Chapter 5, Problem 161CP

a.

Interpretation Introduction

Interpretation: The flow rate of air needs to be calculated which is necessary to deliver the required amount of oxygen.

Concept Introduction:The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

  $\text{PV = nRT}$

Where R is Universal gas constant.

Answer to Problem 161CP

  $\text{8}\text{.6}\times {\text{10}}^3\text{ L air/min}$ .

Explanation of Solution

Given:

The flow rate of methane in combustion chamber is 200. L/min at 1.50 atm and ambient temperature. At 1.00 atm and same temperature, the air is added to the chamber and gases are ignited.

The balanced chemical equation for the possible reaction is:

  ${\text{CH}}_{\text{4}}{\text{ + 2O}}_{\text{2}}\to {\text{ CO}}_{\text{2}}{\text{ + 2H}}_{\text{2}}\text{O}$

The molar amount of methane per minute is calculated using ideal gas equation as:

  $\text{n = }\frac{\text{PV}}{\text{RT}}$

Substituting the values:

  $\begin{array}{l}\text{n = }\frac{\text{(1}\text{.50 atm)(200}\text{. L/min)}}{\left(\text{0}\text{.08206 }\frac{\text{L}\text{.atm}}{\text{K}\text{.mol}}\right)(298\text{ K)}}\\ \text{n = 12}{\text{.268 mol CH}}_4/\min \end{array}$

Now, according to balanced reaction, 1 mole of methane requires two moles of oxygen gas. So, if ${\text{O}}_{\text{2}}$ is at three times then, there are 6 moles of ${\text{O}}_{\text{2}}$ for each mole of ${\text{CH}}_4$ . Thus, the molar amount of ${\text{O}}_{\text{2}}$ is:

  $\left(\text{12}{\text{.268 mol CH}}_4/\min \right)\left(\frac{{\text{6 mol O}}_2}{{\text{1 mol CH}}_4}\right)\text{=73}{\text{.608 mol O}}_2/\text{min}$

The molar amount of air needed for per minute is calculated using mole fraction as:

  $\left(\text{73}{\text{.608 mol O}}_2/\min \right)\left(\frac{1\text{ mol air}}{0.21{\text{ mol O}}_2}\right)=350.51\text{ mol air/min}$

Now, the volume required to deliver this amount of air per minute will be:

  $\begin{array}{l}\text{V = }\frac{\text{nRT}}{\text{P}}\\ \text{V = }\frac{\text{(350}\text{.51 mol air/min)(0}\text{.08206 Latm/molK})(298\text{ K)}}{1.0\text{ atm}}\\ \text{V = 8}\text{.6}\times {\text{10}}^3\text{ L air/min}\end{array}$

b.

Interpretation Introduction

Interpretation: The composition of the exhaust gas needs to be calculated in terms of mole fraction of ${\text{CO, CO}}_{\text{2}}{\text{, O}}_{\text{2}}{\text{,N}}_{\text{2}}{\text{ and H}}_{\text{2}}\text{O}$ .

Concept Introduction:The mole fraction is ratio of number mole of a constituent present in the mixture to the total number of moles of all the constituents present in the mixture.

Answer to Problem 161CP

  ${\chi }_{{\text{O}}_2}\text{=0}\text{.13}$ , ${\chi }_{{\text{CO}}_2}\text{=0}\text{.032}$ , ${\chi }_{\text{CO}}\text{=0}\text{.0017}$ , ${\chi }_{{\text{H}}_2\text{O}}\text{=0}\text{.067}$ and ${\chi }_{{\text{N}}_2}\text{= 0}\text{.77}$ .

Explanation of Solution

Given:

The combustion of methane is not complete under the conditions of part (a) and produces a mixture of ${\text{CO and CO}}_{\text{2}}$ . The amount of 95 % of carbon in the exhaust gas was present in ${\text{CO}}_{\text{2}}$ and the remaining is in $\text{CO}$ .

The complete balanced equations for the two possible process are:

  $\begin{array}{l}{\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_2{\text{+2H}}_2\text{O}\\ {\text{CH}}_4+\frac{3}{2}{\text{O}}_2\to {\text{CO+2H}}_2\text{O}\end{array}$

The components present in the final mixture are:

  ${\text{CO, CO}}_{\text{2}}{\text{, O}}_{\text{2}}{\text{, N}}_{\text{2}}{\text{ and H}}_{\text{2}}\text{O}$

Let the molar amount of methane be $\text{x}$ so, the molar amount of ${\text{O}}_2$ is calculated as:

  $\begin{array}{l}{\text{CO}}_2:(0.950{\text{x mol CH}}_4\text{)}\left(\frac{{\text{2 mol O}}_2}{{\text{1 mol CH}}_4}\right)\text{=1}{\text{.90x mol O}}_2\\ \text{CO}:(0.050{\text{x mol CH}}_4\text{)}\left(\frac{\text{1}{\text{.5 mol O}}_2}{{\text{1 mol CH}}_4}\right)\text{=0}{\text{.075x mol O}}_2\end{array}$

The remaining amount of ${\text{O}}_2$ that is the unreacted amount of ${\text{O}}_2$ will be:

  $\text{6}\text{.00x - 1}\text{.90x - 0}{\text{.075x mol O}}_2=4.03{\text{x mol O}}_2$

The amount of ${\text{H}}_{\text{2}}\text{O}$ produced is twice the amount of consumed ${\text{CH}}_4$ and the molar amount of consumed ${\text{O}}_2$ is equal to the molar amount of produced ${\text{CO}}_2\text{ and CO}$ .

