1. Estimate the temperature rise (DT) of the final solution (assuming it is so well insulated that it loses none of the heat of reaction to the environment).
   • The neutralization reaction is
     H⁺(aq) + OH^-(aq) H₂O(liq)

     whose reaction enthalpy is

     DH° = DH_f°[H₂O(liq)] - DH_f°[H⁺(aq)] - DH_f°[OH^-(aq)]
     DH° = -285.9 - (0.0) - (-230.0) kJ/mol = -55.9 kJ/mol

     That's the heat evolved when one mole each, the amount in 1 L (each) of 1 M acid and base, neutralizes. All of that goes toward heating up the 2 L final solution, which weighs 2 kg. The heat capacity of that solution is

     mC_P = 2000 g × 4.184 J g^-1 (°C)^-1 = 8,368 J (°C)^-1
     mC_P = 8.368 kJ / °C

     So the temperature change, DT, becomes

     DT = -DH° / mC_P = 55.9 kJ/mol / (8.368 kJ/°C) = 6.68 °C

2. Let the original temperatures of the reactant solutions be 25°C. If the solution heat capacities stayed about the same regardless of acid/base concentration, how concentrated would the acid and base have to be to actually raise the temperature of the resultant solution to water's boiling point? (Let them both have the same concentration and again assume no heat lost to the surroundings. Now you see how dangerous it is to titrate even moderately concentrated acids or bases; always dilute them - quantitatively - first.)
   • If we wanted to raise the solution to boiling this way, we'd have to increase its temperature by 75°C (from 25°C to 100°C) not just 6.68°C. So instead of 1M solutions, we should release (75/6.68)=11.2 times as much heat using 11.2 M solutions. Fairly concentrated, all right, but the resultant boiling would splatter hot acid and base on us, a situation to be avoided!

sucrose + H₂O fructose + glucose

Find the standard enthalpy of that reaction from the heats of combustion of sucrose, fructose, and glucose which are 5.6467, 2.8267, and 2.8158 MJ / mol, respectively. (Watch those significant figures! You're going to get small differences of large numbers: the best recipe for losing accuracy.)

• The combustion reactions in question are:
  sucrose + 12 O₂(g) 12 CO₂(g) + 11 H₂O(liq),   DH₁°
  fructose + 6 O₂(g) 6 CO₂(g) + 6 H₂O(liq), DH₂°
  glucose + 6 O₂(g) 6 CO₂(g) + 6 H₂O(liq), DH₃°

  If we reverse the last two reactions (and negate their DH) and add them to the first reaction, all the CO₂ molecules will cancel out and all but one of the waters will too. That will give us:

  sucrose + H₂O fructose + glucose, DH₁° - DH₂° - DH₃°

  for a hydrolysis enthalpy of
  5.6467-2.8267-2.8158 MJ/mol = 4.2×10^-3 MJ/mol = 4.2 kJ/mol

1. sucrose, C₆H₂₂O₁₁(s)
   • 6 C(graphite) + 11 H₂(g) + 5½ O₂(g) C₆H₂₂O₁₁(s)
     (Note that we must use fractional stoichiometric coefficients if we want the reaction to produce on mole of sucrose.)

2. PBr₅(s)
   • P(red) + 2½ Br₂(liq) PBr₅(s)
     (Yes, it's really P₄, but no one uses that.)

3. U(CO)₆(g) (used to separate isotopes by gaseous diffusion)
   • U(s) + 6 C(graphite) + 3 O₂(g) U(CO)₆(g)

4. NH₄NO₃(s)
   • N₂(g) + 2 H₂(g) + 1½ O₂ NH₄NO₃(s)

5. NaCl(s)
   • Na(s) + ½ Cl₂(g) NaCl(s)

(See your periodic table conversion chart for the relationship of atm to psi.)

• n = PV/RT

  P_absolute = P_gauge + P_external = 60 psi (1 atm / 14.7 psi) + 1 atm
  P_absolute = 5.08 atm

  n = 5.08 atm × 1 L / [(0.08206 atm L mol^-1 K^-1)×(298.16 K)]
  n = 0.21 moles air

• n = PV / RT

  P_X = P_total - P_H2O
  P_X = 1.0133 atm - 23.76 torr (1 atm / 760 torr)
  P_X = 0.9820 atm

  n = 0.9820 atm × 1 L / [(0.08206 atm L mol^-1 K^-1)×(298.16 K)]
  n = 0.04014 moles of gas X

  MW_X = m_X / n_X = 1.6035 g / 0.04014 mol = 39.95 g mol^-1 (argon)

(Why didn't he just use a Geiger counter, one wonders?)

