Chapter 21, Problem 21.123QP

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the given “cracking” reaction has to be calculated

Concept Introduction:

Enthalpy of reaction: ( $\Delta \text{H}°$)

The enthalpy of a reaction is calculated by subtracting the heat of formation of reactatns from heat of formation of products.

$\Delta \text{H}°\text{ =}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}$

Answer to Problem 21.123QP

The standard enthalpy change for the given “cracking” reaction is $74.87\text{ kJ}$

Explanation of Solution

Given data:

The given cracking reaction is:

${\text{CH}}_{\text{4}}\text{(g) }\stackrel{}{\to }{\text{C(graphite) + 2H}}_{\text{2}}(g)$

To Calculate: The standard enthalpy change for the given “cracking” reaction.

Using the Appendix C, the heat of formation ( $\Delta \text{H}{°}_f$) of the reactants and products are written under their equation as follows,

$\begin{array}{l}{\text{CH}}_{\text{4}}\text{(g) }\stackrel{}{\to }{\text{C(graphite) + 2H}}_{\text{2}}(g)\\ \Delta \text{H}{°}_f-74.8700\text{ (kJ)}\end{array}$

The $\Delta \text{H}°$ is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[0 - (-74}\text{.87)] kJ}\\ \text{=74}\text{.87 kJ}\end{array}$

Hence, the standard enthalpy change is $74.87\text{ kJ}$

Conclusion

The standard enthalpy change for the given “cracking” reaction is calculated as $74.87\text{ kJ}$

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the given “cracking” reaction has to be calculated

Concept Introduction:

Enthalpy of reaction: ( $\Delta \text{H}°$)

The enthalpy of a reaction is calculated by subtracting the heat of formation of reactatns from heat of formation of products.

$\Delta \text{H}°\text{ =}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}$

Answer to Problem 21.123QP

The standard enthalpy change for the given “cracking” reaction is $137.15\text{ kJ}$

Explanation of Solution

Given data:

The given cracking reaction is:

${\text{C}}_{\text{2}}{\text{H}}_6(g)\stackrel{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{4}}(g){\text{ + 2H}}_{\text{2}}(g)$

To Calculate: The standard enthalpy change for the given “cracking” reaction.

Using the Appendix C, the heat of formation ( $\Delta \text{H}{°}_f$) of the reactants and products are written under their equation as follows,

$\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_6(g)\stackrel{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{4}}(g){\text{ + 2H}}_{\text{2}}(g)\\ \Delta \text{H}{°}_f-84.68\text{ 52}\text{.47}0\text{ (kJ)}\end{array}$

The $\Delta \text{H}°$ is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[52}\text{.47 - (-84}\text{.68)] kJ}\\ \text{=137}\text{.15 kJ}\end{array}$

Hence, the standard enthalpy change is $137.15\text{ kJ}$

Conclusion

The standard enthalpy change for the given “cracking” reaction is calculated as $137.15\text{ kJ}$