Question

From the data below, calculate the total heat (in J) needed to convert 0.100 mol of gaseous ethanol at 300.0°C and 1 atm to liquid ethanol at 25.0°C and 1 atm: b.p. at 1 atm: 78.5°C AH : 40.5 kJ/mol vap Cliquid: 2.45 J/g °C Cgas: 1.43 J/g.°C

Answer 1

n=0. I mole =(0.1846) = 416 gm. ethand ethanol Now at boiling point transition from diq. to gas happens. So to convert ol mol ethano) at 300°c and latm to 78.5° and lato needed heat TH= mcgas XAT (m = mass of ethano =(416x143 X 221'5) Coars = 143 J/goc | 4T = (300 -78.5) – 2210500 = 1.457 KJ To convert i mot ethanol gas to liquid we need 40.5.6K5 So for o'l mol a a a aa 4:05KJ (AH2)

And now to convert liq ethanol at 78.5°c at lalm to 25°c at lahm 1. we need - | 44: C: 4T - 4 x 2 45 x (785 -25) 1 = 0.60063 kJ So Total heal needed to convert ol mol of gaseous ethanol at 300°C and lahm to liquid ethanol at 25c and lalm is- ( 44 + alty + 4H3) = 1457 + 405 + 0'603) KJ = 6'11 KI = 6110 J (Aws).