Chapter 5, Problem 5.161QP

(a)

Interpretation Introduction

Interpretation:

Adding the ${\text{N}}_{\text{2}}$ supplied in air to the power plant running on a fossil fuel having a formula ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}$. The balanced combustion equation for a complex fuel having $\text{120\%}$ stoichiometric combustion needs to be written.

Concept Introduction:

According to law of conservation of mass

$\text{Reactants}\to \text{products}$

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

$\text{No of moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

Answer to Problem 5.161QP

The power plant running on a fossil fuel having a formula ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}$ with added ${\text{N}}_{\text{2}}$ gas the reactants and products of the reactions are balanced.

Explanation of Solution

Given,

Power plant driven by fossil fuel combustion generate substantial greenhouse gases $\left({\text{CO}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}\right)$ and gases that subsidizes to low air quality $\left(\text{e}\text{.g}{\text{SO}}_{\text{2}}\right)$ to evaluate the emission on a regular basis inert nitrogen gas is supplied with air to balance reactions.

The power plant running on a fossil fuel having the formula ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}$

$\text{100}\text{\%}\text{Combustion}$ of ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}$ taken from problem 3.141 as,

$\text{2}{\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S+43}{\text{O}}_{\text{2}}\text{+161}\text{.7}{\text{N}}_{\text{2}}\to \text{36}{\text{CO}}_{\text{2}}\text{(g)+10}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+161}\text{.7}{\text{N}}_{\text{2}}\text{(g)}$

Moles of Oxygen can be calculated as

$\text{No of}{\text{O}}_{\text{2}}\text{ moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

$\text{43}\text{mol}{\text{O}}_{\text{2}}\text{×}\frac{\text{3}\text{.76}\text{mol}{\text{N}}_{\text{2}}}{\text{1}\text{.00}\text{mol}{\text{O}}_{\text{2}}}\text{=161}\text{.7}\text{mol}{\text{N}}_{\text{2}}$

$\text{120}\text{\%}\text{Combustion}$ of ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}$ can calculated as

$\begin{array}{l}\text{1}\text{.2×43=51}\text{.6}\text{mol}\text{of}{\text{O}}_{\text{2}}\\ \text{1}\text{.2×161}\text{.7=194}\text{.04}\text{mol}\text{of}{\text{N}}_{\text{2}}\end{array}$

$\begin{array}{l}\text{Balanced reaction for120\%}\text{combustion}\text{is}\\ \text{2}{\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S+51}\text{.6}{\text{O}}_{\text{2}}\text{+194}\text{.04}{\text{N}}_{\text{2}}\to \text{36}{\text{CO}}_{\text{2}}\text{(g)+10}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+194}{\text{.04N}}_{\text{2}}\text{(g)+8}{\text{.6O}}_{\text{2}}\text{(g)}\end{array}$

Conclusion

The equation of power plant running on a fossil fuel gas is balanced.

(b)

Interpretation Introduction

Interpretation:

When deriving the balanced combustion reaction to effectively convert the $\frac{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ volume to a molar ratio, which law is used to define need to be identified.

Concept Introduction:

Boyle’s law:  According to Boyle’s law, pressure and volume are inversely proportional to each other at a given temperature.

$\begin{array}{l}\text{PV}\text{=}\text{Constant }\mathrm{...}\text{(1)}\\ \text{or}\\ \text{ Vα}\frac{\text{1}}{\text{P}}\end{array}$

Where,

P = Pressure

V = Volume

Charles law: According to Charles law, temperature and volume are directly proportional to each other at a constant pressure.

   $\begin{array}{l}\text{VαT(at}\text{a}\text{constant}\text{pressure) }\mathrm{...}\text{(2)}\\ \text{or}\\ \frac{\text{V}}{\text{T}}\text{=}\text{C}\end{array}$

Where,

T = Temperature

V = Volume.

Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure) }\mathrm{...}\text{(3)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\text{PV}\text{=}\text{nRT}$

Answer to Problem 5.161QP

Avogadro’s law is used to explain the volume to mole ratio for the above given fossil fuel power plant operation.

Explanation of Solution

Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Conclusion

The law used in the operation of fossil fuel power plant is identified.

(c)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water condenses from (b) calculate the volume/volume percentages for gases ${\text{CO}}_{\text{2}}{\text{,N}}_{\text{2}}{\text{ and SO}}_{\text{2}}$.

Concept Introduction:

1. According to law of conservation of mass

$\text{Reactants}\to \text{products}$

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

$\text{No of moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Answer to Problem 5.161QP

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{CO}}_{\text{2}}\text{=15\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{N}}_{\text{2}}\text{=80}\text{.6\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\text{=0}\text{.83\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{O}}_{\text{2}}\text{=3}\text{.6\%}\end{array}$

Explanation of Solution

Chemical balanced equation

$\text{2}{\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S+51}\text{.6}{\text{O}}_{\text{2}}\text{+194}\text{.04}{\text{N}}_{\text{2}}\to \text{36}{\text{CO}}_{\text{2}}\text{(g)+10}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+194}{\text{.04N}}_{\text{2}}\text{(g)+8}{\text{.6O}}_{\text{2}}\text{(g)}$

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

The total gas volume = $\text{36+2+194}\text{.0+8}\text{.6=240}\text{.6}\text{volumes}$

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{CO}}_{\text{2}}\text{=}\left(\frac{\text{36}}{\text{240}\text{.6}}\right)\text{×100=15\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{N}}_{\text{2}}\text{=}\left(\frac{\text{194}\text{.0}}{\text{240}\text{.6}}\right)\text{×100=80}\text{.6\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\text{=}\left(\frac{\text{2}}{\text{240}\text{.6}}\right)\text{×100=0}\text{.83\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{O}}_{\text{2}}\text{=}\left(\frac{\text{8}\text{.6}}{\text{240}\text{.6}}\right)\text{×100=3}\text{.6\%}\end{array}$

Conclusion

The volume/volume ratio percentages of the gases ${\text{CO}}_{\text{2}}{\text{,N}}_{\text{2}}{\text{ and SO}}_{\text{2}}$ are calculated as $\text{15\%,80}\text{.6\%}\text{and}\text{0}\text{.83\%}$ respectively.

