Chapter 6, Problem 6.105P

(a)

Interpretation Introduction

Interpretation:

The heat that is released when $25.0\text{ g}$ of methane burns in excess ${\text{O}}_2$ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions.

The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum n\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.105P

$-1.25\times {10}^3\text{kJ}$ is released when $25.0\text{ g}$ of methane burns in excess ${\text{O}}_2$.

Explanation of Solution

The expression to calculate the moles of ${\text{CH}}_{\text{4}}$ is:

  $\text{Moles}\text{of}{\text{CH}}_{\text{4}}=\left(\frac{{\text{mass of CH}}_{\text{4}}}{{\text{Molar mass of CH}}_{\text{4}}}\right)$        (1)

Substitute $25\text{g}$ for mass of ${\text{CH}}_{\text{4}}$ and $16.04\text{g/mol}$ for molar mass of ${\text{CH}}_{\text{4}}$ in the equation (1).

  $\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_{\text{4}}=\left(\frac{25\text{g}}{16.04\text{g/mol}}\right)\\ =1.5586\text{mol}\end{array}$

The balanced chemical equation for the reaction of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_2$ is as follows:

  ${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{1\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\\ \left\{1\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{4}}\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$        (2)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-285.840\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$, $-74.87\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{4}}\left(g\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (2).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(\text{1}\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(2\text{mol}\right)\left(-285.840\text{kJ/mol}\right)\right\}-\\ \left\{\left(\text{1}\text{mol}\right)\left(-74.87\text{kJ/mol}\right)+0\right\}\end{array}\right]\\ =–802.282\text{ kJ/mol}\end{array}$

The heat released by $1.5586\text{mol}$ of ${\text{CH}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{Heat relased}=\left(1.5586\text{mol}\right)\left(–802.282\text{ kJ/mol}\right)\\ =–1250.4\text{kJ}\\ \approx -1.25\times {10}^3\text{kJ}\text{.}\end{array}$

Conclusion

The standard enthalpy of reaction depends on the standard enthalpy of formation of the product and the standard enthalpy of formation of reactant at the standard conditions.

(b)

Interpretation Introduction

Interpretation:

The temperature of the product mixture is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Molar heat capacity $\left(C_{\text{m}}\right)$ of a substance is the amount of heat needed to raise the temperature of $1\text{mol}$ of a substance by $1\text{K}$.

The formula to calculate heat required is as follows:

  $q=\left(\text{mol}\right)\left(C_{\text{m}}\right)\left(\Delta T\right)$

Here,

$\Delta T$ is the temperature difference.

$q$ is the heat released or absorbed.

$C_{\text{m}}$ is the molar heat capacity of the substance.

Answer to Problem 6.105P

The temperature of the product mixture is $2.24\times {10}^3°\text{C}$.

Explanation of Solution

The formula to calculate the moles of ${\text{CO}}_2$ is as follows:

  $\text{Moles of}{\text{CO}}_2=\text{moles of}{\text{CH}}_{\text{4}}\left(\frac{1\text{mol}{\text{CO}}_2}{1\text{mol}{\text{CH}}_{\text{4}}}\right)$        (3)

Substitute $1.5586\text{mol}$ for moles of ${\text{CH}}_{\text{4}}$ in the equation (3).

  $\begin{array}{c}\text{Moles of}{\text{CO}}_2=1.5586\text{mol}\left(\frac{1\text{mol}{\text{CO}}_2}{1\text{mol}{\text{CH}}_{\text{4}}}\right)\\ =1.5586\text{mol}\end{array}$

The formula to calculate the moles of ${\text{H}}_{\text{2}}\text{O}$ is as follows:

  $\text{Moles of}{\text{H}}_{\text{2}}\text{O}=\text{moles of}{\text{CH}}_{\text{4}}\left(\frac{2\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{CH}}_{\text{4}}}\right)$        (4)

Substitute $1.5586\text{mol}$ for moles of ${\text{CH}}_{\text{4}}$ in the equation (4).

