Chapter 4, Problem 4D.18E

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be determined.

Concept Introduction:

The standard enthalpy of the reaction is the difference of the sum of standard enthalpy of formation of all the products and the sum of standard enthalpy of formation of all the reactants.  The mathematical equation for the calculation of standard enthalpy of the reaction is shown below.

$\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(reac\tan ts\right)$

Explanation of Solution

The given balanced chemical equation is shown below.

    $MgC{O}_3\left(s\right)+2HCl\left(aq\right)\to MgC{l}_{\text{2}}\left(aq\right)+H_{\text{2}}O\left(l\right)+C{O}_2\left(g\right)$

The standard enthalpy of formation of product that are $MgC{l}_{\text{2}}\left(aq\right)$, $C{O}_{\text{2}}\left(g\right)$ and $H_2\text{O}\left(l\right)$ is $-641.32\text{kJ}\cdot {\text{mol}}^{-1}$, $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$ and $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of product is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(\begin{array}{l}{\text{n}}_{MgC{l}_{\text{2}}}\Delta {\text{H}}_f^{\circ }\left(MgC{l}_{\text{2}},aq\right)+{\text{n}}_{C{O}_2}\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)\\ +{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)\end{array}\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{MgC{l}_{\text{2}}}$ is the number of moles of $MgC{l}_{\text{2}}\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(MgC{l}_{\text{2}},aq\right)$ is the standard enthalpy of formation of $MgC{l}_{\text{2}}\left(aq\right)$.
• ${\text{n}}_{C{O}_2}$ is the number of moles of $C{O}_{\text{2}}\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is the standard enthalpy of formation of $C{O}_{\text{2}}\left(g\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_2\text{O}\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_2\text{O}\left(l\right)$.

The value of ${\text{n}}_{MgC{l}_{\text{2}}}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(MgC{l}_{\text{2}},aq\right)$ is $-641.32\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{C{O}_2}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{2}O}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{MgC{l}_{\text{2}}}$, $\Delta {\text{H}}_f^{\circ }\left(MgC{l}_{\text{2}},aq\right)$, ${\text{n}}_{C{O}_2}$, $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(\begin{array}{l}\left(1mol\right)\times \left(-641.32\text{kJ}\cdot {\text{mol}}^{-1}\right)+\\ \left(1mol\right)\times \left(-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =-641.32\text{kJ}-\text{393}\text{.51}\text{kJ}-\text{285}\text{.83}\text{kJ}\\ =-1320.66kJ\end{array}$

The standard enthalpy of formation of reactants that are $MgC{O}_3\left(s\right)$ and $HCl\left(aq\right)$ is $-1095.8\text{kJ}\cdot {\text{mol}}^{-1}$ and $-167.16\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}={\text{n}}_{MgC{O}_3}\Delta {\text{H}}_f^{\circ }\left(MgC{O}_3,s\right)+{\text{n}}_{HCl}\Delta {\text{H}}_f^{\circ }\left(HCl,aq\right)$        (2)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{MgC{O}_3}$ is the number of moles of $MgC{O}_3\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(MgC{O}_3,s\right)$ is the standard enthalpy of formation of $MgC{O}_3\left(s\right)$.
• ${\text{n}}_{HCl}$ is the number of moles of $HCl\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(HCl,aq\right)$ is the standard enthalpy of formation of $HCl\left(aq\right)$.

The value of ${\text{n}}_{MgC{O}_3}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(MgC{O}_3,s\right)$ is $-1095.8\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{HCl}$ is $2mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(HCl,aq\right)$ is $-167.16\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{MgC{O}_3}$, $\Delta {\text{H}}_f^{\circ }\left(MgC{O}_3,s\right)$, ${\text{n}}_{HCl}$ and $\Delta {\text{H}}_f^{\circ }\left(HCl,aq\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}=\left(1mol\right)\times \left(-1095.8\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2mol\right)\times \left(-167.16\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-1095.8\text{kJ}+334.32kJ\\ =-761.48kJ\end{array}$

The standard enthalpy for the given reaction is calculated by the expression shown below.

  $\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(reac\tan ts\right)$        (3)

Where,

• $\Delta \text{H}°$ is standard enthalpy of reaction.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-1320.66kJ$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is $-761.48kJ$.

Substitute the value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ in equation (3).

  $\begin{array}{c}\Delta \text{H}°=\left(-1320.66kJ\right)-\left(-761.48kJ\right)\\ =-1320.66kJ+761.48kJ\\ =-559.18kJ\end{array}$

Thus, the standard enthalpy for the given reaction is $\underset{\_}{-559.18kJ}$.