(d)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water vapor on wet basis from (c) calculate the volume/volume percentages for gases ${\text{CO}}_{\text{2}}{\text{,N}}_{\text{2}}{\text{ and SO}}_{\text{2}}$.

Concept Introduction:

1. According to law of conservation of mass

$\text{Reactants}\to \text{products}$

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

$\text{No of moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Answer to Problem 5.161QP

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{CO}}_{\text{2}}\text{=14}\text{.4\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{H}}_{\text{2}}\text{O=4}\text{.0\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\text{=0}\text{.8\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{N}}_{\text{2}}\text{=77}\text{.4\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{O}}_{\text{2}}\text{=3}\text{.4\%}\end{array}$

Explanation of Solution

Chemical balanced equation

$\text{2}{\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S+51}\text{.6}{\text{O}}_{\text{2}}\text{+194}\text{.04}{\text{N}}_{\text{2}}\to \text{36}{\text{CO}}_{\text{2}}\text{(g)+10}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+194}{\text{.04N}}_{\text{2}}\text{(g)+8}{\text{.6O}}_{\text{2}}\text{(g)}$

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

The total gas volume = $\text{36+2+10+8}\text{.6+194}\text{.0=250}\text{.6}\text{volumes}$

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{CO}}_{\text{2}}\text{=14}\text{.4\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{H}}_{\text{2}}\text{O=4}\text{.0\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\text{=0}\text{.8\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{N}}_{\text{2}}\text{=77}\text{.4\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{O}}_{\text{2}}\text{=3}\text{.4\%}\end{array}$

Conclusion

The volume/volume ratio percentages of the gases ${\text{CO}}_{\text{2}}{\text{,H}}_{\text{2}}{\text{O,SO}}_{\text{2}}{\text{,N}}_{\text{2}}{\text{ and O}}_{\text{2}}$ are calculated as $\text{14}\text{.4\%,4}\text{.00\%,0}\text{.80\%,77}\text{.4\%,}\text{and}\text{3}\text{.4\%}$ respectively.

(e)

Interpretation Introduction

Interpretation:

Among the two fuels ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}\text{and}{\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$ keeping the air quality in mind which one is more likely to be used has to be identified.

Answer to Problem 5.161QP

The fuel ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$ can be used in the port.

Explanation of Solution

The fuel ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$ can be used in the port because of the lower content of sulfur in ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$, it will give lesser concentration of ${\text{SO}}_{\text{2}}$ in the Stack gas emissions.

Conclusion

The fuel lower content of sulfur is identified as ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$.

(f)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water vapor on wet basis from (d) calculate the volume/volume percentages for gases ${\text{SO}}_{\text{2}}$ produced from the fuels ${\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}\text{and}{\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$ assuming that $\text{120\%}$ stoichiometric combustion.

Concept Introduction:

1. According to law of conservation of mass

$\text{Reactants}\to \text{products}$

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

$\text{No of moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Answer to Problem 5.161QP

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\left({\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}\right)\text{=0}\text{.83\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\left({\text{C}}_{\text{32}}{\text{H}}_{\text{12}}\text{S}\right)\text{=0}\text{.55\%}\\ \end{array}$

Explanation of Solution

Chemical balanced equation

$\text{2}{\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S+51}\text{.6}{\text{O}}_{\text{2}}\text{+194}\text{.04}{\text{N}}_{\text{2}}\to \text{36}{\text{CO}}_{\text{2}}\text{(g)+10}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+194}{\text{.04N}}_{\text{2}}\text{(g)+8}{\text{.6O}}_{\text{2}}\text{(g)}$

Similarly for ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$

$\text{2}{\text{C}}_{\text{32}}{\text{H}}_{\text{18}}{\text{S+90O}}_{\text{2}}{\text{+282N}}_{\text{2}}\to \text{64}{\text{CO}}_{\text{2}}\text{(g)+18}{\text{H}}_{\text{2}}\text{O(l)+2}{\text{SO}}_{\text{2}}\text{(g)+282}{\text{N}}_{\text{2}}\text{(g)+15}{\text{O}}_{\text{2}}\text{(g)}$

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

$\begin{array}{l}\frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\left({\text{C}}_{\text{18}}{\text{H}}_{\text{10}}\text{S}\right)\text{=}\left(\frac{\text{1}}{\text{363}}\right)\text{×100=0}\text{.83\%}\\ \frac{\text{volume}}{\text{volume}}\text{\%}\text{of}{\text{SO}}_{\text{2}}\left({\text{C}}_{\text{32}}{\text{H}}_{\text{12}}\text{S}\right)\text{=}\left(\frac{\text{2}}{\text{363}}\right)\text{×1000=0}\text{.55\%}\\ \end{array}$

From volume/volume we can say that ${\text{SO}}_{\text{2}}$ content in the air is reduced when we use the fuel with the formula ${\text{C}}_{\text{32}}{\text{H}}_{\text{28}}\text{S}$.