Imp Quest MTD
Transcript of Imp Quest MTD
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MTD Important questions and answers
(Short answer questions)
1) State any three applications of thermodynamics in metallurgy
Ans:
In the construction of phase diagrams
To understand various chemical reactions
To predict the correct reducing agent in
Ferrous and non ferrous extractive metallurgy
To understand phase transformations during heat treatment of an alloy
To predict new phases in the development of newer materials
) State a) the first law of thermodynamics !) its mathematical equation and c) la!el the terms
Ans:
Statement:
The total energy of an isolated system remains constant, though there are changes from one form of energy to
another.
Mathematical equation:
! " # $
$here ! %hange in internal energy of the closed system
" ! &eat transferred to or from the system$ ! wor' done (y or on the system
") Define a) heat of reaction !) heat of reaction at constant pressure c) heat of reaction at constant #olume
Ans:
a) $eat of %eaction: The amount of heat evolved or a(sor(ed in a chemical reaction when the reaction is
performed according to a (alanced chemical equation at a given temperature and pressure is called heat of
reaction
!) $eat of %eaction at constant &ressure: The heat of reaction at constant pressure at a certain temperature is
defined as the amount of heat evolved or a(sor(ed at constant pressure when the reaction is performed according
to a (alanced chemical equation. This is represented (y &
c) $eat of %eaction at constant #olume: The heat of reaction at constant volume at a certain temperature isdefined as the amount of heat evolved or a(sor(ed at constant volume when the reaction is performed according
to a (alanced chemical equation. This is represented (y or )
') rite any three statements of Second law of thermodynamics
Ans:
*tatement+:
- It is impossi(le to construct a machine, operating in cycles, which will produce wor' continuously (y a(sor(ing
heat from a single heat reservoir.
- /elvin+0lanc' statement of the second law1
*tatement+2:
- &eat does not flow spontaneously from a cold to a hot (ody
- %lausius statement of the second law1
*tatement+3:
- $henever a spontaneous process ta'es place, it is accompanied (y an increase in the total entropy of the universe
That is *univ 45
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) hat information can !e drawn from a positi#e slope change of a line in the *llingham diagram+
Ans:
The slope of the llingham diagram is given (y +6*5
7ut ,S-. S-products/ S-reactants
,S-(ecomes negative when S-reactants0 S-products then only the slope 8/,S
-)(ecomes positive,
This implies that the reactants are undergoing phase transformation which may (e either melting are (oiling
depending upon the value of /,S-if +6*5 is low it is melting if +6*5 is high it is (oiling.
) hy the line of the reaction 2 3 4. 24 is down wards in the *llingham diagram.
Ans:
For the reaction
234..0 24
$e have a solid reacting with a gas to produce two moles of gas, and so there is a su(stantial increase in entropy
so the line slopes sharply downward as shown (elow in the figure.
ntropy change for the reaction is given as
*5! 2*58%9# 2*5
;%4< *5
8921
&ere *5;%4 is assumed negligi(le since ;%4 is solid.
*ince for every one mole of consumption of 92gas two moles of %9 gas is (eing produced
Therefore 2*58%9 4 *5
892
7ut slope of the line is given as ! +*5; 5
&ence the line of the reaction 2% < 52! 2%9 is down wards in the llingham diagram.
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5) Slate Trouton6s rule and its use
Ans:
Trouton7s rule: It states that the molar entropy of vapori=ation of various liquids at their normal (oiling point is almost
same and is a(out >?+>> @ /+mol+9r 2 cal.
I.e. * ! &AT(.p! 2 cal
8se:
The rule is used to estimate the enthalpy of vapori=ation of liquids whose (oiling points are 'nown.
