Chapter 23, Problem 111SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced equations for the combustion of ethane gas and liquid ethanol have to be written and calculate the enthalpy of combustion of each component.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

  Enthalpy of combustion:

• The standard enthalpy of combustion is the enthalpy change that occurs when one mole of substance burns completely under the standard conditions of ${\text{25}}^{\text{0}}\text{C}$ and 1 atm.
• The standard enthalpy of combustion is always negative.

Answer to Problem 111SCQ

$\begin{array}{l}\text{Balanced}\text{equation}\text{for}\text{the}\text{combustion of ethane:}\\ {\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g) + 7O}}_{\text{2}}\text{(g) }\to {\text{ 4CO}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{O(l)}\\ \text{Balanced}\text{equation}\text{for}\text{the}\text{combustion of ethanol}\\ {\text{2C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH + 7O}}_{\text{2}}\to {\text{4CO}}_{\text{2}}{\text{ + 6H}}_{\text{2}}\text{O}\end{array}$

Heat of combustion of ethane - $\text{- 52 kJ/g}$

Heat combustion of ethanol -  $\text{- 30}\text{.6 kJ/g}$.

Explanation of Solution

Let’s write the balanced equation of combustion of ethane:

${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g) + 7O}}_{\text{2}}\text{(g) }\to {\text{ 4CO}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{O(l)}$

Let’s calculate the enthalpy of combustion:

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}\text{ (products) - }{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}\text{ (reactants)}}}\\ {\text{= 4ΔH}}_{\text{f}}^{\text{o}}{\text{ (CO}}_{\text{2}}{\text{(g)) + 6ΔH}}_{\text{f}}^{\text{o}}{\text{ (H}}_{\text{2}}{\text{O(l)) - 7ΔH}}_{\text{f}}^{\text{o}}{\text{ (O}}_{\text{2}}{\text{(g))- 2ΔH}}_{\text{f}}^{\text{o}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g))}\\ \text{= 4 × (-393}\text{.5 kj/mol) + 6×(-285}\text{.83) - 7 (0 kj/mol) - 2×(-83}\text{.85 kj/mol)}\\ \text{= - 3121}\text{.3 kj}\end{array}$

The combustion of one mole of ethane released $\text{- 3121}\text{.3 kj}$ energy.

Let’s calculate the energy released by 1 gm of ethane:

$\begin{array}{c}\text{Relaesed Energy for1gm ethane = }\frac{\text{-1560}\text{.6 kj/mol}}{\text{30 g/mol}}\\ \text{= -52 kJ/g}\end{array}$

Therefore, Heat of combustion of ethane is $\text{- 52 kJ/g}$.

Let’s write the balanced equation of combustion of ethanol:

${\text{2C}}_{\text{2}}{\text{H}}_5{\text{OH (g) + 7O}}_{\text{2}}\text{(g) }\to {\text{ 4CO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O(l)}$

Let’s calculate the enthalpy of combustion:

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}\text{ (products) - }{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}\text{ (reactants)}}}\\ {\text{= 4ΔH}}_{\text{f}}^{\text{o}}{\text{ (CO}}_{\text{2}}{\text{(g)) + 6ΔH}}_{\text{f}}^{\text{o}}{\text{ (H}}_{\text{2}}{\text{O(l)) - 7ΔH}}_{\text{f}}^{\text{o}}{\text{ (O}}_{\text{2}}{\text{(g))- 2ΔH}}_{\text{f}}^{\text{o}}{\text{(C}}_{\text{2}}{\text{H}}_5\text{OH(g))}\\ \text{= 4 × (-393}\text{.5 kj/mol) + 6×(-285}\text{.83) - 7 (0 kj/mol) - 2×(-235}\text{.3 kj/mol)}\\ \text{= - 2818}\text{.4 kj}\end{array}$

The combustion of one mole of ethanol released $\text{- 2818}\text{.4 kj}$ energy.

Let’s calculate the energy released by 1 gm of ethanol:

$\begin{array}{c}\text{Relaesed Energy for1gm ethanol = }\frac{\text{-2818}\text{.4 kj/mol}}{\text{46 g/mol}}\\ \text{= -30}\text{.6 kJ/g}\end{array}$

Therefore, Heat of combustion of ethanol is $\text{- 30}\text{.6 kJ/g}$.

The heat obtained from the combustion of ethanol is less negative than for ethane.

(b)

Interpretation Introduction

Interpretation:

If ethanol is assumed to be partially oxidized ethane, then the effect does this have on the enthalpy of combustion should be determined.

Concept introduction:

Enthalpy of combustion:

• The standard enthalpy of combustion is the enthalpy change that occurs when one mole of substance burns completely under the standard conditions of ${\text{25}}^{\text{0}}\text{C}$ and 1 atm.
• The standard enthalpy of combustion is always negative.

Explanation of Solution

The heat obtained from the combustion of ethanol is less negative than for ethane, so partially oxidizing ethane to form ethanol decreases the amount of energy per gram available from the combustion of the substance.