QJEEM17Ch6295: JEE Main 2017 Chemistry - Class 11 JEEM17Ch6295 Solution: The enthalpy change on freezing of 1 mol of water at 5 o C to…

(1) From 5 $^\circ$ C to 0 $^\circ$ C :

${H_1} = {C_P} \times (\Delta T)$

$= 75.3 \times (278 - 273)$

= 377 J/mol

= 0.377 KJ/mol

(2) Phase change (0 $^\circ$ C to 0 $^\circ$ C) :

${H_2} = - \Delta {H_{fusion}}$

= $-$ 6 KJ/mol

(3) 0 $^\circ$ C to $-$ 5 $^\circ$ C (Ice phase) :

${H_3} = - {C_{P(s)}} \times \Delta T$

$= - 36.8 \times (273 - 268)$

= $-$ 0.184 KJ/mol

$\therefore$ ${H_{Total}} = {H_1} + {H_2} + {H_3}$

$= 0.377 - 6 - 0.184$

= $-$ 5.837 KJ/mol