Chapter 6, Problem 24Q

The combustion of methane can be represented as follows:

a. Use the information given above to determine the value of ∆H for the combustion of methane to form CO₂(g) and 2H₂O(l).

b. What is $\Delta H_{\text{f}}^{°}$ for an element in its standard state? Why is this? Use the figure above to support your answer.

c. How does ∆H for the reaction CO₂(g) + 2H₂O (1) → CH₄(g) + O₂(g) compare to that of the combustion of methane? Why is this?

Interpretation Introduction

Interpretation:

Enthalpy change of combustion reaction of methane has to be calculated.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
• In the equation enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.

${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
• The enthalpy change of formation reaction equal to reeves sign of combustion reaction.
• Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

$\text{ΔH}\left(\text{total}\right)\text{= }{\displaystyle \sum \text{ΔH}}\left(\text{number of steps}\right)\mathrm{......}\text{(2)}$

Rules:

• The sign of ΔH reversed when the reaction is reversed.
• The quantities of reactants and products are directly proportional to the magnitude of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.
• The standard enthalpy of formation of homo atomic molecules is zero.

Answer to Problem 24Q

Enthalpy change $\Delta {\rm H}$ of methane combustion reaction is $-\text{891}\text{kJ}$

Explanation of Solution

To calculate the enthalpy change $\Delta {\rm H}$ of methane combustion reaction

$\begin{array}{l}\text{Reactants}\to \text{Standard State Elements}\\ \text{ΔΗ}\text{=}{\text{ΔΗ}}_{{\text{CH}}_{\text{4(g)}}}{\text{+ΔΗ}}_{{\text{O}}_{\text{2(g)}}}\\ \text{=}\text{75+}\text{0}\\ \text{=}\text{75}\text{kJ}\end{array}$

$\begin{array}{l}\text{Standard State Elements}\to \text{Reactants}\\ \text{ΔΗ}\text{=}{\text{ΔΗ}}_{{\text{CO}}_{\text{2(g)}}}{\text{+ΔΗ}}_{{\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}}\\ \text{=}\text{-394-}\text{572}\\ \text{=}\text{-966}\text{k}\end{array}$

$\begin{array}{l}\text{Reactants }\to \text{Products}\\ \text{ΔΗ}\text{=}\text{75-966}\\ \text{=}-\text{891}\text{kJ}\end{array}$

• According to Hess's Law, the enthalpy of formation values of reactants is subtracted from produce to gives enthalpy change $\Delta {\rm H}$ of reaction.
• The given enthalpy of formation and standard enthalpy of formation values are plugged in the above equation to give an enthalpy change $\Delta {\rm H}$ of methane combustion reaction.
• Enthalpy change $\Delta {\rm H}$ of methane combustion reaction is $-\text{891}\text{kJ}$

Interpretation Introduction

Interpretation:

Support of standard enthalpy of formation each elements involving in the reaction are should be given.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
• In the equation enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.

${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
• The enthalpy change of formation reaction equal to reeves sign of combustion reaction.
• Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

$\text{ΔH}\left(\text{total}\right)\text{= }{\displaystyle \sum \text{ΔH}}\left(\text{number of steps}\right)\mathrm{......}\text{(2)}$

Rules:

• The sign of $\text{ΔH}$ reversed when the reaction is reversed.
• The quantities of reactants and products are directly proportional to the magnitude of $\text{ΔH}$ in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of $\text{ΔH}$ is multiplied by the same integer.
• The standard enthalpy of formation of homo atomic molecules is zero.

Answer to Problem 24Q

The enthalpy change of a reaction is calculated by addition and canceling each side of same chemicals of enthalpy of formation values of reactants and enthalpy change of the reactants to product.

Hence the given standard enthalpy of formation is giving a more support to calculation of enthalpy change $\Delta {\rm H}$ of given reaction.

Explanation of Solution

To explain the support of standard enthalpy of formation each elements involving in the reaction

• According to Hess's Law, the enthalpy of formation values of reactants is subtracted from produce to gives enthalpy change $\Delta {\rm H}$ of reaction and standard enthalpy of formation of element is zero.
• The enthalpy change of a reaction is calculated by addition and canceling each side of same chemicals of enthalpy of formation values of reactants and enthalpy change of the reactants to product.
• Hence the given standard enthalpy of formation is giving a more support to calculation of enthalpy change $\Delta {\rm H}$ of given reaction.

Interpretation Introduction

Interpretation:

The enthalpy change of formation of methane should be compared with combustion of methane.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
• In the equation enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.

${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
• The enthalpy change of formation reaction equal to reeves sign of combustion reaction.
• Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

$\text{ΔH}\left(\text{total}\right)\text{= }{\displaystyle \sum \text{ΔH}}\left(\text{number of steps}\right)\mathrm{......}\text{(2)}$

Rules:

• The sign of ΔH reversed when the reaction is reversed.
• The quantities of reactants and products are directly proportional to the magnitude of $\text{ΔH}$ in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of $\text{ΔH}$ is multiplied by the same integer.
• The standard enthalpy of formation of homo atomic molecules is zero.

Answer to Problem 24Q

The formation of methane reaction is equal to combustion reaction of methane. So the enthalpy change of formation reaction equal to reeves sign of combustion reaction and it is $\text{891}\text{kJ}$

Explanation of Solution

To comparatively explain the enthalpy change of combustion and formation of methane

The formation reaction of methane is,

${\text{CO}}_{\text{2(g) }}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{ (1)}}\to {\text{CH}}_{\text{4(g)}}{\text{ + O}}_{\text{2(g)}}$

$\begin{array}{l}\text{Reactants}\to \text{Standard State Elements}\\ \text{=}\text{394+}\text{572}\\ \text{=}\text{966}\text{k}\end{array}$

$\begin{array}{l}\text{Standard State Elements}\to \text{Reactants}\\ \text{ΔΗ}\text{=}{\text{ΔΗ}}_{{\text{CO}}_{\text{2(g)}}}{\text{+ΔΗ}}_{{\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}}\\ \text{ΔΗ}\text{=}{\text{ΔΗ}}_{{\text{CH}}_{\text{4(g)}}}{\text{+ΔΗ}}_{{\text{O}}_{\text{2(g)}}}\\ \text{=}\text{75+}\text{0}\\ \text{=}-\text{75}\text{kJ}\end{array}$

$\begin{array}{l}\text{Reactants }\to \text{Products}\\ \text{ΔΗ}\text{=}\text{966}-75\\ \text{=}\text{891}\text{kJ}\end{array}$

• According to Hess's Law, the enthalpy change of formation reaction is exactly equal to enthalpy change of combustion reaction.
• The given enthalpy of formation and standard enthalpy of formation values are plugged in the above equation to give an enthalpy change $\Delta {\rm H}$ of methane formation reaction.
• Hence the formation of methane reaction is equal to combustion reaction of methane. So the enthalpy change of formation reaction equal to reeves sign of combustion reaction and it is $\text{891}\text{kJ}$