Chapter 4, Problem 4D.10E

(a)

Interpretation Introduction

Interpretation:

The mass of carbon dioxide produce on the combustion of mixture of natural gas each minute has to be determined.

Concept Introduction:

The combustion of a substance takes place in the presence of oxygen.  The products which are generated upon the combustion of any organic compound are carbon dioxide gas and water along with the large amount of energy in the form of heat and light.

Answer to Problem 4D.10E

The mass of carbon dioxide produced each minute is $\underset{\_}{7.7\times 10^{-5}g}$.

Explanation of Solution

The balanced chemical equation that represents the combustion of one mole of methanegas is shown below.

    $C{H}_4\left(g\right)+O_{\text{2}}\left(g\right)\to C{O}_{\text{2}}\left(g\right)+2H_{2}O\left(l\right)$

The combustion of one mole of methane gas produces one mole of carbon dioxide gas.  Therefore, $9.3kmol$ of $C{H}_4\left(g\right)$ will produce $9.3kmol$ of $C{O}_{\text{2}}\left(g\right)$.

The balanced chemical equation that represents the combustion of one mole of ethane gas is shown below.

    $C_2{H}_6\left(g\right)+\frac{7}{2}{O}_{\text{2}}\left(g\right)\to 2C{O}_{\text{2}}\left(g\right)+3H_{2}O\left(l\right)$

The combustion of one mole of ethane gas produces two mole of carbon dioxide gas.  Therefore, $3.1kmol$ of $C_2{H}_6\left(g\right)$ will produce $3.1kmol\times 2=6.2kmol$ of $C{O}_{\text{2}}\left(g\right)$.

The balanced chemical equation that represents the combustion of one mole of propane gas is shown below.

    $C_3{H}_8\left(g\right)+5O_{\text{2}}\left(g\right)\to 3C{O}_{\text{2}}\left(g\right)+4H_{2}O\left(l\right)$

The combustion of one mole of propane gas produces three mole of carbon dioxide gas.  Therefore, $0.40kmol$ of $C_3{H}_8\left(g\right)$ will produce $0.40kmol\times 3=1.20kmol$ of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$.

The balanced chemical equation that represents the combustion of one mole of butane gas is shown below.

    $C_4{H}_{10}\left(g\right)+\frac{13}{2}{O}_{\text{2}}\left(g\right)\to 4C{O}_{\text{2}}\left(g\right)+5H_{2}O\left(l\right)$

The combustion of one mole of butane gas produces four mole of carbon dioxide gas.  Therefore, $0.20kmol$ of $C_4{H}_{10}\left(g\right)$ will produce $0.20kmol\times 4=0.80kmol$ of $C{O}_{\text{2}}\left(g\right)$.

The total number of moles of carbon dioxide gas produced by the combustion of methane, ethane, propane and butane will get by adding the individual amount of carbon dioxide produced by each gas as shown below.

  $\begin{array}{c}{n}_{\text{total}}\left({\text{CO}}_{\text{2}},\text{g}\right)=9.3\text{kmol}+6.2\text{kmol}+1.20\text{kmol}+0.80\text{kmol}\\ =\text{17}\text{.5}\text{kmol}\end{array}$

Therefore, the total number of moles of carbon dioxide gas produced by the combustion of natural gas mixture is $\text{17}\text{.5}\text{kmol}$.

The mass of carbon dioxide gas produced by the mixture of natural gas is calculated by the relation shown below.

  $m=n\times M_{\text{w}}$        (1)

Where,

• $n$ is the number of moles.
• $m$ is the mass.
• $M_{\text{w}}$ is the molar mass.

The value of $n$ for $C{O}_{\text{2}}\left(g\right)$ is $\text{17}\text{.5}\text{kmol}$.

The value of $M_{\text{w}}$ for $C{O}_{\text{2}}\left(g\right)$ is $44.0\text{g}\cdot {\text{mol}}^{-1}$.

Substitute the values of $n$ and $M_{\text{w}}$ for $C{O}_{\text{2}}\left(g\right)$ in the equation (1).

  $\begin{array}{c}m=\text{17}\text{.5}\text{kmol}\times 44.0\text{g}\cdot {\text{mol}}^{-1}\times \frac{1000\text{mol}}{1\text{kmol}}\\ =7.7\times {10}^{-5}\text{g}\end{array}$

Thus, the mass of carbon dioxide produced each minute is $\underset{\_}{7.7\times 10^{-5}g}$.

(b)

Interpretation Introduction

Interpretation:

The heat released by the combustion of natural gas each minute has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4D.10E

The heat produced per minute is $\underset{\_}{1.46\times 10^{7}kJ}$.

Explanation of Solution

The thermochemical equation for the combustion of methane, ethane, propane and butane are shown below.

