Chapter 8, Problem 8.128QP

(a)

Interpretation Introduction

Interpretation: The enthalpy changes of the given reactions are to be calculated by using average bond energies.

Concept introduction: The energy which is required in a molecule to break one mole of a covalent bond in its gaseous phase is known as bond energy of that bond. It is expressed in terms of enthalpy change denoted by $\left(\Delta H\right)$.

To determine: The enthalpy change of the given reaction.

Answer to Problem 8.128QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{-76kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{H}}_{\text{2}}\text{O}\left(g\right)+\text{CO}\left(g\right)$

The bond enthalpy is calculated by using the formula,

$\Delta H_{\text{rxn}}={\displaystyle \sum \Delta H_{\text{bond}\text{breaking}}-{\displaystyle \sum \Delta H_{\text{bond}\text{forming}}}}$

Average bond energy of a $\text{C}=\text{O}$ bond is $743\text{kJ/mol}$.

Average bond energy of a $\text{H}-\text{H}$ bond in a hydrogen molecule is $436\text{kJ/mol}$.

Average bond energy of a $\text{O}-\text{H}$ bond in water is $463\text{kJ/mol}$.

Average bond energy of a $\text{C}\equiv \text{O}$ bond is $1072\text{kJ/mol}$.

Substitute the values of average bond energies of $\text{C}=\text{O}$, $\text{H}-\text{H}$, $\text{O}-\text{H}$ and $\text{C}\equiv \text{O}$ bond in the above formula to calculate the change in enthalpy of the reaction.

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(2\times \text{C}=\text{O}+1\times \text{H}-\text{H}\right)-\left(2\times \text{O}-\text{H}+1\times \text{C}\equiv \text{O}\right)\\ =\left(2\times 743\text{kJ/mol}+1\times 436\text{kJ/mol}\right)-\left(2\times 463\text{kJ/mol}+1\times 1072\text{kJ/mol}\right)\\ =\left(1486+436\text{kJ/mol}\right)-\left(926+1072\text{kJ/mol}\right)\\ =1922-1998\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{-76kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{-76kJ/mol}$.

(b)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Answer to Problem 8.128QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{80kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }2\text{NO}\left(g\right)$

The bond enthalpy is calculated by using the formula,

$\Delta H_{\text{rxn}}={\displaystyle \sum \Delta H_{\text{bond}\text{breaking}}-{\displaystyle \sum \Delta H_{\text{bond}\text{forming}}}}$

Average bond energy of a $\text{N}\equiv \text{N}$ bond in a nitrogen molecule is $941\text{kJ/mol}$.

Average bond energy of a $\text{O}=\text{O}$ bond is $495\text{kJ/mol}$.

Average bond energy of a $\text{N}\equiv \text{O}$ bond is $678\text{kJ/mol}$.

Substitute the values of average bond energies of $\text{N}\equiv \text{N}$, $\text{O}=\text{O}$ and $\text{N}\equiv \text{O}$ bond in the above formula to calculate the change in enthalpy of the reaction.

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(1\times \text{N}\equiv \text{N}+1\times \text{O}=\text{O}\right)-\left(2\times \text{N}\equiv \text{O}\right)\\ =\left(1\times 941\text{kJ/mol}+1\times 495\text{kJ/mol}\right)-\left(2\times 678\text{kJ/mol}\right)\\ =\left(941+495\text{kJ/mol}\right)-\left(1356\text{kJ/mol}\right)\\ =1436-1356\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{80kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{80kJ/mol}$.

(c)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Answer to Problem 8.128QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{61kJ/mol}$.

Explanation of Solution

The given reaction is,

$\text{C}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)\stackrel{}{\to }2\text{CO}\left(g\right)$

The bond enthalpy is calculated by using the formula,

$\Delta H_{\text{rxn}}={\displaystyle \sum \Delta H_{\text{bond}\text{breaking}}-{\displaystyle \sum \Delta H_{\text{bond}\text{forming}}}}$

The enthalpy of sublimation of graphite, $\text{C}\left(s\right)$ is $719\text{kJ/mol}$.

Average bond energy of a $\text{C}=\text{O}$ bond is $743\text{kJ/mol}$.

Average bond energy of a $\text{C}\equiv \text{O}$ bond is $1072\text{kJ/mol}$.

Substitute the values of average bond energies of $\text{C}-\text{C}$, $\text{C}=\text{O}$ and $\text{C}\equiv \text{O}$ bond in the above formula to calculate the change in enthalpy of the reaction.

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(1\times \text{C}-\text{C}+2\times \text{C}=\text{O}\right)-\left(2\times \text{C}\equiv \text{O}\right)\\ =\left(1\times 719\text{kJ/mol}+2\times 743\text{kJ/mol}\right)-\left(2\times 1072\text{kJ/mol}\right)\\ =\left(719+1486\text{kJ/mol}\right)-\left(2144\text{kJ/mol}\right)\\ =2205-2144\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{61kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{61kJ/mol}$.

Conclusion

1. a. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{-76kJ/mol}$.
2. b. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{80kJ/mol}$.
3. c. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{61kJ/mol}$.