(d) two points; one for half-neutralized; one for mmol calcs.
pH = 7.49 therefore [H+] = 3.2 x 10¯8
pH = pKa , or [H+] = Ka.
So [OCl¯] / [HOCl] = 1 , or solution must be half neutralized.
initial mmol HOCl = 50.0 x 0.20 = 10.0 mmol
mmol NaOH required = 10.0 ÷ 2 = 5.0 mmol
(e) one point
From equation, 1 mol H+ produced for each 1 mole of HOCl produced, thus [H+] = [HOCl] = 0.065 therefore pH = 1.19
(a) two points; one for line of answer
- 232.7 J/K = S° (C2H6) - (261.4 + 200.9) J./K
S° (C2H6) = 229.6 J/K
units ignored; 1 point earned for 98.9 J/K; 1 point lost if stoichiometry is not implied in process
(b) three points total; one point each portion; any value for T (e.g., 273 K or 298 K) is allowable:
[delta]H° = (- 84.7 kJ) - (226.7 kJ) = -311.4kJ
= - 311.4kJ - (298 K) (- 0.2327kJ/K)
= - 311.4 kJ + 69.3 kJ
= - 242.1 kJ
Negative [delta]G° therefore reaction is spontaneous, or Keq > 1 therefore reaction is spontaneous, or products are favored at equilibrium.
(c) two points
ln K = 242.1 ÷ [(8.31 x 10¯3) (298)] = 97.7
K = 3 x 1042 (1,2,or 3 significant figures acceptable)
(d) two points; first point earned for correct substitution and correct number of bonds, second point earned for setting equal to [delta]Hrxn and correct calculation of answer; no points earned for "extrapolation" techniques to find carbon-carbon triple bond energy; E* is the energy of the carbon-carbon triple bond.