Chapter 4, Problem 4D.20E

Interpretation Introduction

Interpretation:

The standard reaction enthalpy for the formation of anhydrous aluminum bromidehas to be determined.

Concept Introduction:

Hess’s law is defined as the law which states that the enthalpy of a reaction does not depend on the path followed by reaction.  It depends only on the standard enthalpy of formations for estimating the enthalpy of a reaction.  The heat of combustion is the energy produced when one mole of substance reacts with oxygen.

Explanation of Solution

The aim reaction representing the formation of anhydrous aluminum bromide is shown below.

    $2Al\left(s\right)+3B{r}_{\text{2}}\left(l\right)\to 2AlB{r}_3\left(s\right)$

The given reactions along with their reaction enthalpies are shown below.

    $\begin{array}{c}\left(a\right)2Al\left(s\right)+6HBr\left(aq\right)\to 2AlB{r}_3\left(aq\right)+3H_2\left(g\right)\Delta {\text{H}}^{\circ }=-1061kJ\\ \left(b\right)HBr\left(g\right)\to HBr\left(aq\right)\Delta {\text{H}}^{\circ }=-81.15kJ\\ \left(c\right)H_2\left(g\right)+B{r}_2\left(l\right)\to 2HBr\left(g\right)\Delta {\text{H}}^{\circ }=-72.80kJ\\ \left(d\right)AlB{r}_3\left(s\right)\to AlB{r}_3\left(aq\right)\Delta {\text{H}}^{\circ }=-368kJ\end{array}$

The desired reactants for the formation of anhydrous aluminum bromide are aluminum metal and bromine gas. 

As per the aim reaction, the required stoichiometric coefficient of $B{r}_{\text{2}}\left(l\right)$ is $3$. Therefore, reaction (c) is multiplied with $3$.  The modified reaction (c) is shown below.

  $\begin{array}{l}3H_2\left(g\right)+3B{r}_2\left(l\right)\to 6HBr\left(g\right)\Delta {\text{H}}^{\circ }=3\times \left(-72.80kJ\right)\\ =-218.4kJ\end{array}$

$HBr\left(g\right)$ is not coming in the aim reaction.  Therefore, to eliminate it, the required stoichiometric coefficient of $HBr\left(g\right)$ is $6$ in reaction (b).  Thus, reaction (b) is multiplied with $6$.  The modified reaction (b) is shown below.

    $\begin{array}{l}6HBr\left(g\right)\to 6HBr\left(aq\right)\Delta {\text{H}}^{\circ }=6\times \left(-81.15kJ\right)\\ =-486.9kJ\end{array}$

The required stoichiometric coefficient of $AlB{r}_3\left(s\right)$ is $2$. Therefore, reaction (d) is multiplied with $2$.  The modified reaction (d) is shown below.

    $\begin{array}{l}2AlB{r}_3\left(s\right)\to 2AlB{r}_3\left(aq\right)\Delta {\text{H}}^{\circ }=2\times \left(-368kJ\right)\\ =-736kJ\end{array}$

As anhydrous aluminum bromide is the desired product.  Therefore, the above reaction is reversed.  At this time the sign of the change in enthalpy of the reaction is also gets reversed.  The further modified reaction (d) is shown below.

    $2AlB{r}_3\left(aq\right)\to 2AlB{r}_3\left(s\right)\Delta {\text{H}}^{\circ }=+736kJ$

Add the reaction (a), modified reaction (b), modified reaction (c) and modified reaction (d) to get the aim reaction for the formation of anhydrous aluminum bromide.  The addition result is shown below.

    $\begin{array}{l}\underset{\_}{\begin{array}{c}2Al\left(s\right)+6HBr\left(aq\right)\to 2AlB{r}_3\left(aq\right)+3H_2\left(g\right)\Delta {\text{H}}^{\circ }=-1061kJ\\ 6HBr\left(g\right)\to 6HBr\left(aq\right)\Delta {\text{H}}^{\circ }=-486.9kJ\\ 3H_2\left(g\right)+3B{r}_2\left(l\right)\to 6HBr\left(g\right)\Delta {\text{H}}^{\circ }=-218.4kJ\\ 2AlB{r}_3\left(aq\right)\to 2AlB{r}_3\left(s\right)\Delta {\text{H}}^{\circ }=+736kJ\end{array}}\\ 2Al\left(s\right)+3B{r}_{\text{2}}\left(l\right)\to 2AlB{r}_3\left(s\right)\Delta {\text{H}}^{\circ }=\left(\begin{array}{l}-1061kJ+\left(-486.9kJ\right)+\\ \left(-218.4kJ\right)+736kJ\end{array}\right)\\ =-1030.3kJ\end{array}$

Thus, the standard reaction enthalpy for the formation of anhydrous aluminum bromide is $\underset{\_}{-1030.3kJ}$.