Chapter 5, Problem 161CP

Methane (CH₄) gas flows into a combustion chamber at a rate of 200. L/min at 1.50 atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited.

a. To ensure complete combustion of CH₄ to CO₂(g) and H₂O(g), three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent O₂ and 79 mole percent N₂, calculate the flow rate of air necessary to deliver the required amount of oxygen.

b. Under the conditions in part a, combustion of methane was not complete as a mixture of CO₂(g) and CO(g) was produced. It was determined that 95.0% of the carbon in the exhaust gas was present in CO₂. The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of CO, CO₂, O₂, N₂, and H₂O. Assume CH₄ is completely reacted and N₂ is unreacted.

Interpretation Introduction

Interpretation: The flow rate of air is needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$ if the combustion of ${\text{CH}}_{\text{4}}$ is producing $\text{CO}\text{and}{\text{CO}}_{\text{2}}$ are needed to be determined.

Concept introduction:

• Balanced equation of a reaction is written according to law of conservation of mass.

• According to ideal gas law: the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

$\frac{\text{PV}}{\text{n}}\text{=}\text{RT}$, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}{\text{V}}_1}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_2}{{\text{n}}_{\text{2}}}$

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Equation for mole fraction of a molecule in a mixture of molecules is,

• $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{total}}\text{) }}$

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

• Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the flow rate of air needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$ if the combustion of ${\text{CH}}_{\text{4}}$ is producing $\text{CO}\text{and}{\text{CO}}_{\text{2}}$.

Answer to Problem 161CP

The flow rate of air needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$ is $\text{8}\text{.7}\times {10}^3\text{L/min}$.

Explanation of Solution

To find: the balanced chemical equation of the given reaction.

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

The reactants of the given reaction are ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$.

The products of the given reaction are ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$.

Therefore,

The chemical equation of the reaction given is,

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

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction will be,

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

Hence,

The balanced chemical equation for the given reaction is, ${\text{CH}}_{\text{4}}\text{+}2{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$.

Assumes the number of moles of ${\text{CH}}_{\text{4}}$ is as x.

To find: the number of moles of required ${\text{O}}_{\text{2}}$ and air in the given reaction.

The number of moles of required ${\text{O}}_{\text{2}}$ is $\text{6x}\text{mol}$.

The number of moles of required air is $\text{29x}\text{mol}$.

The balanced chemical equation for the given reaction is, ${\text{CH}}_{\text{4}}\text{+}2{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$.

Here, the combustion of 1 mole of ${\text{CH}}_{\text{4}}$ requires 2 mole of ${\text{O}}_{\text{2}}$.

But to ensure the completion of combustion of ${\text{CH}}_{\text{4}}$, three times as much oxygen is necessary is required. Therefore,

If the number of moles of ${\text{CH}}_{\text{4}}$ is as x, then the number of moles of required ${\text{O}}_{\text{2}}$ is,

$\text{3}\text{×}\text{x}\text{mol}{\text{CH}}_{\text{4}}\text{×}\frac{\text{2}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\text{=}\text{6x}\text{mol}{\text{O}}_{\text{2}}$

The mole percent of ${\text{O}}_{\text{2}}$ in the air is 21.

That means,

${\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.21}{\text{n}}_{\text{air}}$

Therefore,

The number of moles of required air is,

${\text{n}}_{\text{air}}\text{=}\frac{\text{6x}\text{mol}}{\text{0}\text{.21}}\text{=}\text{29x}\text{mol}$

Hence,

The number of moles of required ${\text{O}}_{\text{2}}$ is $\text{6x}\text{mol}$.

The number of moles of required air is $\text{29x}\text{mol}$.

To determine: the flow rate of air needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$.

The flow rate of air needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$ is $\text{8}\text{.7}\times {10}^3\text{L/min}$.

The volume of ${\text{CH}}_{\text{4}}$ is given as $\text{200}\text{L/min}$.

