Chapter 5, Problem 111SCQ

Isomers are molecules with the same elemental composition but a different atomic arrangement. Three isomers with the formula C₄H₈ are shown in the models below. The enthalpy of combustion (Δ_cH°) of each isomer, determined using a calorimeter, is as follows:

1. (a) Draw an energy level diagram relating the energy content of the three isomers to the energy content of the combustion products, CO₂(g) and H₂O(ℓ).
2. (b) Use the Δ_cH° data in part (a), along with the enthalpies of formation of CO₂(g) and H₂O(ℓ) from Appendix L, to calculate the enthalpy of formation for each of the isomers.
3. (c) Draw an energy level diagram that relates the enthalpies of formation of the three isomers to the energy of the elements in their standard states.
4. (d) What is the enthalpy change for the conversion of cis-2-butene to trans-2-butene?

Interpretation Introduction

Interpretation:

The energy level diagram relating the energy content of the three isomers has to be determined

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ a compound directly from its component elements in their standard states. And is given by

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

The energy level diagram is given below

Figure 1

Interpretation Introduction

Interpretation:

The enthalpy of formation of ${\text{CO}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}$ has to be calculated.

Concept Introduction:

Heat energy required to raise the temperature of 1g substance by 1K.Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ of a compound directly from its component elements in their standard states. And is given by

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Given reaction is:

  ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{+6O}}_{\text{2}}\to {\text{4CO}}_{\text{2}}{\text{+4H}}_{\text{2}}\text{O}$

For cis-2-butene

Using the formula ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

$\text{(1mol×}\text{-2709}\text{.8}\text{kJ/mol)}\text{=(4mol×-393}\text{.5kJ/mol)+(4mol×-285}\text{.83}{\text{kJ/mol)-(1mol×Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{)}$$\begin{array}{l}\text{-2709}\text{.8}\text{=}\text{-2717}\text{.40}\text{-}{\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{=-2717}\text{.40}\text{-}\left(\text{-2709}\text{.8}\right)\text{=}\text{-7}\text{.6}\text{kJ/mol}\end{array}$

For trans-2-butene

Using the formula ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

$\text{(1mol×}\text{-2706}\text{.6}\text{kJ/mol)}\text{=(4mol×-393}\text{.5kJ/mol)+(4mol×-285}\text{.83}{\text{kJ/mol)-(1mol×Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{)}$$\begin{array}{l}\text{-2706}\text{.6}\text{=}\text{-2717}\text{.40}\text{-}{\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{=-2717}\text{.40}\text{-}\left(\text{-2706}\text{.6}\right)\text{=}-\text{10}\text{.8}\text{kJ/mol}\end{array}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$=$-\text{10}\text{.8}\text{kJ/mol}$

For 1-butene

Using the formula ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

$\text{(1mol×}\text{-2716}\text{.8}\text{kJ/mol)}\text{=(4mol×-393}\text{.5kJ/mol)+(4mol×-285}\text{.83}{\text{kJ/mol)-(1mol×Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{)}$$\begin{array}{l}\text{-2716}\text{.8}\text{=}\text{-2717}\text{.40}\text{-}{\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}\text{=-2717}\text{.40}\text{-}\left(\text{-2716}\text{.8}\right)\text{=}-0.6\text{kJ/mol}\end{array}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$=$-0.6\text{kJ/mol}$

the enthalpy of formation for each of the isomers found out.

Interpretation Introduction

Interpretation:

The energy level diagram based on the enthalpy of formation has to be determined.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K. Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, $\text{ΔT}$= change in temperature.

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

The energy level diagram based on the enthalpy of formation is:

  

     Figure 2

Interpretation Introduction

Interpretation:

The enthalpy change for the conversion of $\text{cis-2-butene}$ to $\text{trans-2-butene}$ has to be calculated.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K. Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, $\text{ΔT}$= change in temperature.

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Form the question the values given are:

$\begin{array}{l}\text{cis-2-butene}\text{ΔH}\left(\text{kJ/mol}\right)\text{=}\text{-2709}\text{.8}\\ \text{trans-2-butene}\text{ΔH}\left(\text{kJ/mol}\right)\text{=}\text{-2706}\text{.6}\end{array}$

Enthalpy change of cis-2-butence to trans-2-butene

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ (trans-2-butene) - ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$(cis-2-butence)

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-2706}\text{.6}\text{kJ/mol-}\left(\text{-2709}\text{.8}\text{kJ/mol}\right)\text{=}\text{-3}\text{.3}\text{kJ/mol}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-3}\text{.3kJ/mol}$

So, the enthalpy change for a conversion of cis-2-butence to trans-2-butene is

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-3}\text{.3kJ/mol}$