Chapter 9, Problem 9.56PAE

9.56 Using heats of formation tabulated in Appendix E, calculate the heats of reaction for the following.

(a) ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{5}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2CO}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\mathcal{l}\right)$

(b) ${\text{PCl}}_{\text{3}}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{PCl}}_{\text{5}}\left(\text{g}\right)$

(c) ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(\text{g}\right)$

(d) ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+\text{2Al}\left(\text{s}\right)\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+\text{2Fe}\left(\mathcal{l}\right)$

Interpretation Introduction

Interpretation:

The heat of reaction from the heat of formation should be calculated.

Concept Introduction:

The heat of formation is the change in enthalpy when one mole of a substance formed at constant condition (temperature and pressure).

The heat of reaction is the change in enthalpy at constant pressure.

Hess’s law: states change in enthalpy of all the changes occurred in the reaction.

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

Answer to Problem 9.56PAE

Solution: The heat of reaction is $\text{1297}\text{.83 kJ/mole}$.

Explanation of Solution

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

$\text{ΔH =}\left[\text{-786 – 285}\text{.83}\right]-\left[\text{226}\right]\text{ = -1297}\text{.83}\text{kJ/mol}$

Interpretation Introduction

Interpretation:

The heat of reaction from the heat of formation should be calculated.

Concept Introduction:

The heat of formation is the change in enthalpy when one mole of a substance formed at constant condition (temperature and pressure).

The heat of reaction is the change in enthalpy at constant pressure.

Hess’s law: states change in enthalpy of all the changes occurred in the reaction.

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

Answer to Problem 9.56PAE

Solution: The heat of reaction is $\text{-119}\text{.8 kJ/mole}$.

Explanation of Solution

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

$\Delta \text{H= }\Delta {\text{H}}_{PC{l}_{{}_5}}-[\Delta {\text{H}}_{PC{l}_{{}_3}}+\Delta {\text{H}}_{C{l}_2}]$

$\begin{array}{l}\text{ΔH= -151}\text{.8 - [-32}\text{.8 +0]}\\ \text{ = -151}\text{.8 +32 = -119}\text{.8 kJ/mol}\end{array}$

Interpretation Introduction

Interpretation:

The heat of reaction from the heat of formation should be calculated.

Concept Introduction:

The heat of formation is the change in enthalpy when one mole of a substance formed at constant condition (temperature and pressure).

The heat of reaction is the change in enthalpy at constant pressure.

Hess’s law: states change in enthalpy of all the changes occurred in the reaction.

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

Answer to Problem 9.56PAE

Solution: The heat of reaction is $\text{466 kJ/mole}$.

Explanation of Solution

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

$\Delta \text{H= }[\Delta {\text{H}}_{C_2{H}_{{}_4}}+\Delta {\text{H}}_{H_{2}O}]-\Delta {\text{H}}_{C_2{H}_{5}OH}$

$\text{ΔH = 277 – }\left[\text{52+ }\left(\text{-241}\right)\right]\text{ = 466 kJ/mol}$

Interpretation Introduction

Interpretation:

The heat of reaction from the heat of formation should be calculated.

Concept Introduction:

The heat of formation is the change in enthalpy when one mole of a substance formed at constant condition (temperature and pressure).

The heat of reaction is the change in enthalpy at constant pressure.

Hess’s law: states change in enthalpy of all the changes occurred in the reaction.

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

Answer to Problem 9.56PAE

Solution: The heat of reaction is $\text{-852 kJ/mole}$.

Explanation of Solution

$\Delta \text{H=}{\displaystyle \sum \Delta {\text{H}}_{f\text{ Product }}-{\displaystyle \sum \Delta {\text{H}}_{f\text{ Reactant}}}}$

$\text{ΔH = -1676 -}\left[\text{-824}\right]\text{ = -852 kJ/mole}$