Chapter 14, Problem 14.65P

Interpretation Introduction

Interpretation:

The value of $\Delta H{°}_{\text{rxn}}$ for combustion of oxalic acid in $\text{kJ/mol}$ has to be determined. Also, value of $\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ has to be determined.

Concept Introduction:

Enthalpy of reaction can be used to find heat capacity $\left(c_{\text{P}}\right)$. Heat capacity is the value of energy for particular substance that is required to raise substance’s temperature by $1°\text{C}$ or $1\text{K}$. The expression to calculate heat capacity is as follows:

  $c_{\text{P}}=\frac{q_{\text{P}}}{\Delta T}$

Here, $q_{\text{P}}$ input of energy at constant pressure, $\Delta T$ is increase in temperature after input of energy.

The value heat capacity of unit mole substance is known as molar heat capacity and it is represented by $C_{\text{P}}$. The expression of molar heat capacity is as follows:

  $C_{\text{P}}=\frac{q_{\text{P}}}{n\Delta T}$

Here, $n$ is number of moles.

Also, The value heat capacity per gram is known as specific heat capacity and it is represented by $c_{\text{sp}}$. The expression of molar heat capacity is as follows:

  $c_{\text{sp}}=\frac{q_{\text{P}}}{m\Delta T}$

Here, $m$ is mass in grams.

In calorimeter, when heat is released it can only be absorbed by contents of calorimeter. This results in an increase in temperature. Since all heat released is absorbed by calorimeter thus change in enthalpy of reaction is written as follows:

  $\Delta H=-\Delta H_{\text{cal}}$

In bomb calorimeter, the expression of energy change in the reaction is written as follows:

  $\Delta U=-c_{\text{V,cal}}\Delta T$

Answer to Problem 14.65P

The value of $\Delta H{°}_{\text{rxn}}$ for combustion of oxalic acid in $\text{kJ/mol}$ is $-245.5\text{ kJ/mol}$ and value of $\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ is $-827.3{\text{ kJ·mol}}^{\text{–1}}$.

Explanation of Solution

The chemical equation for combustion of oxalic acid is as follows:

  ${\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The expression to calculate number of moles of oxalic acid is as follows:

  $\text{Mole}=\frac{\text{Given mass}}{\text{Molar mass}}$        (1)

Substitute $2.50\text{ g}$ for given mass and $90.03\text{ g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}\text{Mole}=\left(\frac{2.50\text{ g}}{90.03\text{ g/mol}}\right)\\ =0.0278\text{ mol}\end{array}$

The expression to calculate energy of combustion reaction is as follows:

  $\Delta U=-c_{\text{V,cal}}\Delta T$        (2)

Substitute $8.75\text{ kJ}\cdot {\text{K}}^{-1}$ for $c_{\text{P,cal}}$ and $0.780\text{ K}$ for $\Delta T$ in equation (2).

  $\begin{array}{c}\Delta U=-\left(8.75\text{ kJ}\cdot {\text{K}}^{-1}\right)\left(0.780\text{ K}\right)\\ =-6.825\text{ kJ}\end{array}$

Since value of $\Delta H$ is equal to $\Delta U$ in bomb calorimeter, therefore, value of $\Delta H$ is also equal to $-6.825\text{ kJ}$.

The value of $\Delta H{°}_{\text{rxn}}$ for combustion of $0.0278\text{ mol}$ of oxalic acid can be calculated as follows:

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\frac{-6.825\text{ kJ}}{0.0278\text{ mol}}\\ =-245.5\text{ kJ/mol}\end{array}$

The expression of $\Delta H{°}_{\text{rxn}}$ for combustion chemical equation of oxalic acid is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(1\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)\\ +\left(\frac{1}{2}\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (3)

Rearrange equation (3) to calculate value of $\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ as follows:

  $\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(1\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\Delta H{°}_{\text{rxn}}-\left(\frac{1}{2}\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$        (4)

Substitute $-393.5{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$, $-285.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-245.5\text{ kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H{°}_{\text{rxn}}$, $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in equation (4).

  $\begin{array}{c}\Delta H{°}_{\text{f}}\left({\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)=\left[\begin{array}{c}\left[\begin{array}{l}\left(2\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(1\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ -\left(-245.5\text{ kJ}\cdot {\text{mol}}^{-1}\right)-\left(\frac{1}{2}\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-827.3{\text{ kJ·mol}}^{\text{–1}}\end{array}$