Chapter 14.5, Problem 7PPB

(a)

Interpretation Introduction

Interpretation:

The entropy changes associated with the phase transitions of sodium has to be found.

Concept Introduction:

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process.  For all spontaneous process in an isolated system there will be an increase in entropy. Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy changes associated with a phase transition reaction can be found by the following equation.

$\text{ΔS }\text{}\text{=}\text{}\text{}\frac{\text{ΔΗ}}{\text{Τ}}$

Where,

$\Delta {\rm H}$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}$ is the change in entropy in the system

Answer to Problem 7PPB

Solution:

The entropy changes associated with the solid-liquid transitions $\left({\text{ΔS}}^{°}{}_{\text{fus}}\right)$ of sodium is found to be $6.51{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The enthalpy changes associated with the solid-liquid transitions $\left({\text{ΔH}}^{°}{}_{\text{fus}}\right)$ of sodium is found to be $\text{2}{\text{.41kJmol}}^{\text{-1}}$

The melting point of sodium is found to be $97.61°\text{C}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\text{(Na(s))}=51.05{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}\text{(Na(l))}=57.56{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left(\text{Na(s) }\right)\text{=0}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left(\text{Na(l) }\right)\text{=}\text{2}{\text{.41JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{fus}}$

$\text{Na(s)}\to \text{Na(l)}$

${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}\text{=}{\text{S}}^{\text{°}}{\text{(Na(l))- S}}^{\text{°}}\text{(Na(s))}\\ \text{=}\text{(57}{\text{.56JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-(51}{\text{.05 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ \text{=}6.51{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =0.0065{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{fus}}$

$\text{Na(s)}\to \text{Na(l)}$

${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants the given equation.

${\text{ΔH}}^{°}{}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reacttants}}$

   $\begin{array}{l}{\text{= ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{Na(l) - ΔΗ}}_{\text{f}}{}^{\text{°}}\text{Na(s)}\\ \text{= (2}{\text{.41kJmol}}^{\text{-1}}\text{)-(0})\\ \text{= 2}{\text{.41kJmol}}^{\text{-1}}\end{array}$

To calculate melting point of $\text{Na}$

The melting point of ${\text{Br}}_{\text{2}}$ can be calculated as given below.

${\text{Τ}}_{\text{boiling}}\text{}\text{=}\text{}\text{}\frac{{\text{ΔΗ}}^{°}{}_{\text{fus}}}{{\text{ΔS}}_{\text{fus}}}$

$\begin{array}{l}\text{= }\frac{\text{2}{\text{.41kJmol}}^{\text{-1}}}{0.0065{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\\ =\text{370}\text{.76K}\end{array}$

The temperature in Kelvin is converted into degree Celsius by subtracting the value of $273.15$ from  it.

$\begin{array}{l}\text{The meltingpoint of} \text{Na}=\text{}330.71\text{K}\\ =370.761-273.15\\ =97.61°\text{C}\end{array}$

Conclusion

The entropy changes associated with the phase transitions of sodium was found.

(b)

Interpretation Introduction

Interpretation:

The entropy changes associated with the phase transitions of sodium has to be found.

Concept Introduction:

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process.  For all spontaneous process in an isolated system there will be an increase in entropy. Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy changes associated with a phase transition reaction can be found by the following equation.

$\text{ΔS }\text{}\text{=}\text{}\text{}\frac{\text{ΔΗ}}{\text{Τ}}$

Where,

$\Delta {\rm H}$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}$ is the change in entropy in the system

Answer to Problem 7PPB

The entropy changes associated with the liquid-vapour transitions $\left({\text{ΔS}}^{°}{}_{\text{vap}}\right)$ of sodium is found to be $96.1{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The enthalpy changes associated with the liquid-vapour transitions $\left({\text{ΔH}}^{°}{}_{\text{vap}}\right)$ of sodium is found to be $\text{105}{\text{.3kJmol}}^{\text{-1}}$

The boiling point of sodium is found to be $822.7°\text{C}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\text{(Na(g))}=153.7{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}\text{(Na(l))}=57.56{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left(\text{Na(g) }\right)\text{=107}{\text{.7JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left(\text{Na(l) }\right)\text{=}\text{2}{\text{.41JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{vap}}$

$\text{Na(l)}\to \text{Na(g)}$

${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}\text{=}{\text{S}}^{\text{°}}{\text{(Na(g)- S}}^{\text{°}}\text{(Na(l))}\\ \text{=}\text{(153}{\text{.7JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-(57}{\text{.56 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ \text{=}96.14{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =0.0961{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{vap}}$

$\text{Na(l)}\to \text{Na(g)}$

${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants the given equation.

${\text{ΔH}}^{°}{}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reacttants}}$

   $\begin{array}{l}{\text{= ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{Na(g) - ΔΗ}}_{\text{f}}{}^{\text{°}}\text{Na(l)}\\ \text{= (107}{\text{.7kJmol}}^{\text{-1}}\text{)-(2}{\text{.41kJmol}}^{\text{-1}})\\ \text{= 105}{\text{.3kJmol}}^{\text{-1}}\end{array}$

To calculate boiling point of $\text{Na}$

The boiling point of ${\text{Br}}_{\text{2}}$ can be calculated as given below,

${\text{Τ}}_{\text{boiling}}\text{}\text{=}\text{}\text{}\frac{{\text{ΔΗ}}^{°}{}_{\text{vap}}}{{\text{ΔS}}^{°}{}_{\text{vap}}}$

$\begin{array}{l}\text{= }\frac{\text{105}{\text{.3kJmol}}^{\text{-1}}}{0.0961{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\\ =1095.7\text{K}\end{array}$

The temperature in Kelvin is converted into degree Celsius by subtracting the value of $273.15$ from it.

$\begin{array}{l}\text{The boiling point of} \text{Na}=\text{}1095.7\text{K}\\ =370.761-273.15\\ =822.7°\text{C}\end{array}$

Conclusion

The entropy changes associated with the phase transitions of sodium was found.