Chapter 21, Problem 45PS

When boron hydrides burn in air, the reactions are very exothermic

1. (a) Write a balanced equation for the combustion of B₅H₉(g) in air to give B₂O₃(s) and H₂O(g).
2. (b) Calculate the enthalpy of combustion for B₅H₉(g) (Δ_fH° = 73.2 kJ/mol), and compare it with the enthalpy of combustion of B₂H₆ (−2038 kJ/mol). (The enthalpy of formation of B₂O₃(s) is −1271.9 kJ/mol.)
3. (c) Compare the enthalpy of combustion of C₂H₆(g) with that of B₂H₆(g). Which transfers more energy as heat per gram?

Interpretation Introduction

Interpretation:

The balanced chemical equation for the combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ should be written.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation

Answer to Problem 45PS

The balanced chemical equation for the combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ in air is,

  ${\text{2B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)+12{\text{O}}_{\text{2}}\left(\text{g}\right)\to 5{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+9{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Explanation of Solution

Boron hydride of the formula ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ reacts with the oxygen present in air to give oxide of boron. The equation is said to be balanced if it has an equal number of atoms on both product and reactant side. This is done by multiplying the compounds with the stoichiometric coefficient.

The balanced chemical equation for the combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ in air is,

  ${\text{2B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)+12{\text{O}}_{\text{2}}\left(\text{g}\right)\to 5{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+9{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Interpretation Introduction

Interpretation:

The enthalpy of combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ should be determined.

Concept introduction:

Enthalpy of reaction is the energy required or energy released when a chemical reaction occurs. If the process is endothermic then the energy is absorbed on the formation of the product. In such a case the sign of enthalpy change is positive. If the process is exothermic, then the energy is released on the formation of the product. In such a case the sign of enthalpy change is negative. In general, combustion reactions are exothermic in nature.

The standard enthalpy change is expressed as,

${\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)$

Answer to Problem 45PS

The ${\text{Δ}}_{\text{r}}{H}^{\text{°}}$ for the combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ in air is $-4341.2kJ\text{/}mol\text{-}rxn$.

Explanation of Solution

The ${\text{Δ}}_{\text{r}}{H}^{\text{°}}$ for the combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ in air is calculated below.

Given:

 Refer to Appendix L for the values of standard enthalpies.

The standard enthalpy of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ is $73.2\text{ kJ/mol}$.

The standard enthalpy of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $0\text{ kJ/mol}$.

The standard enthalpy of ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ is $-1271.9\text{ kJ/mol}$.

The standard enthalpy of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-241.83\text{ kJ/mol}$.

The balanced reaction is,

  ${\text{2B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)+12{\text{O}}_{\text{2}}\left(\text{g}\right)\to 5{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+9{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The standard enthalpy change is expressed as,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{5 mol B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right]+\\ \left({\text{9 mol H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{2 mol B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)\right]+\\ \left({\text{12 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{5 mol B}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-1271.9\text{ kJ/mol}\right)+\\ \left({\text{9 mol H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-241.83\text{ kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{2 mol B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(73.2\text{ kJ/mol}\right)+\\ \left({\text{12 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(0\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-\text{4341}\text{.2 kJ/mol-rxn}\end{array}$

The enthalpy of combustion of ${\text{B}}_2{\text{H}}_6\left(\text{g}\right)$ is $-\text{2038 kJ/mol-rxn}$.

Thus, the enthalpy of combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$ is double to that ${\text{B}}_2{\text{H}}_6\left(\text{g}\right)$.

Interpretation Introduction

Interpretation:

The enthalpy of combustion of ${\text{C}}_2{\text{H}}_6\left(\text{g}\right)$ should be given and compare the value with the enthapy of combustion of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\left(\text{g}\right)$.

Concept introduction:

Enthalpy of reaction is the energy required or energy released when a chemical reaction occurs. If the process is endothermic then the energy is absorbed on the formation of the product. In such a case the sign of enthalpy change is positive. If the process is exothermic, then the energy is released on the formation of the product. In such a case the sign of enthalpy change is negative. In general, combustion reactions are exothermic in nature.

The standard enthalpy change is expressed as,

${\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)$

Answer to Problem 45PS

The enthalpy of combustion of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $-1428.7kJ\text{/}mol\text{-}rxn$.

The enthalpy of combustion for ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ per gram is $47.6kJ\text{/}g$.

The enthalpy of combustion for ${\text{B}}_2{\text{H}}_6\left(\text{g}\right)$ per gram is $73.6kJ\text{/}g$.

Explanation of Solution

The enthalpy of combustion of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpies.

The standard enthalpy of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $-83.85\text{ kJ/mol}$.

The standard enthalpy of ${\text{O}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The standard enthalpy of ${\text{CO}}_2\left(\text{g}\right)$ is $-393.509\text{ kJ/mol}$.

The standard enthalpy of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-241.83\text{ kJ/mol}$.

The balanced chemical equation is,

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)+\frac{7}{2}{\text{O}}_2\left(\text{g}\right)\to 2{\text{CO}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The standard enthalpy change is expressed as,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{2 mol CO}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{CO}}_2\left(\text{g}\right)\right]+\\ \left({\text{3 mol H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\right]+\\ \left(\text{3}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{2 mol CO}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(-393.509\text{ kJ/mol}\right)+\\ \left({\text{3 mol H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-241.83\text{ kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-83.85\text{ kJ/mol}\right)+\\ \left(\text{3}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-\text{1428}\text{.7 kJ/mol-rxn}\end{array}$

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $30\text{ g/mol}$.

Thus, the enthalpy of combustion for ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ per gram is $47.6\text{ kJ/g}$.

The molar mass of ${\text{B}}_2{\text{H}}_6\left(\text{g}\right)$ is $27.66\text{ g/mol}$.

Thus, the enthalpy of combustion for ${\text{B}}_2{\text{H}}_6\left(\text{g}\right)$ per gram is $73.6\text{ kJ/g}$.