The pressure of ${\text{CH}}_{\text{4}}$ is given as $\text{1}\text{.50}\text{atm}$.

The number of moles of ${\text{CH}}_{\text{4}}$ is x.

The pressure of air is given as $\text{1atm}$.

The number of moles of required air is calculated as $\text{29x}\text{mol}$.

According to ideal gas law, the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

$\frac{\text{PV}}{\text{n}}\text{=}\text{RT}$, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}{\text{V}}_1}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_2}{{\text{n}}_{\text{2}}}$

Take gas-1 as ${\text{CH}}_{\text{4}}$ and gas-2 as air,

Therefore,

The volume of air is,

$\frac{\text{200}\text{L/min}\times \text{1}\text{.50}\text{atm}\times \text{29x}\text{mol}}{\text{1atm}\times \text{x}\text{mol}}=\text{8}\text{.7}\times {10}^3\text{L/min}$

Hence,

The flow rate of air needed to complete the given reaction of combustion of ${\text{CH}}_{\text{4}}$ is $\text{8}\text{.7}\times {10}^3\text{L/min}$.

Interpretation Introduction

Interpretation: The composition of the exhaust gas in terms of mole fraction of ${\text{CO,CO}}_{\text{2}}{\text{,O}}_{\text{2}}{\text{,N}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}$ if the combustion of ${\text{CH}}_{\text{4}}$ is producing $\text{CO}\text{and}{\text{CO}}_{\text{2}}$ are needed to be determined.

Concept introduction:

• Balanced equation of a reaction is written according to law of conservation of mass.

• According to ideal gas law: the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

$\frac{\text{PV}}{\text{n}}\text{=}\text{RT}$, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}{\text{V}}_1}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_2}{{\text{n}}_{\text{2}}}$

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Equation for mole fraction of a molecule in a mixture of molecules is,

• $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{total}}\text{) }}$

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

• Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the composition of the exhaust gas in terms of mole fraction of ${\text{CO,CO}}_{\text{2}}{\text{,O}}_{\text{2}}{\text{,N}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}$ if the    combustion of ${\text{CH}}_{\text{4}}$ is producing $\text{CO}\text{and}{\text{CO}}_{\text{2}}$.

Answer to Problem 161CP

The mole fraction of $\text{CO}$ in the exhaust gas of given reaction is $\text{0}\text{.0017}$.

The mole fraction of ${\text{CO}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.0032}$.

The mole fraction of ${\text{O}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.13}$.

The mole fraction of ${\text{N}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.77}$.

The mole fraction of ${\text{H}}_{\text{2}}\text{O}$ in the exhaust gas of given reaction is $\text{0}\text{.067}$.

Explanation of Solution

To find: the balanced chemical equation of the given reactions.

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

${\text{CH}}_{\text{4}}\text{+}\frac{3}{2}{\text{O}}_{\text{2}}\to \text{CO}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$

The reactants of the given reactions are ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$.

The products of the given reaction are $\text{CO}$, ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$.

Therefore,

The chemical equations of the given reactions are,

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

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

These equations are not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equations of given reactions are,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms then oxygen atoms.

Therefore, the balanced chemical equation of the given reactions will be,

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

${\text{CH}}_{\text{4}}\text{+}\frac{3}{2}{\text{O}}_{\text{2}}\to \text{CO}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$

Hence,

The balanced chemical equations for the given reactions are,

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

${\text{CH}}_{\text{4}}\text{+}\frac{3}{2}{\text{O}}_{\text{2}}\to \text{CO}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$

#assumes the number of moles of ${\text{CH}}_{\text{4}}$ is as x.

To find: the number of moles of produced $\text{CO}$, ${\text{CO}}_{\text{2}}$, ${\text{N}}_{\text{2}}$,${\text{H}}_{\text{2}}\text{O}$ and remaining ${\text{O}}_{\text{2}}$ in the given reactions.

