JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 18)
The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is :
(Given $$\Delta $$fusH = 6 kJ mol$$-$$1 at 0oC,
Cp(H2O, $$\ell $$ = 75.3J mol$$-$$1 K$$-$$1)
Cp(H2O s) =36.8 J mol$$-$$1 K$$-$$1)
(Given $$\Delta $$fusH = 6 kJ mol$$-$$1 at 0oC,
Cp(H2O, $$\ell $$ = 75.3J mol$$-$$1 K$$-$$1)
Cp(H2O s) =36.8 J mol$$-$$1 K$$-$$1)
5.44 kJ mol$$-$$1
5.81 kJ mol$$-$$1
6.56 kJ mol$$-$$1
6.00 kJ mol$$-$$1
Explanation
(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
Comments (0)
