Chapter 22, Problem 158CP

Estimate ΔH for the following reactions using bond energies given in Table 8.5.

3CH₂=CH₂(g) + 3H₂(g) → 3CH₂–CH₃(g)

The enthalpies of formation for C₆H₆(g) and C₆H₁₂ (g) are 82.9 and −90.3 kJ/mol. respectively. Calculate ΔH° for the two reactions using standard enthalpies of formation from Appendix 4. Account for any differences between the results obtained from the two methods.

Interpretation Introduction

Interpretation:

$\Delta H$ for the following reaction has to be estimated –

The estimation has to be done using bond energy values and standard enthalpy of formation values.

Concept Introduction:

Enthalpy is heat content of the system. The value of enthalpy does not depend on the path of a reaction but depend on state of the system. It has a unique value for each state of the system. Thus, enthalpy is a state function. Enthalpy is represented as,

Enthalpy change, denoted by $\text{ΔH}$, refers to heat evolved or absorbed during a reaction. $\text{ΔH}$ can be represented as,

$\text{ΔH = }\sum \text{n }\times \text{D(bonds broken) - }\sum \text{n }\times \text{D (bonds formed)}$

Where $\text{'}\sum \text{'}$ represents sum of terms, ‘D’ represents the bond energy per mole of bonds and ‘n’ represents the number of moles of the bond.

Energy required to break the existing bonds carries positive sign as energy is supplied to the system and the energy released in formation of new bonds carries negative as energy is removed from the system.

Answer to Problem 158CP

$\Delta H$ for the first reaction is estimated using bond energy values is $-141kJ/mol$

$\Delta H$ for the second reaction is estimated using bond energy values is $-141kJ/mol$

Explanation of Solution

The given reaction is,

In the above reaction, Hydrogen atoms are added to the $\text{C=C}$ bond.  In this process $3\text{ C-C}$ bonds are broken, $3\text{ H-H}$ bonds are broken and $\text{6 C-H}$ bonds are formed.  The bond dissociation energy of $\text{C-C}$ and $\text{H-H}$ bond are $347kJ/mol\text{ and }432kJ/mol$ respectively.  The bond energy of $\text{C-H}$ bond is $413kJ/mol$.

Energy required to break the existing bonds carries positive sign as energy is supplied to the system and the energy released in formation of new bonds carries negative as energy is removed from the system.

Accordingly, the enthalpy change for the above reaction is calculated as –

$\begin{array}{l}{\text{ΔH = 3 × D}}_{\text{C-C}}{\text{ + 3 D}}_{\text{H-H}}{\text{ - 6× D}}_{\text{C-H}}\\ \text{ = }\left(\text{3 mol }\times \text{347 kJ/mol}\right)\text{ + }\left(3\text{ mol}\times 432\text{ kJ/mol}\right)-\left(6\text{ mol}\times 413\text{ kJ/mol}\right)\\ \text{ = -141 kJ/mol}\end{array}$

Enthalpy of formation of ethylene, hydrogen and ethane are $227\text{ kJ/mol, 0 kJ/mol and -84}\text{.7 kJ/mol}$ respectively.  Accordingly, the enthalpy change for the above reaction is calculated as –

$\begin{array}{l}{\text{ΔH}}^{°}{\text{ = ΔH}}_{\text{f}}^{\text{°}}{\text{(products) - ΔH}}_{\text{f}}^{\text{°}}\text{(reactants)}\\ \text{ = 3 mol}\left(\text{-84}\text{.7 kJ/mol}\right)\text{ - }\left(\text{3 mol×227 kJ/mol + 3 mol }\times \text{0 kJ/mol}\right)\\ \text{ = -254}\text{.1 kJ - 227 kJ = -935}\text{.1 kJ}\end{array}$

The term “bond energy” and “enthalpy of formation” are not similar.  Bond energy refers to energy released or absorbed when isolated elements combine to form a bond.  Enthalpy of formation refers to the energy released or absorbed during formation of a compound from other compounds.

Thus, enthalpy of formation of each of the compound on the reactant and product side is widely different from bond energy of the same compounds. Hence enthalpy of reaction calculated using these two different parameters is not same.

The another reaction is,

In the above reaction, Hydrogen atoms are added to the $\text{C=C}$ bond.  In this process $3\text{ C-C}$ bonds are broken, $3\text{ H-H}$ bonds are broken and $\text{6 C-H}$ bonds are formed.  The bond dissociation energy of $\text{C-C}$ and $\text{H-H}$ bond are $347kJ/mol\text{ and }432kJ/mol$ respectively.  The bond energy of $\text{C-H}$ bond is $413kJ/mol$.

Accordingly, the enthalpy change for the above reaction is calculated as –

$\begin{array}{l}{\text{ΔH = 3 × D}}_{\text{C-C}}{\text{ + 3 D}}_{\text{H-H}}{\text{ - 6× D}}_{\text{C-H}}\\ \text{ = }\left(\text{3 mol }\times \text{347 kJ/mol}\right)\text{ + }\left(3\text{ mol}\times 432\text{ kJ/mol}\right)-\left(6\text{ mol}\times 413\text{ kJ/mol}\right)\\ \text{ = -141 kJ/mol}\end{array}$

Enthalpy of formation of benzene, hydrogen and cyclohexane are $\text{82}\text{.9 kJ/mol, 0 kJ/mol and -90}\text{.3 kJ/mol}$ respectively.  Accordingly, the enthalpy change for the above reaction is calculated as –

$\begin{array}{l}{\text{ΔH}}^{°}{\text{ = ΔH}}_{\text{f}}^{\text{°}}{\text{(products) - ΔH}}_{\text{f}}^{\text{°}}\text{(reactants)}\\ \text{ = 1 mol}\left(\text{-90}\text{.3 kJ/mol}\right)\text{ - }\left(\text{1 mol×82}\text{.9 kJ/mol + 3 mol }\times \text{0 kJ/mol}\right)\\ \text{ = -90}\text{.3 kJ - 82}\text{.9 kJ = -173}\text{.2 kJ}\end{array}$

Conclusion

As we discussed earlier, enthalpy of formation of each of the compound on the reactant and product side is widely different from bond energy of the same compounds. Hence enthalpy of reaction calculated using these two different parameters is not same.