The number of moles of produced $\text{CO}$ in the given reaction is $\text{0}\text{.05}\text{x}\text{mol}$.

The number of moles of produced ${\text{CO}}_{\text{2}}$ in the given reaction is $\text{0}\text{.95}\text{x}\text{mol}$.

The number of moles of ${\text{N}}_{\text{2}}$ in the given reaction is $\text{23}\text{x}\text{mol}$.

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ in the given reactions is $\text{2x}\text{mol}$.

The number of moles remaining ${\text{O}}_{\text{2}}$ in the given reactions is $\text{4}\text{.03x}\text{mol}$.

• The carbon containing reactant is only ${\text{CH}}_{\text{4}}$, but the carbon containing products are ${\text{CO}}_{\text{2}}$ and $\text{CO}$.

  The produced ${\text{CO}}_{\text{2}}$ containing 95% of carbon in the reactant ${\text{CH}}_{\text{4}}$.

  The number of moles of ${\text{CH}}_{\text{4}}$ is taken as x mole.

  Therefore,

  The number of moles of produced${\text{CO}}_{\text{2}}$ in the given reaction is $\text{0}\text{.95}\text{x}\text{mol}$.

  So the remaining $5\%$  of carbon is contained by $\text{CO}$.

  Therefore,

  The number of moles of produced$\text{CO}$ in the given reaction is $\text{0}\text{.05}\text{x}\text{mol}$.

• The number of moles of required air in the given reaction is calculated as $\text{29x}\text{mol}$.

  The 79 mole percent of air is ${\text{N}}_{\text{2}}$.

  That means,

  ${\text{n}}_{{\text{N}}_{\text{2}}}\text{=}\text{0}\text{.79}{\text{n}}_{\text{air}}=0.79\times \text{29x}\text{mol}\approx 23\text{x}\text{mol}$

  The number of moles of${\text{N}}_{\text{2}}$ in the given reaction is $\text{23}\text{x}\text{mol}$.

• The number of moles of ${\text{CH}}_{\text{4}}$ is taken as x mole.

  The balanced chemical equations for the given reactions are,

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

${\text{CH}}_{\text{4}}\text{+}\frac{3}{2}{\text{O}}_{\text{2}}\to \text{CO}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}$

Therefore,

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ in the given reactions is,

$\text{2}\text{×}\text{x}\text{mol}{\text{CH}}_{\text{4}}\text{×}\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\times 1\text{mol}{\text{CH}}_{\text{4}}}\text{=}2\text{x}\text{mol}{\text{H}}_{\text{2}}\text{O}$

The number of moles of produced${\text{H}}_{\text{2}}\text{O}$ in the given reactions is $\text{2x}\text{mol}$.

• The number of moles of produced $\text{CO}$ in the given reaction is $\text{0}\text{.05}\text{x}\text{mol}$.

  The number of moles of produced ${\text{CO}}_{\text{2}}$ in the given reaction is $\text{0}\text{.95}\text{x}\text{mol}$.

  The reacted ${\text{O}}_{\text{2}}$ in the given reaction is,

  $\text{0}\text{.95}\text{x}\text{mol}{\text{CO}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}{\text{O}}_{\text{2}}}{1\text{mol}{\text{CO}}_{\text{2}}}\text{=}\text{1}\text{.9x}\text{mol}{\text{O}}_{\text{2}}$

  $\text{0}\text{.05x}\text{mol}\text{CO}\text{×}\frac{1.5\text{mol}{\text{O}}_{\text{2}}}{1\text{mol}\text{CO}}\text{=}0.075\text{x}\text{mol}{\text{O}}_{\text{2}}$

  That is,

  $\text{1}\text{.9x}\text{mol}\text{+}\text{0}\text{.075x}\text{mol}\text{=}\text{1}\text{.975x}\text{mol}$.

• The number of moles of required ${\text{O}}_{\text{2}}$ in the given reaction is calculated as $\text{6x}\text{mol}$.

