Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

Given data:

$\text{n}\text{CO}\text{+}\left(\text{2n+1}\right){\text{H}}_{\text{2}}\to {\text{C}}_{\text{n}}{\text{H}}_{\text{2n+2}}\text{+}\text{n}{\text{H}}_{\text{2}}\text{O}$

The balanced equation when n=1 is given below:

$\text{CO}\text{+}3{\text{H}}_{\text{2}}\to {\text{CH}}_4\text{+}{\text{H}}_{\text{2}}\text{O}$

Bonds broken:

  $\begin{array}{l}\text{1}\text{mol}\text{C}\equiv \text{O}\text{bond}\text{=}\text{1}\left(\text{1073}\text{kJ}\right)\text{=}\text{1073}\text{kJ}\\ \\ \text{3}\text{mol}\text{H}-\text{H}\text{bond}\text{=}\text{3}\left(\text{436}\text{kJ}\right)\text{=}\text{1308}\text{kJ}\\ \\ \text{Total}\text{energy}\text{absorbed}\text{=}\text{1}\left(\text{1073}\text{kJ}\right)\text{=}\text{2381}\text{kJ}\end{array}$

Bonds formed:

  $\begin{array}{l}\text{4}\text{mol}\text{C-H}\text{bond}\text{=}\text{4}\left(\text{416}\text{kJ}\right)\text{=}\text{1664}\text{kJ}\\ \\ \text{2}\text{mol}\text{O}\text{-H}\text{bond}\text{=}\text{2}\left(\text{467}\text{kJ}\right)\text{=}\text{934}\text{kJ}\\ \\ \text{Total}\text{energy}\text{released=}\text{2598}\text{kJ}\end{array}$

The total energy change is given below:

  $\left(\text{+2381}\text{kJ}\right)\text{+}\left(\text{-}\text{2598}\text{kJ}\right)\text{=}\text{-217}\text{kJ}$

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

Given data:

$\text{n}\text{CO}\text{+}\left(\text{2n+1}\right){\text{H}}_{\text{2}}\to {\text{C}}_{\text{n}}{\text{H}}_{\text{2n+2}}\text{+}\text{n}{\text{H}}_{\text{2}}\text{O}$

The balanced equation when n=1 is given below:

$\text{CO}\text{+}3{\text{H}}_{\text{2}}\to {\text{CH}}_4\text{+}{\text{H}}_{\text{2}}\text{O}$

The given reaction is $\text{n}\text{CO}\text{+}\left(\text{2n+1}\right){\text{H}}_{\text{2}}\to {\text{C}}_{\text{n}}{\text{H}}_{\text{2n+2}}\text{+}\text{n}{\text{H}}_{\text{2}}\text{O}$.

From the equation it is clear that there will always be n $\text{C}\equiv \text{O}$ triple bonds to break $\left(\text{2n+1}\right)\text{H}-\text{H}$ to break. Also, the number of $\text{C-H}$ bond forming will be $\left(\text{2n+2}\right)$ and the number of $\text{O-H}$ bond will be $\text{2n}$ and also the number of $\text{C-C}$ bond forming will be $\text{n-1}$.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.