Chapter 6, Problem 106AE

(a)

Interpretation Introduction

Interpretation: The given reactions should be explained via the concept of work in thermodynamic.

Concept Introduction:

Work:

• The work is a transfer of energy from one system to another to without change of entropy.
• In gases process the work is meant by the volume change at constant pressure it is,

  $\begin{array}{l}\text{W=}\text{-PΔV }\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{pressure}\\ \text{ΔV}\text{is}\text{volume}\text{change}\end{array}$

  In gas gaseous reaction:

  $\text{Δn}\text{α}\text{ΔV}$

• The number of moles change $\text{Δn}$ is directly proportional to volume change $\text{ΔV}$.
• The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is positive the work (W) is done by the system.

  The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is negative the work (W) is done on the system.

• In a reaction the number of moles of reactants and product is respect with the work is done by or on the system.

  $\begin{array}{l}\text{1)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{=}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{no}\text{work}\text{done}\\ \text{2)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{>}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\\ \text{3)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{<}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\end{array}$

Explanation of Solution

• The balanced equation is:

                                          $\text{a)}{\text{N}}_{\text{2(g)}}\to {\text{N}}_{\text{2(l)}}$

• In (a) the volume of products is decreases compare to reactants so the work is done on the system.

(b)

Interpretation Introduction

Interpretation: The given reactions should be explained via the concept of work in thermodynamic.

Concept Introduction:

Work:

• The work is a transfer of energy from one system to another to without change of entropy.
• In gases process the work is meant by the volume change at constant pressure it is,

  $\begin{array}{l}\text{W=}\text{-PΔV }\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{pressure}\\ \text{ΔV}\text{is}\text{volume}\text{change}\end{array}$

  In gas gaseous reaction:

  $\text{Δn}\text{α}\text{ΔV}$

• The number of moles change $\text{Δn}$ is directly proportional to volume change $\text{ΔV}$.
• The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is positive the work (W) is done by the system.

  The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is negative the work (W) is done on the system.

• In a reaction the number of moles of reactants and product is respect with the work is done by or on the system.

  $\begin{array}{l}\text{1)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{=}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{no}\text{work}\text{done}\\ \text{2)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{>}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\\ \text{3)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{<}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\end{array}$

Explanation of Solution

                        

• The balanced equation is:

                       $\text{b)}{\text{CO}}_{\text{(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(g)}}\to {\text{H}}_{\text{2(g)}}{\text{+CO}}_{\text{2(g)}}$

• In (b) no change in number of moles and volume of reactants and products so there is no work done.

(c)

Interpretation Introduction

 Interpretation: The given reactions should be explained via the concept of work in thermodynamic.

Concept Introduction:

Work:

• The work is a transfer of energy from one system to another to without change of entropy.
• In gases process the work is meant by the volume change at constant pressure it is,

  $\begin{array}{l}\text{W=}\text{-PΔV }\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{pressure}\\ \text{ΔV}\text{is}\text{volume}\text{change}\end{array}$

  In gas gaseous reaction:

  $\text{Δn}\text{α}\text{ΔV}$

• The number of moles change $\text{Δn}$ is directly proportional to volume change $\text{ΔV}$.
• The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is positive the work (W) is done by the system.

  The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is negative the work (W) is done on the system.

• In a reaction the number of moles of reactants and product is respect with the work is done by or on the system.

  $\begin{array}{l}\text{1)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{=}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{no}\text{work}\text{done}\\ \text{2)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{>}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\\ \text{3)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{<}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\end{array}$

Explanation of Solution

Explanation:                

• The balanced equation is:

                $\text{c)}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2(s)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{3Ca(OH)}}_{\text{2(s)}}{\text{+2PH}}_{\text{3(g)}}$

• In (c) the Phosphane gas was produced so the volume is increases therefore the work is done by the system on surrounding.

(d)

Interpretation Introduction

Interpretation: The given reactions should be explained via the concept of work in thermodynamic.

Concept Introduction:

Work:

• The work is a transfer of energy from one system to another to without change of entropy.
• In gases process the work is meant by the volume change at constant pressure it is,

  $\begin{array}{l}\text{W=}\text{-PΔV }\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{pressure}\\ \text{ΔV}\text{is}\text{volume}\text{change}\end{array}$

  In gas gaseous reaction:

                                      $\text{Δn}\text{α}\text{ΔV}$

• The number of moles change $\text{Δn}$ is directly proportional to volume change $\text{ΔV}$.
• The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is positive the work (W) is done by the system.

  The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is negative the work (W) is done on the system.

• In a reaction the number of moles of reactants and product is respect with the work is done by or on the system.

  $\begin{array}{l}\text{1)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{=}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{no}\text{work}\text{done}\\ \text{2)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{>}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\\ \text{3)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{<}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\end{array}$

Explanation of Solution

• The balanced equation is:

                $\text{d)}{\text{2CH}}_{\text{3}}{\text{OH}}_{\text{(l)}}{\text{+3O}}_{\text{2(g)}}\to {\text{2CO}}_{\text{2}}{}_{\text{(g)}}{\text{+4H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

• In (d) the volume of products is decreases compare to reactants so the work is done on the system.

(e)

Interpretation Introduction

Interpretation: The given reactions should be explained via the concept of work in thermodynamic.

Concept Introduction:

Work:

• The work is a transfer of energy from one system to another to without change of entropy.
• In gases process the work is meant by the volume change at constant pressure it is,

  $\begin{array}{l}\text{W=}\text{-PΔV }\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{pressure}\\ \text{ΔV}\text{is}\text{volume}\text{change}\end{array}$

  In gas gaseous reaction:

                                      $\text{Δn}\text{α}\text{ΔV}$

• The number of moles change $\text{Δn}$ is directly proportional to volume change $\text{ΔV}$.
• The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is positive the work (W) is done by the system.

  The number of moles change $\text{Δn}$ or volume change $\text{ΔV}$ is negative the work (W) is done on the system.

• In a reaction the number of moles of reactants and product is respect with the work is done by or on the system.

  $\begin{array}{l}\text{1)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{=}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{no}\text{work}\text{done}\\ \text{2)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{>}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\\ \text{3)}{\displaystyle \sum {\text{n}}_{\text{reactants}}}\text{<}{\displaystyle \sum {\text{n}}_{\text{products}}}\text{work}\text{done}\text{on}\text{the}\text{system}\end{array}$

Explanation of Solution

• The balanced equation is:

                                    $\text{e)}{\text{I}}_{\text{2(s)}}\to {\text{I}}_{\text{2(g)}}$

• In case (e) the iodine gas was produced so the volume is increases therefore the work is done by the system on surrounding.