Chapter 9, Problem 9.90P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the formation of dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ and ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is to be calculated.

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

$\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 9.90P

$\Delta H_{\text{rxn}}^{°}$ for the formation of dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ and ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is $-326\text{kJ}$ and $-369\text{kJ}$ respectively.

Explanation of Solution

The given chemical equation for the formation of dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ is as follows:

The number of broken bonds is $\text{8 C}-\text{H}$ bond and $\text{1 O=O}$ bonds.

The number of bonds formed is $6\text{ C}-\text{H}$ bonds, $2\text{ C}-\text{O}$ and $2\text{ O}-\text{H}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{8BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{O=O}}\right)-\left({\text{6BE}}_{\text{C}-\text{H}}+{\text{2BE}}_{\text{C}-\text{O}}+{\text{2BE}}_{\text{O}-\text{H}}\right)$ (1)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $498\text{kJ/mol}$ for ${\text{BE}}_{\text{O=O}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\left(\text{8 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(498\text{kJ/mol}\right)\right)-\\ \left(\text{6 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(358\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =-326\text{kJ}\end{array}$

The given chemical equation for the formation of ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is as follows:

The number of broken bonds is $\text{8 C}-\text{H}$ bond and $\text{1 O=O}$ bonds.

The number of bonds formed is $5\text{ C}-\text{H}$ bonds, $1\text{ C}-\text{O}$, $1\text{ C}-\text{C}$ and $3\text{ O}-\text{H}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{8BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{O=O}}\right)-\left({\text{5BE}}_{\text{C}-\text{H}}+{\text{1BE}}_{\text{C}-\text{O}}+{\text{1BE}}_{\text{C}-\text{C}}+{\text{3BE}}_{\text{O}-\text{H}}\right)$ (2)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $498\text{kJ/mol}$ for ${\text{BE}}_{\text{O=O}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$, $347\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{C}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\left(\text{8 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(498\text{kJ/mol}\right)\right)-\\ \left(\text{5 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(358\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\\ \left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =-369\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the formation of dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ and ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is $-326\text{kJ}$ and $-369\text{kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Among the formation reaction of dimethyl ether and ethanol, the more exothermic reaction is to be identified.

Concept introduction:

In the case of a reaction, the change in enthalpy $\left(\Delta H\right)$ is the difference in the energy of the product and reactant. The general expression to calculate $\Delta H$ is,

$\Delta H=H_{\text{Product}}-H_{\text{Reactant}}$ (1)

Here,

$\Delta H$ is the change in enthalpy of the system.

$H_{\text{Product}}$ is the enthalpy of the products.

$H_{\text{Reactant}}$ is the enthalpy of the reactants.

Endothermic reactions are the reactions in which energy in the form of the heat or light is absorbed by the reactant for the formation of the product. $H_{\text{Product}}$ is greater than $H_{\text{Reactant}}$ in the endothermic reactions.

Exothermic reactions are the reactions in which energy in the form of the heat or light is released with the product. $H_{\text{Reactant}}$ is greater than $H_{\text{Product}}$ in the exothermic reactions.

Answer to Problem 9.90P

The formation reaction of ethanol is more exothermic as compared to dimethyl ether.

Explanation of Solution

$\Delta H_{\text{rxn}}^{°}$ for the formation of dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ is $-326\text{kJ}$.

$\Delta H_{\text{rxn}}^{°}$ for the formation of ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is $-369\text{kJ}$.

The value of $\Delta H_{\text{rxn}}^{°}$ for the formation of ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is more negative as compared to dimethyl ether $\left({\text{CH}}_3{\text{OCH}}_3\right)$ so more energy is released in the case of ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ than dimethyl ether. Therefore the formation reaction of ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{O}\mathrm{H}\right)$ is more exothermic.

Conclusion

Exothermic reactions are the reactions in which energy in the form of the heat or light is released with the product. $H_{\text{Reactant}}$ is greater than $H_{\text{Product}}$ in the exothermic reactions.

(c)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the conversion of ethanol to dimethyl ether is to be calculated.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law $\Delta H$ of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. $\Delta H_{\text{overall rxn}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_{\text{n}}$

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so $\Delta H$ of an overall reaction can be calculated by the addition or subtraction of the individual steps whose $\Delta H$ is known.

Answer to Problem 9.90P

$\Delta H_{\text{rxn}}^{°}$ for the conversion of ethanol to dimethyl ether is $43\text{kJ}$.

Explanation of Solution

The enthalpy change of the following reaction is $\Delta H_1^{°}$.

${\text{2CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CH}}_3{\text{OCH}}_3\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (1)

The enthalpy change of the following reaction is $\Delta H_2^{°}$.

${\text{2CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (2)

Reverse the equation (2).

${\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\to {\text{2CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)$ (3)

The enthalpy change for the reaction (3) is calculated as,

$\Delta H_3^{°}=-\Delta H_2^{°}$

Add equation (1) and (3).

${\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)\to {\text{CH}}_3{\text{OCH}}_3\left(g\right)$ (4)

The enthalpy change of the final reaction (4) is $\Delta H_{\text{rxn}}^{\text{°}}$.

The expression to calculate $\Delta H_{\text{rxn}}^{\text{°}}$ is as follows:

$\Delta H_{\text{rxn}}^{\text{°}}=\Delta H_1^{°}+\Delta H_3^{°}$ (5)

Substitute $-326\text{kJ}$ for $\Delta H_1^{°}$ and $+369\text{kJ}$ for $\Delta H_3^{°}$ in the equation (5).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{°}}=-326\text{kJ}+369\text{kJ}\\ =43\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the conversion of ethanol to dimethyl ether is $43\text{kJ}$.