  $\begin{array}{c}\text{Moles of}{\text{H}}_{\text{2}}\text{O}=1.5586\text{mol}\left(\frac{2\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{CH}}_{\text{4}}}\right)\\ =3.1172\text{mol}\end{array}$

The formula to calculate the moles of ${\text{O}}_2$ is as follows:

  $\text{Moles of}{\text{O}}_2=\text{moles of}{\text{CH}}_{\text{4}}\left(\frac{2\text{mol}{\text{O}}_2}{1\text{mol}{\text{CH}}_{\text{4}}}\right)$        (5)

Substitute $1.5586\text{mol}$ for moles of ${\text{CH}}_{\text{4}}$ in the equation (5).

  $\begin{array}{c}\text{Moles of}{\text{O}}_2=1.5586\text{mol}\left(\frac{2\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{CH}}_{\text{4}}}\right)\\ =3.1172\text{mol}\end{array}$

The mole fraction of ${\text{O}}_2$ and ${\text{N}}_2$ is $0.79$ and $0.21$.

The formula to calculate the moles of ${\text{N}}_2$ is as follows:

  $\text{Moles of}{\text{N}}_2=\left(\text{moles of}{\text{O}}_2\right)\left(\frac{0.79\text{mol}{\text{N}}_2}{0.21\text{mol}{\text{O}}_2}\right)$        (6)

Substitute $3.1172\text{mol}$ for moles of ${\text{O}}_2$ in the equation (6).

  $\begin{array}{c}\text{Moles of}{\text{N}}_2=\left(3.1172\text{mol}\right)\left(\frac{0.79\text{mol}{\text{N}}_2}{0.21\text{mol}{\text{O}}_2}\right)\\ =11.72661\text{mol}\end{array}$

The formula to calculate the final temperature is as follows:

  $q=\left[\begin{array}{l}\left({\text{mol}}_{{\text{CO}}_2}\right)\left(C_{{\text{CO}}_2}\right)\left(T_{\text{final}}-T_{\text{initial}}\right)+\left({\text{mol}}_{{\text{H}}_{\text{2}}\text{O}}\right)\left(C_{{\text{H}}_{\text{2}}\text{O}}\right)\left(T_{\text{final}}-T_{\text{initial}}\right)\\ +\left({\text{mol}}_{{\text{N}}_2}\right)\left(C_{{\text{N}}_2}\right)\left(T_{\text{final}}-T_{\text{initial}}\right)\end{array}\right]$        (7)

Rearrange the equation (7) to calculate $T_{\text{final}}$ as follows:

  $T_{\text{final}}=\left(\frac{q}{\left[\begin{array}{l}\left({\text{mol}}_{{\text{CO}}_2}\right)\left(C_{{\text{CO}}_2}\right)+\left({\text{mol}}_{{\text{H}}_{\text{2}}\text{O}}\right)\left(C_{{\text{H}}_{\text{2}}\text{O}}\right)\\ +\left({\text{mol}}_{{\text{N}}_2}\right)\left(C_{{\text{N}}_2}\right)\end{array}\right]}+T_{\text{initial}}\right)$        (8)

Substitute $1.5586\text{mol}$ for ${\text{mol}}_{{\text{CO}}_2}$, $57.2\text{ J/mol°C}$ for $C_{{\text{CO}}_2}$, $3.1172\text{mol}$ for ${\text{mol}}_{{\text{H}}_{\text{2}}\text{O}}$, $36.0\text{ J/mol°C}$ for $C_{{\text{H}}_{\text{2}}\text{O}}$, $11.72661\text{mol}$ for ${\text{mol}}_{{\text{N}}_2}$, $30.5\text{ J/mol°C}$ for $C_{{\text{N}}_2}$, $0.0°\text{C}$ for $T_{\text{initial}}$ and $1250.4\text{ kJ}$ for $q$ in the equation (8).

  $\begin{array}{c}{T}_{\text{final}}=\left(\frac{1250.4\text{ kJ}\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{\left[\begin{array}{l}\left(1.5586\text{mol}\right)\left(57.2\text{ J/mol°C}\right)+\left(3.1172\text{mol}\right)\left(36.0\text{ J/mol°C}\right)\\ +\left(11.72661\text{mol}\right)\left(30.5\text{ J/mol°C}\right)\end{array}\right]}+0.0°\text{C}\right)\\ =2236.72°\text{C}\\ \approx 2.24\times {10}^3°\text{C}\text{.}\end{array}$

Conclusion

The amount of species in the reaction depends on the stoichiometry of the reaction by the relationship between the reactants and products.