Question

The enthalpy change on freezing of 1 mol of water at $5^{o}C$ to ice at $-5^{o}C$ is:

$[{Given:△}_{fus}H=6kJ{mol}^{-1}at{0}^{o}C,C_p(H_{2}O,l)=75.3J{mol}^{-1}{K}^{-1},C_p(H_{2}O,s)$$=36.8J{mol}^{-1}{K}^{-1}]$

• A. $-6.56kJ{mol}^{-1}$
• B. $-5.81kJ{mol}^{-1}$
• C. $-6.00kJ{mol}^{-1}$
• D. $-5.44kJ{mol}^{-1}$

Answer 1

The correct option is A $-6.56kJ{mol}^{-1}$

For questions like these, go step wise
I] Water (liquid) at $5C\to$ water(l) at $0C$
$△t=5k{C}_p=75.3J{mol}^{-1}{K}^{-1}molesof{H}_{2}O=1mol△H=m{c}_p△t=1\times 75.3\times 5=376.5J$
II] Water (liquid) at $0C\to$ water(l) at $0C$
$Latentheat\left(L\right)=△H_{fus}L=6KJ{mol}^{-1}moles\left(H_{2}O\right)=1△H=ml=6000J\times 1=6000J$
III] Water (liquid) at $0C\to$ water(l) at $-5C$
$△t=5k{C}_p=36.8J{mol}^{-1}{K}^{-1}moles\left(H_{2}O\right)=1△H=m{c}_p△T=1\times 36.8\times 5=184J$
Summation of all $△H$
$=376.5+6000+184=6560.5J=6.56kJ$Since cooling of water is an exothermic reaction, $△H$ is always negative
$△H=-6.56kJ$