9) Define: a) fugacity !) acti#ity
Ans:
a) ;ugacity: itis a measure of escaping tendency of a su(stance to prefer one phase 8liquid, solid, gas
over another. It is represented (y BfC
Fugacity appers as pressure (ut not really the pressure it is also 'nown as pseudo pressure or fictitious
pressure (ut at low pressures fugacity approaches pressure mathematically it is expressed as
!) Acti#ity: the activity of a su(stance is defined as the ratio of fugacity in the given state to the
fugacity in the standard 8or pure state. Mathematically it is expressed as shown (elowa . f Magnesium alloy contains =1 atom? aluminium 2alculate the composition of the
alloy in wt? The atomic weights of Al and Mg are =9 and '" respecti#ely
Solution:
,55FF
FF X
MgXatwtMgatomAlXatwtAlatom
AlXatwtAlatomofAlWt
+
=
,5532.2G:H.I,,558I>.2JH.I,
I>.2JH.I,F X
XX
XofAlWt
+
=
. ="
Therefore $t of Mg ! 55+2.3
!55
1-) Define rate of a reaction and e@press it mathematically
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Kate of a reaction can (e defined as the decrease in concentration of reactants or increase in concentration of
products with respect to time. Mathematically it can (e expressed as shown (elow:
Kate of a reaction !
If concentration is expressed (y BxC then the change in concentration is expressed (y BdxC, Then dx is positive for
the products since the concentration of products increases with time, whereas for reactants dx is negative since the
concentration of reactants decrease with time.
Then the rate of reaction can also (e expressed as
Kate !
$here dx +++ the change in concentration of products
dT+++ the change in time9r
Kate !
$here dx +++ the change in concentration of reactants
dT+++ the change in time
11) Define a) System !) Surrounding c) !oundary with a neat setch
Ans:
System: L *ystem is that part of the universe, which is under thermodynamic study
Surroundings: verything external to the system is called surroundings
Boundary: The real or imaginary surface separating the system from the surroundings is called the
(oundary
*@ample: L gas entrapped in a cylinder fitted with piston as illustrated (elow
%hange in concentration of reactants or products
Time
dx
dT
+dx
dT
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1) A System a!sor!s cal of heat from the surroundings at constant pressure then it performs a wor of '19-
C calculate the change in internal energy of the system
Solution:
Lccording to First law of thermodynamics d8 . E >
&ere ". 2.H ' cal ! 2H55 N G.> @ ! 5GH5 @
$ ! ?H5 @ ,Oow d) ! 5GH5 # G>5
. 5- C
1") Define a) *@othermic reaction !) *ndothermic reaction and c) state sign con#entions
Ans:
a) *@othermic reaction: The reactions, which are accompanied (y the evolution of heat, are 'nown as exothermic
reactions. In exothermic reactions the enthalpy of products is less than that of the reactants.
!) *ndothermic reaction: The reactions, which are accompanied (y the a(sorption of heat, are 'nown as endothermicreactions. In endothermic reactions, the enthalpy of products is more than that of the reactants. For these reactions & !
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That is *univ 45O.7: any one statement can (e sufficient
( " ! Tds
$here " ! amount of heat transferred
T ! Temperature in /elvin
ds ! %hange in entropy
1) State any three limitations and three uses of *llingham diagrams
Ans:
The limitations of llingham diagram are:
It is applica(le only to su(stances in their pure states that is pure metals and pure oxides or sulphides, (ut pure
su(stances are rare in metal extractions therefore the information derived from the diagram cannot (e directly applied
to actual reactions
2 %ompounds are assumed to (e stoichiometric which often is not true
3 The diagram shows the direction in which equili(rium lies (ut it does not specify the conditions under which it would
(e reached
G Oo information on the 'inetics of the reactions can (e o(tained from the diagrams
O.7: any three limitations can (e sufficient
The uses of the *llingham diagram:
To determine the reducing agent for reducing a given metallic oxide to metalP
2 To determine the partial pressure of oxygen that is in equili(rium with a metal oxide at a given temperature, this
will help us to predict the temperatures at which a metal is sta(le and the temperatures over which it will
spontaneously oxidi=e
3 Qetermine the ratio of car(on monoxide to car(on dioxide 8%9A%92ratio that will (e a(le to reduce the oxide to
metal at a given temperature.
G To determine the relative sta(ilities of oxides and sulphides etc.
