Question

One mole of liquid naphthalene at its freezing point of 80 ˚C is brought into contact with a very large ice-water bath, which remains at 0 ˚C as the naphthalene freezes and cools to 0˚C. The enthalpy of fusion of naphthalene is 19.2 kJ mol-1 and ~ C p for solid naphthalene is 180 J K-1 mol-1. Calculate the decrease in entropy for the naphthalene and the increase in entropy for the overall system (naphthalene + ice-water bath).

Answer 1

Hl chages of maphhalone steps: follashy The can be e rikem in the coc) Naptalone (,20)ajtkaoe Naphthalene Sep 1 19.2 x 103 353 19.2 kJ.mot\ xI mol Nova, 4S J.K S4-39 J-K (80+ 213) K and ASm Ln 233 K -46. 26 J.K- The decrease in for the naphthalene = (54.39+46-2) JK Now, the amount of heat tramsfered frohe ice-bath to 80-2 K 19.2 xI0 J mo 1 mol x 180 J mo X"! K'x x Inmol (19 200 144 0 o) 33600 J 33600 J . rInoieabe im of dhe ice-batt = 123. 08 J-K* 23 K for the overale egtem 123-08-100-65) 3- . The increase n An - 22. 43 J. K