Question

The enthalpy change on freezing of 1 mol of water at $5^{0}C$ is ice at $-5^{0}C$ is :

(Given${\Delta }_{fus}H=6kJmo{l}^{-1}at{0}^{0}C$,

$C_p(H_{2}O,1)=75.3Jmo{l}^{-1}{K}^{-1}$

$C_p(H_{2}O,s)=36.8Jmo{l}^{-1}{K}^{-1})$)

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

Answer 1

The correct option is C $6.56kJmo{l}^{-1}$

In order to calculate the enthalpy change for $H_{2}Oat{5}^{0}C$, we need to calculate the enthalpy change of all the transformation involved in the process.

(a) Energy change of 1 mol, $H_{2}O\left(l\right)at{5}^{0}C\to$$1mol{H}_{2}O\left(l\right),0^{0}C$

(b) Energy change of 1 mol, $H_{2}O\left(l\right)at{0}^{0}C\to 1mol{H}_{2}O\left(s\right)\left(ice\right),0^{0}C$

(c) Energy change of 1mol, $\text{ice (s), at}{0}^{0}C\to 1mol,ice\left(s\right),-5^{0}C$

Total $\Delta H=C_p\left[H_{2}O\right(l\left)\right]\Delta T+\Delta Hfreezing+C_p\left[H_{2}O\right(s\left)\right]\Delta T$

$=\left(75.3Jmo{l}^{-1}{K}^{-1}\right)(0-5)K+(-6\times {10}^{3}Jmo{l}^{-1})+\left(36.8Jmo{l}^{-1}{K}^{-1}\right)(-5-0)K$

$\Delta H=-6.56kJmo{l}^{-1}$

$\Delta H=6.56kJmo{l}^{-1}$