O.7: any three uses can (e sufficient
1) a) Define *llingham diagram !) write its principle of construction and c) draw the line for the reaction 2 3 4
. 24
Ans:
a)The llingham diagram is a plot (etween the standard free energy change 8R 5 of a reaction as a function ofthe temperature 8T
!)The principle of construction is ,F- . ,$-/T ,S-
$here 6R5 ! *tandard free energy change
6&5 ! standard enthalpy change
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6*5! standard entropy change
c)
15) The #apour pressure pG (measured in mm $g) of liquid arsenicG is gi#en !y log p . / '-
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Ans:
a L solution is a homogeneous mixture composed of two or more su(stances. In the study of solutions, it is
customary to designate the component present in larger proportion as the solvent and the one in smaller
proportion as the solute.
( quation for calculating wt from atom
,55FF
FF X
BXatwtBatomAXatwtAatom
AXatwtAatomofAWt +=
*imilarly,
,55FF
FF X
BXatwtBatomAXatwtAatom
BXatwtBatomofBWt
+=
c quation for calculating atom from wt
,55AFAF
AFF X
atwtBBwtatwtAAwt
atwtAAwtofAatom
+=
*imilarly,
,55AFAF
AFF X
atwtBBwtatwtAAwt
atwtBBwtofBatom
+=
-) State the effect of temperature on rate of reaction with Arrhenius rate equation
Ans:
*ffect of Temperature:The rate of a reaction is increased considera(ly (y a little increase in temperature. The effect of
temperature on the rate of the reaction can (e explained (y considering the Lrrhenius rate equation, which is given (elow
Kate ! L e+"AKT
$here L+++0re exponential constant
"+++Lctivation energy
K+++)niversal gas constant
T+++Temperature in /
From the a(ove equation, we can conclude that the rate of a reaction increases exponentially with temperature.
1) a) Define $omogeneous systemG gi#e one e@ample !) Define heterogeneous systemG gi#e one e@ample
Ans:
a) $omogeneous system:L system, which consists of *ingle 0hase and )niform composition throughout, is called a
&omogeneous system
*@ample:L pure *olid or Siquid, L mixture of gases, $ater < salt
O.7: any one example can (e sufficient
!) $eterogeneous system:L system, which consists of More than one 0hases and %omposition, is not uniform is called
&eterogeneous system
*@ample:L Mixture two or more immisci(le solids,
L Mixture of two or more immisci(le liquids,
Ice in contact with water
Ll+2 *i
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Rrey cast Iron
0earlite
Metal # *lag + Ras
) Define a) $eat 2apacity !) $eat capacity at 2onstant &ressure c) $eat capacity at 2onstant olume
Ans:
a) $eat 2apacity
- The quantity of heat required to raise the temperature of a (ody (y 5
- If the temperature of the (ody is raised from Tto T2(y passing an amount of heat ", the heat capacity % of the
(ody is given (y
% ! " A 8T2+T
89r
dT
QC=
!) $eat capacity at 2onstant olume
- It is the amount of heat required to raise the temperature of a (ody (y 5at constant volume
- It is denoted (y %v
- Lt constant volume heat capacity can (e given as
- dT
qVVC =
- $here qvis heat a(sor(ed at constant volume
c) $eat capacity at 2onstant olume
- It is the amount of heat required to raise the temperature of a (ody (y 5at constant 0ressure
- It is denoted (y %p
- Lt constant 0ressure heat capacity can (e given as
-
dTC
QP
P=
- $here "0is heat a(sor(ed at constant pressure
") calculate the heat of formation of J&2l0 at =9 K from the following data:J&0 3 "(2l). L&2l" N$
-=9 . /19 calO L&2l" 3 (2l) . J&2l0 N$
-=9 . /""9 cal
Solution:
The required reaction is: ;04 < HA2 %l 2! ;0%lH4
Set
2;04 < 38%l2! 2V0%l3W &5
2> ! +H.> 'cal+++++++ 8
V0%l3W < 8%l2 ! ;0%lH4 &52> ! +33.> 'cal++++++++ 82
Multiply eqn 82 with B2C (oth sides
2V0%l3W < 28%l2 ! 2;0%lH4 &5
2> ! +J?.J 'cal+++ 83
Oow add eqn 8 and 83
2;04 < H8%l2! 2;0%lH4&52> ! +>3.G 'cal+++++++ 8G
Qivide eqn 8G with B2C
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;04 < HA2 8%l2! ;0%lH4 &52> ! +G.? 'cal++++++ 8H
The a(ove reaction is same as that of the required reaction
Therefore the heat of formation of ;0%lH4 at 2> / is ! /'15 cal (Ans)
') a) write the mathematical relation !etween entropy and heat !) define spontaneous process and gi#e one
e@ample
Ans:
a) Mathematical relation !etween entropy and heat: For a reversi(le change ta'ing place at a fixed
temperature 8T, the change in entropy 8* is equal to heat energy a(sor(ed or evolved divided (y thetemperature 8T
That is,
S.
