Chapter 6, Problem 54E

For the following reactions at constant pressure, predict if ΔH > ΔE, ΔH < ΔE, or ΔH = ΔE.

a. 2HF(g) → H₂(g) + F₂(g)

b. N₂(g) + 3H₂(g) → 2NH₃(g)

c. 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Interpretation Introduction

Interpretation:

For the given set of reactions it should be predicted that which is greater either $\text{ΔE or ΔH}$.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. ${\text{ΔH = Σ}}_{\text{nP}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(P)}}}^{{}^{\circ }}-{\text{Σ}}_{\text{nR}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(R)}}}^{\circ }$.

Considering the following equation we can identify whether $\text{ΔE or ΔH}$ is greater for the given reaction.

$\begin{array}{l}{\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}\\ {\text{where, Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\text{.}\\ \text{ΔH}=\text{Enthaply change}\\ \text{ΔE}\text{=}\text{Energy change}\\ \text{R}\text{=}\text{Gas constant}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Hess law: It states that the total enthalpy of the reaction is equal to the sum of the enthalpy change for individual steps involved in the reaction.

To predict: Whether $\text{ΔH}$ or $\text{ ΔE}$ greater for the given set of reactions.

Answer to Problem 54E

For the given reaction $\text{ΔH = ΔE}$.

Explanation of Solution

Analysing the given reaction shows that there are two gaseous reactants that gives rise to 2 gaseous products therefore, ${\text{Δn}}_{\text{g}}$ is calculated as follows,

$\begin{array}{l}{\text{Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\\ \text{=}\text{2-2=0}\end{array}$

For the given reaction ${\text{Δn}}_{\text{g}}$ is 0 hence using the given equation ${\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}$ we found that $\text{ΔH = ΔE}$.

Interpretation Introduction

Interpretation:

For the given set of reactions it should be predicted that which is greater either $\text{ΔE or ΔH}$.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. ${\text{ΔH = Σ}}_{\text{nP}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(P)}}}^{{}^{\circ }}-{\text{Σ}}_{\text{nR}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(R)}}}^{\circ }$.

Considering the following equation we can identify whether $\text{ΔE or ΔH}$ is greater for the given reaction.

$\begin{array}{l}{\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}\\ {\text{where, Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\text{.}\\ \text{ΔH}=\text{Enthaply change}\\ \text{ΔE}\text{=}\text{Energy change}\\ \text{R}\text{=}\text{Gas constant}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Hess law: It states that the total enthalpy of the reaction is equal to the sum of the enthalpy change for individual steps involved in the reaction.

To predict: Whether $\text{ΔH}$ or $\text{ ΔE}$ greater for the given set of reactions.

Answer to Problem 54E

For the given reaction $\text{ΔH < ΔE}$.

Explanation of Solution

Analysing the given reaction shows that there are 4 gaseous reactants that gives rise to 2 gaseous products therefore, ${\text{Δn}}_{\text{g}}$ is calculated as follows,

$\begin{array}{l}{\text{Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\\ \text{=}\text{2-4}\text{=}-2\end{array}$

For the given reaction ${\text{Δn}}_{\text{g}}$ is -2 hence using the given equation ${\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}$ we found that $\text{ΔH < ΔE}$.

Interpretation Introduction

Interpretation:

For the given set of reactions it should be predicted that which is greater either $\text{ΔE or ΔH}$.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. ${\text{ΔH = Σ}}_{\text{nP}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(P)}}}^{{}^{\circ }}-{\text{Σ}}_{\text{nR}}{\text{ΣH}}_{{{}_{\text{f}}}_{\text{(R)}}}^{\circ }$.

Considering the following equation we can identify whether $\text{ΔE or ΔH}$ is greater for the given reaction.

$\begin{array}{l}{\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}\\ {\text{where, Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\text{.}\\ \text{ΔH}=\text{Enthaply change}\\ \text{ΔE}\text{=}\text{Energy change}\\ \text{R}\text{=}\text{Gas constant}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Hess law: It states that the total enthalpy of the reaction is equal to the sum of the enthalpy change for individual steps involved in the reaction.

To predict: Whether $\text{ΔH}$ or $\text{ ΔE}$ greater for the given set of reactions.

Answer to Problem 54E

For the given reaction $\text{ΔH >ΔE}$.

Explanation of Solution

Analysing the given reaction shows that there are 9 gaseous reactants that gives rise to 10 gaseous products therefore, ${\text{Δn}}_{\text{g}}$ is calculated as follows,

$\begin{array}{l}{\text{Δn}}_{\text{g}}\text{= number of gaseous products - number of gaseous reactants}\\ \text{=}10\text{-9}\text{=}1\end{array}$

For the given reaction ${\text{Δn}}_{\text{g}}$ is -2 hence using the given equation ${\text{ΔH = ΔE+Δn}}_{\text{g}}\text{RT}$ we found that $\text{ΔH >ΔE}$.