Chapter 8, Problem 8.127QP

(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 8.127QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{863kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{N}}_{\text{2}}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{2NH}}_{\text{3}}\left(g\right)$

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

Average bond energy of a $\text{N}-\text{H}$ bond in ammonia is $388\text{kJ/mol}$.

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

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(3\times \text{H}-\text{H}+\text{2}\times \text{N}\equiv \text{N}\right)-\left(6\times \text{N}-\text{H}\right)\\ =\left(3\times 436\text{kJ/mol}+2\times 941\text{kJ/mol}\right)-\left(6\times 388\text{kJ/mol}\right)\\ =\left(1308+1882\text{kJ/mol}\right)-\left(2328\text{kJ/mol}\right)\\ =3190-2328\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{863kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{863kJ/mol}$.

(b)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Answer to Problem 8.127QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{98kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{N}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{H}}_2{\text{NNH}}_2\left(g\right)$

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

Average bond energy of a $\text{N}-\text{H}$ bond is $388\text{kJ/mol}$.

Average bond energy of a $\text{N}-\text{N}$ bond is $163\text{kJ/mol}$.

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

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(2\times \text{H}-\text{H}+1\times \text{N}\equiv \text{N}\right)-\left(4\times \text{N}-\text{H}+1\times \text{N}-\text{N}\right)\\ =\left(2\times 436\text{kJ/mol}+1\times 941\text{kJ/mol}\right)-\left(4\times 388\text{kJ/mol}+1\times 163\text{kJ/mol}\right)\\ =\left(872+941\text{kJ/mol}\right)-\left(1552+163\text{kJ/mol}\right)\\ =1813-1715\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{98kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{98kJ/mol}$.

(c)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Answer to Problem 8.127QP

Solution

The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{93kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is,

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

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 in a hydrogen molecule is $495\text{kJ/mol}$.

Average bond energy of a $\text{N}-\text{O}$ bond is $201\text{kJ/mol}$.

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

$\begin{array}{c}\Delta H_{\text{rxn}}=\left(1\times \text{O}=\text{O}+2\times \text{N}\equiv \text{N}\right)-\left(2\times \text{N}\equiv \text{N}+2\times \text{N}-\text{O}\right)\\ =\left(1\times 495\text{kJ/mol}+2\times 941\text{kJ/mol}\right)-\left(2\times 941\text{kJ/mol}+2\times 201\text{kJ/mol}\right)\\ =\left(495+1882\text{kJ/mol}\right)-\left(1882+402\text{kJ/mol}\right)\\ =2377-2284\text{kJ/mol}\end{array}$

Simplify the above equation,

$\Delta H_{\text{rxn}}=\underset{\_}{93kJ/mol}$

Hence, the enthalpy change of the given reaction is $\underset{\_}{93kJ/mol}$.

Conclusion

1. a. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{863kJ/mol}$.
2. b. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{98kJ/mol}$.
3. c. The enthalpy change $\left(\Delta H_{\text{rxn}}\right)$ of the given reaction is $\underset{\_}{93kJ/mol}$.