ET
!) Spontaneous process: L process, which proceeds on its own accord, without any outside assistance, is
termed a spontaneous or natural process or irreversi(le process.
*@amples Kolling stone from high level to low level
2 &eat flow from high temperature to low temperature
3 Ras flow from high pressure to low pressure
O.7: any one example can (e sufficient
) a) Define *llingham diagram !) State any three uses of *llingham diagrams
Ans:
aThe llingham diagram is a plot (etween the standard free energy change 8R 5 of a reaction as a function of
the temperature 8T( The uses of the llingham diagram:
H To determine the reducing agent for reducing a given metallic oxide to metalP
J To determine the partial pressure of oxygen that is in equili(rium with a metal oxide at a given temperature, this
will help us to predict the temperatures at which a metal is sta(le and the temperatures over which it will
spontaneously oxidi=e
? Qetermine the ratio of car(on monoxide to car(on dioxide 8%9A%92ratio that will (e a(le to reduce the oxide to
metal at a given temperature.
> To determine the relative sta(ilities of oxides and sulphides etc
) a) write the principle of construction of *llingham diagram and !) draw the lines for the reactions 2 3 4.
24O 2 3 4. 24Ans:
a)The principle of construction is ,F- . ,$-/T ,S-
$here 6R5 ! *tandard free energy change
6&5 ! standard enthalpy change
6*5! standard entropy change
!)
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5) a) write the !asic idea that the phase equili!ria pro#ides !) write the 2lapeyron equation for solid > liquidequili!rium of a single component system and la!el the sym!ols
Ans:
a)0hase equili(ria, a (ranch of chemical thermodynamics. The (asic idea the phase equili(ria provide is to
understand how we predict what phases will (e sta(le in a system at a given temperature, pressure, and
composition, and how those sta(ility relations will change as the temperature, pressure and compositions are
varied.
!) ;or solid P liquid, equili(rium the %lapeyron eqn is:
d&
.
N$fusion
dT T N$here d0Xchange in pressure &fusionXSatent heat of fusion
TXmelting point dTXchange in melting point
U X change in volume of liquid and volume of solid 8 i.e., Ul# Us
9) Define: a) fugacity !) acti#ityAns:
a) ;ugacity: itis a measure of escaping tendency of a su(stance to prefer one phase 8liquid, solid, gas over another. It isrepresented (y BfC
Fugacity appers as pressure (ut not really the pressure it is also 'nown as pseudo pressure or fictitious pressure (ut at low
pressures fugacity approaches pressure mathematically it is expressed as
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!) Acti#ity: the activity of a su(stance is defined as the ratio of fugacity in the given state to the fugacity in the standard
8or pure state.