  The reacted ${\text{O}}_{\text{2}}$ in the given reaction is $\text{1}\text{.975x}\text{mol}$.

  Therefore,

  The number of moles remaining${\text{O}}_{\text{2}}$ in the given reactions is,

  $\text{6x}\text{mol}-\text{1}\text{.975x}\text{mol}\approx 4.03\text{x}\text{mol}$

To determine: the compositions of the exhaust gas in terms of mole fraction of ${\text{CO,CO}}_{\text{2}}{\text{,O}}_{\text{2}}{\text{,N}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}$ if the    combustion of ${\text{CH}}_{\text{4}}$ is producing $\text{CO}\text{and}{\text{CO}}_{\text{2}}$.

The mole fraction of $\text{CO}$ in the exhaust gas of given reaction is $\text{0}\text{.0017}$.

The mole fraction of ${\text{CO}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.0032}$.

The mole fraction of ${\text{O}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.13}$.

The mole fraction of ${\text{N}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.77}$.

The mole fraction of ${\text{H}}_{\text{2}}\text{O}$ in the exhaust gas of given reaction is $\text{0}\text{.067}$.

The number of moles of produced $\text{CO}$ in the given reaction is calculated as $\text{0}\text{.05}\text{x}\text{mol}$.

The number of moles of produced ${\text{CO}}_{\text{2}}$ in the given reaction is calculated as $\text{0}\text{.95}\text{x}\text{mol}$.

The number of moles of ${\text{N}}_{\text{2}}$ in the given reaction is calculated as $\text{23}\text{x}\text{mol}$.

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ in the given reactions is calculated as $\text{2x}\text{mol}$.

The number of moles remaining ${\text{O}}_{\text{2}}$ in the given reactions is calculated as $\text{4}\text{.03x}\text{mol}$.

These molecules are presented in the exhaust gas.

Therefore,

The total number of moles of in the exhaust gas is,

$\text{0}\text{.05}\text{x}\text{mol}\text{+}\text{0}\text{.95}\text{x}\text{mol}\text{+}\text{23}\text{x}\text{mol}\text{+}\text{2x}\text{mol}\text{+}\text{4}\text{.03x}\text{mol}\text{=}\text{30x}\text{mol}$.

Equation for mole fraction of a molecule in a mixture of molecules is,

$\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{total}}\text{) }}$

Therefore,

The mole fraction of $\text{CO}$ in the exhaust gas is,

$\frac{\text{0}\text{.05x}\text{mol}}{\text{30x}\text{mol}}=0.0017$

The mole fraction of ${\text{CO}}_2$ in the exhaust gas is,

$\frac{\text{0}\text{.95x}\text{mol}}{\text{30x}\text{mol}}=0.032$

The mole fraction of ${\text{N}}_{\text{2}}$ in the exhaust gas of given reaction is

$\frac{\text{23}\text{x}\text{mol}}{\text{30x}\text{mol}}=0.77$

The mole fraction of ${\text{H}}_{\text{2}}\text{O}$ in the exhaust gas of given reaction is

$\frac{\text{2x}\text{mol}}{\text{30x}\text{mol}}=0.067$

The mole fraction of ${\text{O}}_2$ in the exhaust gas is,

$\frac{\text{4}\text{.03x}\text{mol}}{\text{30x}\text{mol}}=0.13$

Hence,

The mole fraction of $\text{CO}$ in the exhaust gas of given reaction is $\text{0}\text{.0017}$.

The mole fraction of ${\text{CO}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.0032}$.

The mole fraction of ${\text{O}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.13}$.

The mole fraction of ${\text{N}}_{\text{2}}$ in the exhaust gas of given reaction is $\text{0}\text{.77}$.

The mole fraction of ${\text{H}}_{\text{2}}\text{O}$ in the exhaust gas of given reaction is $\text{0}\text{.067}$.