Question

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- 402.T mmH3 Total = Pa-Roate, Show your work for the calculation of Al S n , and normal boiling point. (2 pts experimental, 1 p each for literature and percen eror,6 ps cakulation box for sorrest calculation) 21 Howpor for water experimental literature percent error Pexcent10-3.xloo) Calculation fr ΔΗ0wapor for water 4 0 oo0 ASo vapor for water experimental literature percent error Recent 9- 189 Calculation for Δ.Soupor for water BT Barometric Pressure (mm Hg)pt Data Table pt ach blank Temperature of WaterC) Volume of Air Bubble (mL) 640 m 630 mi 525 ml 1.35 ml 4 30 ml 425 ml 4os mu Boo'C To o C 53 ОС 50.0 C 2.5 c 12o m Calculations: (os pt for each block) r ePr (am)n Poser S) Tae ) Tocafar ( 53k 233x 402.7326.80 0.349 343 x2. 131.3294 o39-094 33 k 242x 2152023 -1252 1.135 333 K 300x lo 111 84 021 1563 338 Kİ2.96x10° |5486-1 1809 | 0.24 323 K 3loxo 621.3111 o13 -2ol 215 5K3.63xlo 7 5164213. o.28 1-1.212 90 usation to increase until the mperature drops, evaporation asing the rate of condn low causing a he equilibrium on

Answer 1

The plot of ln(P_water) versus 1/T can be drawn as follows.

ln(P_water) = -4462.52556*(1/T) + 11.80159

The actual equation: ln(P_water) = (-$\Delta$H^o/R)(1/T) + $\Delta$S^o/R

Here, $\Delta$H^o/R = 4462.52556 K

i.e. $\Delta$H^o = 4462.52556 K * 8.314 J/mol.K = 37.101 kJ/mol

And $\Delta$S^o/R = 11.80159

i.e. $\Delta$S^o = 11.80159*8.314 J/mol.K = 98.118 J/mol.K

Now, the percentage error in $\Delta$H^o = {(44 - 37.101)/44}*100 = 15.68%

And the percentage error in $\Delta$S^o = {(118.9 - 98.118)/118.9}*100 = 17.48%