Question

Question: The molar enthalpy of fusion of ice at ${\mathbf{0}}^{\mathbf{°}}\mathbf{\text{C}}$is$\mathbf{6}\mathbf{.02}\mathbf{}{\mathbf{\text{kJmol}}}^{\mathbf{-}\mathbf{1}}$; the molar heat capacity of under cooled water is$\mathbf{75}\mathbf{.3}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{-}\mathbf{1}}$.

(a) One mole of under cooled water at$\mathbf{-}{\mathbf{10}}^{\mathbf{°}}\mathbf{\text{C}}$is induced to crystallize in a heat-insulated vessel. The result is a mixture of ice and water at${\mathbf{0}}^{\mathbf{°}}\mathbf{\text{C}}$. What fraction of this mixture is ice?

(b) Calculate ΔSfor the system.

Short answer

a) Number of moles present in the mixture is $0.125\text{mol}$.

b) The change in entropy of the system $\Delta S=2.75{\text{JK}}^{-1}$.

Step-by-step solution

Given data

The molar enthalpy of fusion of ice at$0^{°}\text{C}$ is $6.02\text{kJ}$${\text{mol}}^{-1}$; the molar heat capacity of undercooled water is $75.3\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The initial temperature is$-{10}^{°}\text{C}$ and the final temperature is $0^{°}\text{C}$.

Concept used

Entropy Change is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic system. It is related to the conversion of heat or enthalpy in work.

Calculate the heat absorbed by the ice

(a)

Initial temperature:

$T_1=-{10}^{°}\text{C}=-10+273.15=263.15\text{K}$

Final temperature:

$T_2=0^{°}\text{C}=0+273.15=273.15\text{K}$

The heat absorption has been calculating as follows:

$q=n{\text{c}}_p\Delta T$ ........ (1)

Where, n-number of moles, c_p - denotes capacity, and ΔT - the change in entropy.

$\begin{array}{c}q=1\text{mol}\times 75.3{\text{JK}}^{-1}{\text{mol}}^{-1}\times \left(T_2-T_1\right)\\ =1\text{mol}\times 75.3{\text{JK}}^{-1}{\text{mol}}^{-1}\times (273.15\text{K}-263.15\text{K})\\ =753\text{J}\end{array}$

$q=\Delta H_{fus}$

Calculate the number of moles in the mixture

The number of moles has been calculated as follows:

$\Delta H_{fus}=n\Delta H_{fus\left(ice\right)}$ ......... (2)

$\Delta H_{fus\left(ice\right)}$is $6.02{k_{Jmol}}^{-1}$.

$\Delta H_{fus}=753\text{J}$.

Substitute these values in equation (2)

$\begin{array}{c}753\text{J}=n\times 6.02{\text{kJmol}}^{-1}\\ n=\frac{753\text{J}}{6.02\text{k}\cdot {\text{Jmol}}^{-1}}\\ =\frac{753.\text{J}}{6.02\times {10}^3{\text{Jmol}}^{-1}}\\ =0.125\text{mol}\end{array}$

Calculate the change in entropy

(b)

The change in entropy has been calculated as follows:

$\Delta S=\frac{\Delta H_{fus}}{\Delta T}$ .......... (3)

$\Delta T=273K$.

$\Delta H_{fus}=753\text{J}$.

Substitute these values in equation (3)

$\begin{array}{c}\Delta S=\frac{753.\text{J}}{273.15\text{K}}\\ =2.75{\text{JK}}^{-1}\end{array}$