Question

Calculate the change in entropy when 1 mol of solid iodine at a temperature of 360 K is heated at constant pressure to produce liquid iodine at a temperature of 410 K. The constant pressure molar heat capacity of solid iodine is 54.44 J $K^{-1}$ $mo{l}^{-1}$ and liquid iodine is 80.67 J $K^{-1}$ $mo{l}^{-1}$.
The melting point of iodine is 387 K and molar enthalpy of fusion of iodine is 7.87 kJ/mol.

• A. 8.6 $J{K}^{-1}$ $mo{l}^{-1}$
• B. 28.9 $J{K}^{-1}$ $mo{l}^{-1}$
• C. 20.3 $J{K}^{-1}$ $mo{l}^{-1}$
• D. 11.7 $J{K}^{-1}$ $mo{l}^{-1}$

Answer 1

The correct option is B 28.9 $J{K}^{-1}$ $mo{l}^{-1}$

Given,
enthalpy of fusion of iodine= 7.87 kJ/mol
heat capacity of solid iodine $=54.44J{K}^{-1}mo{l}^{-1}$
heat capacity of liquid iodine $=80.67J{K}^{-1}mo{l}^{-1}$

Change in entropy in heating iodine from 360 K to 387 K
$=C_{p}ln\frac{T_2}{T_1}$
$=54.44ln\frac{387}{360}$
$=3.93J{K}^{-1}mo{l}^{-1}$

Change in entropy of fusion of solid iodine $=\frac{\Delta H}{T}$
$=\frac{7870}{387}=20.3J{K}^{-1}mo{l}^{-1}$

Change in entropy in liquid iodine from 387 K to 410 K
$=C_{p}ln\frac{T_2}{T_1}$
$=80.67ln\frac{410}{387}$
$=4.65J{K}^{-1}mo{l}^{-1}$

$\text{Total change in entropy}=3.93+20.3+4.65=28.9J{K}^{-1}mo{l}^{-1}$