ha 164 oc 35 ac 36 •c HO 138 oc Ha 186 oc 89 ac 69 •c 158 oc Ha 202 oc Increasing surface area gives rise to increased Van Der Waals interactions Here's the question: How, exactly do intermolecular forces increase as molecular weight increases? Well, the key force that is acting here are Van der Waals dispersion forces, which are proportional to surface area. So as you increase the length of the chain, you also increase the surface area, which means that you increase the ability of individual molecules to attract each other.

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