Chapter 8, Problem 1DE

Interpretation Introduction

Interpretation:

The facts related to the resonance stabilization of benzene, heat of hydrogenation and fuel value of benzene and acetylene are to be determined; also the experiment to prove the existence of resonance in cyclooctatetraene is to be determined.

Concept Introduction:

The enthalpy change of the combustion reaction is calculated by the difference of sum of the enthalpy change of products and reactants.

(a)

To determine:

The comparison of heat of hydrogenation of benzene and acetylene and the greater fuel value out of $1.0\text{mol}$ benzene and $3.0\text{mol}$ acetylene; also the explanation of stability of benzene on the basis of these calculations.

Explanation of Solution

The heat of hydrogenation of acetylene is $-1299.53\text{kJ/mol}$ and the fuel value of $3.0\text{mol}$ acetylene is $-149.73\text{kJ/g}$.

The heat of hydrogenation of benzene is $-3301.39\text{kJ/mol}$ and the fuel value of benzene is $-42.29\text{kJ/g}$.

The fuel value of acetylene is higher than benzene which is consistent with the stability of benzene.

Interpretation Introduction

Step 1:

To determine:

The heat of hydrogenation and fuel value of $3.0\text{mol}$ acetylene.

Answer to Problem 1DE

Solution:

The heat of hydrogenation of acetylene is $-1299.53\text{kJ/mol}$ and the fuel value of $3.0\text{mol}$ acetylene is $-149.73\text{kJ/g}$.

Explanation of Solution

The balanced chemical equation for the combustion of acetylene is,

${\text{C}}_2{\text{H}}_2\left(g\right)+\frac{5}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(2\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{C}}_2{\text{H}}_2\left(g\right)}^{\circ }+\frac{5}{2}\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{C}}_2{\text{H}}_2\left(g\right)}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^0=\left(2\times -393.5\text{kJ/mol}+1\times -285.83\text{kJ/mol}\right)\\ -\left(1\times 226.7\text{kJ/mol}+\frac{5}{2}\times 0.0\text{kJ/mol}\right)\\ =-1299.53\text{kJ/mol}\end{array}$

The fuel value of acetylene is calculated by dividing its standard enthalpy change by the mass.

$\begin{array}{c}\text{Fuel}\text{value}\text{of}{\text{C}}_2{\text{H}}_2=\frac{-1299.53\text{kJ/mol}}{26.036\text{g/mol}}\times 3.0\text{mol}\\ =-149.73\text{kJ/g}\end{array}$

Interpretation Introduction

Step 2:

To determine:

The heat of hydrogenation and fuel value of $1.0\text{mol}$ benzene.

Answer to Problem 1DE

Solution:

The heat of hydrogenation of benzene is $-3301.39\text{kJ/mol}$ and the fuel value of benzene is $-42.29\text{kJ/g}$.

Explanation of Solution

The balanced chemical equation for the combustion of benzene is,

${\text{C}}_6{\text{H}}_6\left(g\right)+\frac{15}{2}{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(6\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+3\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{C}}_6{\text{H}}_6\left(l\right)}^{\circ }+\frac{15}{2}\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{CH}}_4\left(g\right)}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^0=\left(6\times -393.5\text{kJ/mol}+3\times -285.83\text{kJ/mol}\right)\\ -\left(1\times 82.9\text{kJ/mol}+\frac{15}{2}\times 0.0\text{kJ/mol}\right)\\ =-3301.39\text{kJ/mol}\end{array}$

The fuel value of benzene is calculated by dividing its standard enthalpy change by the mass.

$\begin{array}{c}\text{Fuel}\text{value}\text{of}{\text{C}}_6{\text{H}}_6=\frac{-3301.39\text{kJ/mol}}{78.06\text{g/mol}}\times 1.0\text{mol}\\ =-42.29\text{kJ/g}\end{array}$

Since, the fuel value of benzene is less than that of acetylene. Therefore, benzene ring is especially stable and not gets easily oxidized. Thus, the calculations of heat of hydrogenation and fuel value are consistent with the extra stabilization of benzene.

Conclusion

The heat of hydrogenation of acetylene is $-1299.53\text{kJ/mol}$ and the fuel value of $3.0\text{mol}$ acetylene is $-149.73\text{kJ/g}$.

