Chapter 2, Problem 2.4PR

(a)

Interpretation Introduction

Interpretation:

Change in molar enthalpy of carbon dioxide has to be calculated for the temperature range ${\text{15}}^{\text{ο}}\text{C}$ to ${\text{37}}^{\text{ο}}\text{C}$.

Concept introduction:

Enthalpy:

Enthalpy is a property of a thermodynamic system that is equal to the sum of the internal energy of the system and the product of pressure and volume.  For a closed system where transfer of matter between system and surroundings is prohibited, for the processes that occur at constant pressure, the heat absorbed or released equals to the change in enthalpy.

It is denoted as $\text{H}$. The molar enthalpy is defined as the enthalpy per mole. It is denoted as ${\text{H}}_{\text{m}}$.

From thermodynamics,

$\text{H}\text{=}\text{U}\text{+}\text{PV}$

Where,

$\text{U}\text{=}$ internal energy of the system

H is the enthalpy energy

P is the Pressure

V is the volume.

Specific heat capacity at constant pressure:

Specific heat capacity at constant pressure can be defined as the amount of energy requires increasing the temperature of a substance by ${\text{1}}^{\text{ο}}\text{C}$.

It is denoted as ${\text{C}}_{\text{p}}$. The molar specific heat capacity is defined as the specific heat capacity per mole and it is denoted by ${\text{C}}_{\text{p,m}}$.

  ${\text{C}}_{\text{p}}={\left(\frac{\partial \text{H}}{\partial \text{T}}\right)}_{\text{p}}$

Answer to Problem 2.4PR

The molar enthalpy change for carbon dioxide at the temperature range of ${\text{15}}^{\text{ο}}\text{C}$ to ${\text{37}}^{\text{ο}}\text{C}$ is $\text{818}\text{.251}\text{J}{\text{mol}}^{\text{-1}}$.

Explanation of Solution

According to the question the heat capacity of substance is often reported in the form,

  ${\text{C}}_{\text{p,m}}\text{(T)}\text{=}\text{a}\text{+bT}{\text{+c/T}}^{\text{2}}$

Now the change of the molar enthalpy has to be determined.

From thermodynamics it can be stated that,

  ${\text{C}}_{\text{p}}={\left(\frac{\partial \text{H}}{\partial \text{T}}\right)}_{\text{p}}$

Now, combining the two equations,

  ${\left(\frac{\partial {\text{H}}_{\text{m}}}{\partial \text{T}}\right)}_{\text{p}}=\text{a}\text{+bT}{\text{+c/T}}^{\text{2}}$

Temperature range is given as ${\text{15}}^{\text{ο}}\text{C}$ to ${\text{37}}^{\text{ο}}\text{C}$.

$\begin{array}{l}{37}^{\text{ο}}\text{C}\text{=}\text{(37}\text{+}\text{273)}\text{K}\\ \\ \text{=}310\text{K}\end{array}$

$\begin{array}{l}{15}^{\text{ο}}\text{C}\text{=}\text{(15}\text{+}\text{273)}\text{K}\\ \\ \text{=}288\text{K}\end{array}$

Now applying integration over the temperature range ${\text{15}}^{\text{ο}}\text{C}$ to ${\text{37}}^{\text{ο}}\text{C}$ and considering the change in enthalpy as $\Delta \text{H}$ for carbon dioxide,

