Question

1)

(a) The vapour pressure of ethanol (C2H5OH) at 15 °C was found to be 5866 Pa, and at 40 °C it was measured to be 21332 Pa. Calculate the enthalpy of vaporisation for ethanol.

(b) Draw the Lewis structures for both an ethanol molecule and a water molecule then label and annotate your drawings to illustrate what intermolecular interaction(s) may exist between the two different molecules.

Answer 1

- solution Ti = 15°C + 273 = 288k, P= 5866 Pa Tz = 40°c +273 = 313kg P2 =21332 pa R = 8.314 J 'moet sHwap (ethance) = ? - we know that, 2. 303 R strap / is dog eng - Strana (t t ) dogle-dog e, Allohen ato te) .. log (21332). Jag (5869) = strane prava ( 2936 312x] 2.303 XR AHVAR 1998 K 213 2.30 3 x 8.114 . : ( ... 4.3290 - 3.7683 = ДНуар 19.147 3 13 k 288 8 313 :. 0.5607 = 1 Dhvap / 25 19.1477 I 90144 0.5607 x 19,147 x 90144 -: AH vap cethanol) = Hvoere. = 38.710,44 Imol "Va (e Pa )

Lewis structure of ethanol (CH₃ CH₂OH) c→ 47 x2 = 8e 0-6e- = Ge- H léx6 = 6e 200- H H I 4-5-6-8-H & I-U- I-U- Hico H I (octet Rule) 2 I Lewis structure of water (H₂O) H → 10*2=28 096e* x1 =665 | F-5-H = H2O: H. Cocter Rule) » Intermolecular interaction between The two different molecule (ethanol two - H HH 0-H . water (ethanol) +5 H Ö-H water non > Intermolecolar between ethanol and Water