• v_He / v_UF6 = ( MW_UF6 / MW_He )^½ or
  MW_UF6 / MW_He = ( v_He / v_UF6 )² = 9.337² = 87.18
  MW_UF6 = 87.18 MW_He = 87.18×4.003 g/mol = 349.0 g/mol

  AW_U = 349.0 g/mol - 6×AW_F = 349.0 g/mol - 6(19.00)
  AW_U = 235.0 g/mol tada

• DE_photon = -DE_atom = + R_H( n_final^-2 - n_initial^-2 ) = hn = hc / l

  And the Balmer (visible) series all have n_final = 2. So the first member of the series would have n_initial = 3, and the fourth such transition would have n_initial = 6.

  l = hc / [R_H(n_final^-2 - n_initial^-2)
  l = 6.626×10^-34 J s × 2.998×10⁸ m s^-1 / [2.18×10^-18 J (6^-2 -2^-2)]
  l = 4.10×10^-7 m = 0.410 mm

• From our Periodic Table, element 112 is the last shown and ends the transition metals of the 7th row; hence it has a [Rn]7s²5f¹⁴6d¹⁰7p⁰ configuration. The latest element known is 118 which is the last 7th row element and must have the next Noble Gas configuration, [Rn]7s²5f¹⁴6d¹⁰7p⁶ which we'll all have to get used to as the [Uuo] noble gas. Right.

  Still no 5g electrons. Indeed after 7p⁶ we'd expect to see 8s¹ and 8s² which would be elements 119 and 120, but then what?

  Enters now the handy-dandy electron configuration cheat table given in class:
  

      /
     1s 
    /  /
      /  /
     2s 2p
    /  /  /
      /  /  /
     3s 3p 3d
    /  /  /  /
      /  /  /  /
     4s 4p 4d 4f
    /  /  /  /  /
      /  /  /  /  /
     5s 5p 5d 5f 5g
    /  /  /  /  /  /
      /  /  /  /  /  /
     6s 6p 6d 6f 6g 6h
    /  /  /  /  /  /  /
      /  /  /  /  /  /  /
     7s 7p 7d 7f 7g 7h 7i
    /  /  /  /  /  /  /  /
      /  /  /  /  /  /  /  /
     8s 8p 8d 8f 8g 8h 8i 8k
    /
  

  The red / shows the filling order after element 118 finishes off the 7p⁽⁶⁾ row. It proceed through the 8s² element 120 (Ubn) and then reaches back to the next diagonal beginning with the 5g¹ element 121 (or Ubu). That means Ubu will be the first element to have 5g valence electron(s) and the new chemistry that will go with them. (Unfortunately, it'll only live a few microseconds before decaying radioactively. And after 121, we won't see any new chemistry until 6h¹ at 169!)

• Al has electronic structure [Ne] 3s² 3p¹ which means that Al³⁺ has the electronic structure of [Ne], its nearest noble gas configuration.

  But Ga has the structure [Ar] 4s² 3d¹⁰ 4p¹ and still follows Al in its preference for Ga³⁺. That suggests that it unloads the same electrons (both s and the p) to give its Ga(III) ion the configuration [Ar] 3d¹⁰ which is not a noble gas configuration. Rather, we're going to have to fall back on the aufbau principles and Hund's rules that say that ½-filled and filled shells (of any kind) offer special stability. So the 3d¹⁰ is the reason that Ga is happy as Ga³⁺.

1. Radius(Al³⁺) vs. Radius(B³⁺)
   • Al lies directly below B in the Periodic Table. As we go down a column, the valence electron wave functions have one more node per row to accommodate and have to be larger as a consequence. The same will be true of their ions. But these are ions, and noble gas configuration ions at that. So these ions compare as would [Ne] vs [He] in electronic structure, albeit with that extra 3+ charge to hunker things down. Still since Ne lies beneath He, the Ne-like thing will be larger than the He-like one. So all of that makes B³⁺ smaller than Al³⁺.

2. Radius(K) vs. Radius(Ca⁺)
   • K and Ca⁺ are isoelectronic; they have the same number of electrons. But those electrons see one extra proton in Ca⁺ and are pulled closer to their nucleus, making Ca⁺ smaller than K.

3. 1st I.E.(Cl) vs. 2nd I.E.(K)
   • All second ionization are larger than any first ionization energy because a second ionization is trying to pull an electron away from a positively charged ion while first ionizations are removing an electron from a neutral atom. So 1st I.E. (Cl) is smaller than 2nd I.E. (K).

4. 1st I.E.(Na) vs. 1st I.E.(Mg)
   • Ionization energies increase across the rows because each electron added fails to completely cancel out each proton in the effective charge that all the other valence electrons see. So since Mg follows Na in the 3rd row, Na has the smaller 1st I.E.

5. 1st E.A.(O) vs. 2nd E.A.(O)
   • First electron affinities are much more impressive than second electron affinities because in the latter, you're trying to add a negative charge to a negative ion rather than a neutral atom! The negatives repel, reducing the stability of O^2- ion. So 1st E.A. (O) is lower (more negative) than 2nd E.A. (O).