  $\begin{array}{c}C{H}_4\left(g\right)+O_{\text{2}}\left(g\right)\to C{O}_{\text{2}}\left(g\right)+2H_{2}O\left(l\right),\Delta H_{\text{c}}°=890kJ\cdot mo{l}^{-1}\\ C_2{H}_6\left(g\right)+\frac{7}{2}{O}_{\text{2}}\left(g\right)\to 2C{O}_{\text{2}}\left(g\right)+3H_{2}O\left(l\right),\Delta H_{\text{c}}°=1560kJ\cdot mo{l}^{-1}\\ C_3{H}_8\left(g\right)+5O_{\text{2}}\left(g\right)\to 3C{O}_{\text{2}}\left(g\right)+4H_{2}O\left(l\right),\Delta H_{\text{c}}°=2220kJ\cdot mo{l}^{-1}\\ C_4{H}_{10}\left(g\right)+\frac{13}{2}{O}_{\text{2}}\left(g\right)\to 4C{O}_{\text{2}}\left(g\right)+5H_{2}O\left(l\right),\Delta H_{\text{c}}°=2874kJ\cdot mo{l}^{-1}\end{array}$

The total energy in the form of heat produced by the combustion of natural gas is caudated as shown below.

  $\Delta H°=\left(\begin{array}{l}\left(9.3\text{kmol}\right)\Delta H_{\text{c}}°\left({\text{CH}}_4,\text{g}\right)+\left(3.1\text{kmol}\right)\Delta H_{\text{c}}°\left({\text{C}}_2{\text{H}}_6,\text{g}\right)+\\ \left(0.40\text{kmol}\right)\Delta H_{\text{c}}°\left({\text{C}}_3{\text{H}}_8,\text{g}\right)+\left(0.20\text{kmol}\right)\Delta H_{\text{c}}°\left({\text{C}}_4{\text{H}}_{10},\text{g}\right)\end{array}\right)$        (2)

Where,

• $\Delta H_{\text{c}}°\left({\text{CH}}_4,\text{g}\right)$ is the enthalpy of combustion of $C{H}_4\left(g\right)$.
• $\Delta H_{\text{c}}°\left({\text{C}}_2{\text{H}}_6,\text{g}\right)$ is the enthalpy of combustion of $C_2{H}_6\left(g\right)$.
• $\Delta H_{\text{c}}°\left({\text{C}}_3{\text{H}}_8,\text{g}\right)$ is the enthalpy of combustion of $C_3{H}_8\left(g\right)$.
• $\Delta H_{\text{c}}°\left({\text{C}}_4{\text{H}}_{10},\text{g}\right)$ is the enthalpy of combustion of $C_4{H}_{10}\left(g\right)$.

The value of $\Delta H_{\text{c}}°\left({\text{CH}}_4,\text{g}\right)$ is $890kJ\cdot mo{l}^{-1}$.

The value of $\Delta H_{\text{c}}°\left({\text{C}}_2{\text{H}}_6,\text{g}\right)$ is $1560kJ\cdot mo{l}^{-1}$.

The value of $\Delta H_{\text{c}}°\left({\text{C}}_3{\text{H}}_8,\text{g}\right)$ is $2220kJ\cdot mo{l}^{-1}$.

The value of $\Delta H_{\text{c}}°\left({\text{C}}_4{\text{H}}_{10},\text{g}\right)$ is $2874kJ\cdot mo{l}^{-1}$.

Substitute the value of $\Delta H_{\text{c}}°\left({\text{CH}}_4,\text{g}\right)$, $\Delta H_{\text{c}}°\left({\text{C}}_2{\text{H}}_6,\text{g}\right)$, $\Delta H_{\text{c}}°\left({\text{C}}_3{\text{H}}_8,\text{g}\right)$ and $\Delta H_{\text{c}}°\left({\text{C}}_4{\text{H}}_{10},\text{g}\right)$ in equation ().

    $\begin{array}{c}\Delta H°=\left(\begin{array}{l}\left(9.3\text{kmol}\right)\times \left(890\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(3.1\text{kmol}\right)\times \left(1560\text{kJ}\cdot {\text{mol}}^{-1}\right)+\\ \left(0.40\text{kmol}\right)\times \left(2220\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(0.20\text{kmol}\right)\times \left(2874\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(9.3\times {10}^3\text{mol}\right)\times \left(890\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(3.1\times {10}^3\text{mol}\right)\times \left(1560\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(0.40\times {10}^3\text{mol}\right)\times \left(2220\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(0.20\times {10}^3\text{mol}\right)\times \left(2874\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =8.277\times {10}^6\text{kJ}+4.836\times {10}^6\text{kJ}+8.88\times {10}^5\text{kJ}+5.748\times {10}^5\text{kJ}\\ =1.45758\times {10}^7\text{kJ}\end{array}$

Thus, after rounding to three significant figures, the heat produced per minute is $\underset{\_}{1.46\times 10^{7}kJ}$.