JEE Main On 8 April 2017 Question 28
Question: The enthalpy change on freezing of 1 mol of water at $ 5{}^\circ C $ to ice at $ -5{}^\circ C $ is : (Given $ {\Delta _{fus}}H=6kJ,mo{l^{-1}} $ at $ 0{}^\circ C $ , $ C _{p}(H _2O,\ell )=75.3J,mo{l^{-1}}{K^{-1}}, $ $ C _{p}(H _2O,s)=36.8J,mo{l^{-1}}{K^{-1}}) $ [JEE Online 08-04-2017]
Options:
A) $ 6.00,kJ,mo{l^{-1}} $
B) $ 5.81,kJ,mo{l^{-1}} $
C) $ 5.44,kJ,mo{l^{-1}} $
D) $ -6.56,kJ,mo{l^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
For questions like these, go step wiseI]
Water (liquid) at 5C→ water(l) at 0C
△t=5k
Cp=75.3J$mol^{−1}K^{−1}$
molesof$H_2O$=1mol
△H=mcp△t=1×75.3×5=376.5J
II] Water (liquid) at 0C→ water(l) at 0C
Latentheat(L)=△Hfus
L=6KJmol−1moles($H_2O$)=1△H=ml=6000J×1=6000J
III] Water (liquid) at 0C→ water(l) at −5C
△t=5kCp=36.8J$mol^{−1}K^{−1}$
moles($H_2O$)=1
△H=mcp△T=1×36.8×5=184J
Summation of all △H=376.5+6000+184=6560.5J=6.56kJ
Since cooling of water is an exothermic reaction, △H is always negative
△H=−6.56kJ