The heat of hydrogenation of benzene is $-3301.39\text{kJ/mol}$ and the fuel value of benzene is $-42.29\text{kJ/g}$.

The fuel value of acetylene is higher than benzene which is consistent with the stability of benzene.

Interpretation Introduction

(b)

To determine:

The heat of hydrogenation and fuel value of toluene.

Answer to Problem 1DE

Solution:

The heat of hydrogenation of toluene is $-3461.82\text{kJ/mol}$ and the fuel value of toluene is $-37.57\text{kJ/g}$.

Explanation of Solution

The balanced chemical equation for the combustion of benzene is,

${\text{C}}_6{\text{H}}_5{\text{CH}}_3\left(g\right)+8{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(6\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+4\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{C}}_6{\text{H}}_5{\text{CH}}_3\left(g\right)}^{\circ }+8\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{C}}_6{\text{H}}_5{\text{CH}}_3\left(g\right)}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^0=\left(6\times -393.5\text{kJ/mol}+4\times -285.83\text{kJ/mol}\right)\\ -\left(1\times -42.5\text{kJ/mol}+6\times 0.0\text{kJ/mol}\right)\\ =-3461.82\text{kJ/mol}\end{array}$

The fuel value of benzene is calculated by dividing its standard enthalpy change by the mass.

$\begin{array}{c}\text{Fuel}\text{value}\text{of}{\text{C}}_6{\text{H}}_6=\frac{-3461.82\text{kJ/mol}}{92.14\text{g/mol}}\\ =-37.57\text{kJ/g}\end{array}$

Conclusion

The heat of hydrogenation of toluene is $-3461.82\text{kJ/mol}$ and the fuel value of toluene is $-37.57\text{kJ/g}$.

Interpretation Introduction

(c)

To determine:

The explanation of stabilization of benzene by using the given values of heat of hydrogenation.

Answer to Problem 1DE

Solution:

The heat of hydrogenation for benzene is more than cyclohexane because benzene is more stable and it requires more energy to break its bonds.

Explanation of Solution

Given

The heat of hydrogenation of benzene to make cyclohexane is $208\text{kJ/mol}$.

The heat of hydrogenation of cyclohexene to make cyclohexane is $120\text{kJ/mol}$.

The heat of hydrogenation is the energy required to break the carbon-hydrogen bonds in the molecule. More is the heat of hydrogenation; more is the stability of the molecule. Since, the heat of hydrogenation of benzene is more than that of the cyclohexene. Therefore, the benzene is more stable than that of the cyclohexene.

Conclusion

The heat of hydrogenation for benzene is more than cyclohexane because benzene is more stable and it requires more energy to break its bonds.

Interpretation Introduction

(d)

To determine:

The explanation of resonance and stability of benzene on the basis of bond lengths and bond angles.

Answer to Problem 1DE

Solution:

The bond lengths and bond angle in benzene are all equal which confirms the existence of phenomenon of resonance in the molecule.

Explanation of Solution

The pi electrons present in the benzene ring is delocalized among all the carbon atoms of the ring. Due to this the bond lengths and bond angles of all carbon-carbon bonds are equal which proves the delocalized bonding in the ring. Therefore, the bond lengths and bond angles are sufficient to decide the existence of resonance and stability in the structure.

Conclusion

The bond lengths and bond angle in benzene are all equal which confirms the existence of phenomenon of resonance in the molecule.

Interpretation Introduction

(e)

To determine:

The experiment to prove the existence of resonance in cyclo octatetraene.

Answer to Problem 1DE

Solution:

The NMR spectrum of cyclo octatetraene proves that this molecule does not exhibit the phenomenon of resonance.

Explanation of Solution

The NMR spectrum of the given sample is the measure of number of hydrogen nuclei in the molecule. The number of peaks present in the NMR spectrum decides the types of nucleus present in the structure. The given structure of cyclooctatetraene contains two types of protons present in the structure having two different bond lengths. Therefore, the pheneomenon of resonance does not exhibit in this structure.

Conclusion

The NMR spectrum of cyclooctatetraene proves that this molecule does not exhibit the phenomenon of resonance.