Given that,

  $\begin{array}{l}\text{a}\text{=}\text{44}\text{.22}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{b}\text{=}\left(\text{8}{\text{.79×10}}^{\text{-3}}\right)\text{J}{\text{K}}^{\text{-2}}{\text{mol}}^{\text{-1}}\\ \text{c}\text{=}\text{-}\left(\text{8}{\text{.62×10}}^{\text{5}}\right)\text{J}\text{K}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\displaystyle \int {\text{dH}}_{\text{m}}\text{=}{\displaystyle \int \text{(a}\text{+bT}{\text{+c/T}}^{\text{2}}\text{)}\text{dT}}}\\ \\ {\displaystyle \underset{{\text{H}}_{\text{1}}}{\overset{{\text{H}}_{\text{2}}}{\int }}{\text{dH}}_{\text{m}}\text{=}\text{a}{\displaystyle \underset{\text{288K}}{\overset{\text{310K}}{\int }}\text{dT}}}\text{+}\text{b}{\displaystyle \underset{\text{288K}}{\overset{\text{310K}}{\int }}\text{TdT}}\text{+}\text{c}{\displaystyle \underset{\text{288K}}{\overset{\text{310K}}{\int }}\frac{\text{dT}}{{\text{T}}^{\text{2}}}}\\ \\ \left({\text{H}}_{\text{2,m}}\text{-}{\text{H}}_{\text{1,m}}\right)\text{=}\text{a}{\left[\text{T}\right]}_{\text{288K}}^{\text{310K}}\text{+}\text{b}{\left[\frac{{\text{T}}^{\text{2}}}{\text{2}}\right]}_{\text{288K}}^{\text{310K}}\text{-}\text{c}{\left[\frac{\text{1}}{\text{T}}\right]}_{\text{288K}}^{\text{310K}}\\ \\ {\text{ΔH}}_{\text{m}}\text{=}\left(\text{44}\text{.22}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{×}\left(\text{310-288}\right)\text{K}\text{+}\text{(8}{\text{.79×10}}^{\text{-3}}{\text{JK}}^{\text{-2}}{\text{mol}}^{\text{-1}}\text{)×}\left(\frac{{\text{(310)}}^{\text{2}}\text{-}{\text{(288)}}^{\text{2}}}{\text{2}}\right){\text{K}}^{\text{2}}\\ \text{-}\text{(-8}{\text{.62×10}}^{\text{5}}{\text{JKmol}}^{\text{-1}}\text{)×}\left(\frac{\text{1}}{\text{310}}\text{-}\frac{\text{1}}{\text{288}}\right){\text{K}}^{\text{-1}}\\ \\ {\text{ΔH}}_{\text{m}}\text{=}\text{818}\text{.251}\text{J}{\text{mol}}^{\text{-1}}\end{array}$

Hence the molar enthalpy change for carbon dioxide at the temperature range of ${\text{15}}^{\text{ο}}\text{C}$ to ${\text{37}}^{\text{ο}}\text{C}$ is $\text{818}\text{.251}\text{J}{\text{mol}}^{\text{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The change of molar enthalpy of a substance over a limited temperature range using the expression of part (a) has to be plotted.

Concept introduction:

Enthalpy:

Enthalpy is a property of a thermodynamic system that is equal to the sum of the internal energy of the system and the product of pressure and volume.  For a closed system where transfer of matter between system and surroundings is prohibited, for the processes that occur at constant pressure, the heat absorbed or released equals to the change in enthalpy.

It is denoted as $\text{H}$. The molar enthalpy is defined as the enthalpy per mole. It is denoted as ${\text{H}}_{\text{m}}$.

From thermodynamics,

$\text{H}\text{=}\text{U}\text{+}\text{PV}$

Where,

$\text{U}\text{=}$ Internal energy of the system

H is the enthalpy energy of system

P is the Pressure of the system

V is the volume of the system.

Specific heat capacity at constant pressure:

Specific heat capacity at constant pressure can be defined as the amount of energy required to increase the temperature of a substance by ${\text{1}}^{\text{ο}}\text{C}$.

It is denoted as ${\text{C}}_{\text{p}}$. The molar specific heat capacity is defined as the specific heat capacity per mole and it is denoted by ${\text{C}}_{\text{p,m}}$.

  ${\text{C}}_{\text{p}}={\left(\frac{\partial \text{H}}{\partial \text{T}}\right)}_{\text{p}}$

Explanation of Solution

The expression for the change of enthalpy of a substance that has been determined is,

  ${\text{ΔH}}_{\text{m}}\text{=}\text{a}{\text{(T}}_{\text{2}}\text{-}{\text{T}}_{\text{1}}\text{)}\text{+}\text{b}\frac{{\text{(T}}_{\text{2}}^{\text{2}}\text{-}{\text{T}}_{\text{1}}^{\text{2}}\text{)}}{\text{2}}\text{-}\text{c}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Now according to question molar enthalpy has to be plotted as function of temperature to see the change within a temperature limit.

Putting the values of the temperature, the values of enthalpy at $\text{310}\text{K}$ and $\text{288}\text{K}$ are $\text{16}\text{.91}\text{kJ}{\text{mol}}^{\text{-1}}$ and $\text{16}\text{.83kJ}{\text{mol}}^{\text{-1}}$ respectively.

The graph is plotted below,

                    Figure-1

Hence from the plot it has been found that with increase in temperature the enthalpy value increases.