Now, by using the composition of air, the amount of ${\text{N}}_2$ can be related to amount of ${\text{O}}_2$ as:

  $(6.00{\text{x mol O}}_2\text{)}\left(\frac{0.79{\text{ mol N}}_2}{\text{0}{\text{.21 mol O}}_2}\right)=22.6{\text{x mol N}}_2\simeq 23{\text{x mol N}}_2$

The total molar amount will be:

  $\begin{array}{l}\text{4}{\text{.03x mol O}}_2+0.950{\text{x mol CO}}_2+0.050\text{x CO+0}{\text{.075x mol O}}_2+2.00{\text{x mol H}}_2{\text{O+23x mol N}}_2\\ =\text{ 30 x mol}\end{array}$

How, the mole fraction of each gas is calculated by dividing number of moles of each by total mole amount as:

  $\begin{array}{l}{\chi }_{{\text{O}}_2}\text{=}\frac{4.03\text{x mol}}{\text{30}\text{.x mol}}\\ {\chi }_{{\text{O}}_2}\text{=0}\text{.13}\end{array}$

  $\begin{array}{l}{\chi }_{{\text{CO}}_2}\text{=}\frac{0.950\text{x mol}}{\text{30}\text{.x mol}}\\ {\chi }_{{\text{CO}}_2}\text{=0}\text{.032}\end{array}$

  $\begin{array}{l}{\chi }_{\text{CO}}\text{=}\frac{0.050\text{x mol}}{\text{30}\text{.x mol}}\\ {\chi }_{\text{CO}}\text{=0}\text{.0017}\end{array}$

  $\begin{array}{l}{\chi }_{{\text{H}}_2\text{O}}\text{=}\frac{2.00\text{x mol}}{\text{30}\text{.x mol}}\\ {\chi }_{{\text{H}}_2\text{O}}\text{=0}\text{.067}\end{array}$

  $\begin{array}{l}{\chi }_{{\text{N}}_2}\text{=}\frac{23\text{x mol}}{\text{30}\text{.x mol}}\\ {\chi }_{{\text{N}}_2}\text{=0}\text{.77}\end{array}$

c.

Interpretation Introduction

Interpretation: The partial pressure of each gas in the part (b) needs to be calculated when the total pressure of exhaust gas is 1.00 atm.

Concept Introduction:The formula used to calculate partial pressure is:

  ${\text{P}}_{\text{A}}{\text{ = X}}_{\text{A}}{\text{P}}_{\text{total}}$ - (a)

Where ${\text{P}}_{\text{A}}$ is partial pressure of component A, ${\text{X}}_{\text{A}}$ is mole fraction of A and ${\text{P}}_{\text{total}}$ total pressure of the gas.

Answer to Problem 161CP

  ${\text{P}}_{{\text{O}}_2}\text{ = }0.13\text{ atm}$ , ${\text{P}}_{{\text{CO}}_2}\text{ = }0.\text{032 atm}$ , ${\text{P}}_{\text{CO}}\text{ = }0.\text{0017 atm}$ , ${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{ = 0}\text{.067 atm}$ and ${\text{P}}_{{\text{N}}_{\text{2}}}\text{ = 0}\text{.77 atm}$ .

Explanation of Solution

The mole fraction of each gas as calculated in (b) part is:

  ${\chi }_{{\text{O}}_2}\text{=0}\text{.13}$ , ${\chi }_{{\text{CO}}_2}\text{=0}\text{.032}$ , ${\chi }_{\text{CO}}\text{=0}\text{.0017}$ , ${\chi }_{{\text{H}}_2\text{O}}\text{=0}\text{.067}$ and ${\chi }_{{\text{N}}_2}\text{= 0}\text{.77}$ .

The partial pressure of each gas is calculated using equation (a) as:

  $\begin{array}{l}{\text{P}}_{{\text{O}}_2}\text{ = }\left(\text{0}\text{.13}\right)\text{(1}\text{.00 atm)}\\ {\text{P}}_{{\text{O}}_2}\text{ = }0.13\text{ atm}\end{array}$

  $\begin{array}{l}{\text{P}}_{{\text{CO}}_2}\text{ = }\left(\text{0}\text{.032}\right)\text{(1}\text{.00 atm)}\\ {\text{P}}_{{\text{CO}}_2}\text{ = }0.\text{032 atm}\end{array}$

  $\begin{array}{l}{\text{P}}_{\text{CO}}\text{ = }\left(\text{0}\text{.0017}\right)\text{(1}\text{.00 atm)}\\ {\text{P}}_{\text{CO}}\text{ = }0.\text{0017 atm}\end{array}$

  $\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{ = }\left(\text{0}\text{.067}\right)\text{(1}\text{.00 atm)}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{ = 0}\text{.067 atm}\end{array}$

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}}\text{ = }\left(\text{0}\text{.77}\right)\text{(1}\text{.00 atm)}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{ = 0}\text{.77 atm}\end{array}$