Chapter 10, Problem 121QRT

(a)

Interpretation Introduction

Interpretation:

The chemical equation of hydrogenation reaction of $\text{4-octyne}$ has to by state with help of Lewis structures.

Concept Introduction:

The addition of ${\text{H}}_{\text{2}}$ in presence of catalyst is known as catalytic hydrogenation reaction. This is a syn addition.  Alkyne is reduced to alkane.  In this reaction two weak pi bond of alkyne and sigma bond between ${\text{H}}_{\text{2}}$ is broken and four $\text{C}-\text{H}$ bonds are formed.

Explanation of Solution

The hydrogenation reaction of $\text{4-octyne}$ gives an alkene, $\text{4-octene}$, in the first step.  The alkyne has two $\pi$ bond in it.  One $\pi$ bond of $cis\text{-4-octyne}$ gets brake to form $\text{4-octene}$. The corresponding chemical reaction is shown below.

Figure 1

The hydrogenation reaction of $cis\text{-4-octyne}$ gives an alkene, octane, in the second step.  The alkene has a $\pi$ bond in it.  The $\pi$ bond of $cis\text{-4-octyne}$ gets brake to form octane. The corresponding chemical reaction is shown below.

Figure 2

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the overall hydrogenation reaction of $\text{4-octyne}$ has to be calculated.

Concept Introduction:

The amount of energy which is produced or absorbed during the reaction is the enthalpy change of a chemical reaction.  When bonds get break, some amount of energy is absorbed by the system.  When bonds are formed, some amount of energy is produced by the system.

Explanation of Solution

The overall reaction hydrogenation reaction of $\text{4-octyne}$ is shown below.

Figure 3

The reactant $\text{4-octyne}$ has $14$ $\text{C}-\text{H}$ bonds, $6$ $\text{C}-\text{C}$ bonds and a $\text{C}\equiv \text{C}$ bond.

The reactant hydrogen gas has a $\text{H}-\text{H}$ bond.

The product octane has $18$ $\text{C}-\text{H}$ bonds and $7$ $\text{C}-\text{C}$ bonds.

The bond enthalpy for $\text{C}-\text{H}$ is $416\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{C}\equiv \text{C}$ is $813\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{C}-\text{C}$ is $356\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{H}-\text{H}$ is $436\text{kJ}/\text{mol}$.

The standard enthalpy change for the overall reaction is calculated by the formula as shown below.

  $\Delta H°=\left(\begin{array}{l}{\displaystyle \sum \left[\left(\text{number}\text{of}\text{bonds}\right)\times D\left(\text{bonds}\text{broken}\right)\right]}\\ -{\displaystyle \sum \left[\left(\text{number}\text{of}\text{bonds}\right)\times D\left(\text{bonds}\text{formed}\right)\right]}\end{array}\right)$

Where,

• $\Delta H°$ is standard enthalpy change for the overall reaction.
• $D$ is the bond enthalpy.

The above equation can rewritten for hydrogenation reaction of $\text{4-octyne}$ as shown below.

  $\Delta H°=\left[\begin{array}{l}\left(14\times D\left(\text{C}-\text{H}\right)+6\times D\left(\text{C}-\text{C}\right)+D\left(\text{C}\equiv \text{C}\right)\right)\\ +2\times D\left(\text{H}-\text{H}\right)-\left(18\times D\left(\text{C}-\text{H}\right)+7\times D\left(\text{C}-\text{C}\right)\right)\end{array}\right]$

Substitute the value of bond enthalpy for $\text{C}-\text{H}$, bond enthalpy for $\text{C}\equiv \text{C}$, bond enthalpy for $\text{C}-\text{C}$, and bond enthalpy for $\text{H}-\text{H}$ in the above equation.

  $\begin{array}{c}\Delta H°=\left[\begin{array}{l}\left(14\times \left(416\text{kJ}/\text{mol}\right)+6\times \left(356\text{kJ}/\text{mol}\right)+\left(813\text{kJ}/\text{mol}\right)\right)\\ +2\times \left(436\text{kJ}/\text{mol}\right)-\left(\left(18\times \left(416\text{kJ}/\text{mol}\right)+7\times \left(356\text{kJ}/\text{mol}\right)\right)\right)\end{array}\right]\\ =\left[5824+2136+813+872-7488-2492\right]\text{kJ}/\text{mol}\\ =\left[9645-9980\right]\text{kJ}/\text{mol}\\ =\underset{\_}{-335kJ/mol}\end{array}$

The standard enthalpy change for the overall hydrogenation reaction of $\text{4-octyne}$ is $\underset{\_}{-335kJ/mol}$.