Mathematically it is expressed as shown (elow
a . f
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Lccording to %harlesC Saw + U Y T 8Lt constant 0 and m
%om(ining the a(ove two laws we can write U Y TA0
9r 0UAT ! /
$here / is a constant
/ value can (e found as follows:
++++++++++++++ 8
$here05is atm pressure
U5is 22.G lit
T5is 2?35/
If the a(ove values are su(stituted in the a(ove equation 8 we get / !5 .5>2 lit+atm
This value of B/C i.e. 5.5>2 lit+atm is constant and is same for every gas. &ence B/C is replaced (y a new
constant called )niversal gas constant. This is represented (y BKC Oow the ideal gas equation ta'es the following
form for +mole of gas
&
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&K represents the sum of enthalpies of reactantsTherefore & ! &7+&L
Oow differentiating 83 on (oth sides (y T
7ut %p!
There fore
Oow
The a(ove equation is called /irchoffCs equation
') Deri#e the free energy relationships
Ans:
Lccording to first law of thermodynamicsE. d8 3
Lccording to second law of thermodynamics dS . E
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9r
Lgain according to the definition of free energy R ! &+T*
Qifferentiating the a(ove equation dR ! d& + Td* + *dT
9r dR ! d 8)
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Que to translational, vi(rational and rotational motions of molecules, atoms and electrons etc of the
system
Internal energy is a state property and extensive in nature.
!) &roof of ds . - for a re#ersi!le process
Set sand s2denote the entropies of (loc' and (loc' 2 respectively, and let Tand T2denote the temperatures of
the (loc's. Set a small quantity qof heat flows from (loc' to (loc' 2, and then the entropy change is given as
under:
E
,2 T
Q
T
Qds
=
1
,,
,2 TTQds =
12,
2,
TT
TTQds
=
For a reversi(le process TT2that is the difference in T and T2 is negligi(le such that for practical purposes we
can ta'e T !T2
Then the entropy change d*rev! q 8T +T2A T T2! 5
Therefore for reversi(le process entropy change dS . -
) Deri#e the 2lapeyron equation for Hiquid > apour equili!rium of a single component system
Ans:
%onsider the phase transformation of Siquid to Uapour of a one+component system such as pure metal
Siquid 8S \ Uapour 8U 8
*ince the S and U, phases are in equili(rium with each other. Therefore, the change in the Ri((s free energy for
the transition given (y quation 8 is =ero.
That is, R ! 5. 82
RU] RS! 5 83
RU ! RS 8G
If the temperature and pressure are altered slightly such that the free energy changes slightly (y dR without
distur(ing the equili(rium
Then 8RU< dRU ! 8 RS< dRS 8H
From 8G RU ! RS
T1
T2
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dRU! dRS 8J
7ut we 'now from dR ! ] *dT < Udp
Therefore dRU ! ] *U dT < UU dp 8?
dRS ! ] *SdT < USdp 8>
From 8J ] *U dT < UU d0 ! ] *SdT < USd0 8
Therefore 8*U] *S dT ! 8UU+ US d0 85
9r * dT ! U d0 8
9r d0AdT ! *A U 82
7ut * ! &vapAT
$here &vap is latent heat of Uaporisation
T is the temperature where phase transformation ta'es place that is (oiling point
There fore
d&.
N$(1")
dT T NThe a(ove equation is called 2lapeyron *quation
5) a) Define the equili!rium constant for the reaction aA 3 !B . c2 3 dD
!) 2alculate the equili!rium constant for the reaction:
JQi40 3 ($) . JQi0 3 ($-)
At 5- 2 from the following data:
JQi0 3 U (-) . JQi40O NF4. /''"1 3 =9'' T C
($) 3 U (-) . ($-)O NF4. /'- 3 '9T C
Ans:
a) aA + bB cC + dD
The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a, (
are the no. of moles of reactants and c, d are the no. of moles of products
The equili(rium constant 8/c is defined as given (elow
!) 2alculation of equili!rium constant:
Set
;Oi4 < ^ 852 ! ;Oi94P R9! +2GG32 < >.GG T @++++++++++++++++8
8&2 < ^ 852 ! 8&25P R9! +2GJ252 < HG.>T @ +++++++++++++++++++++82
9n *u(tracting 8 from 82
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;Oi94 < 8&2 ! ;Oi4 < 8&25P R9! +>>+G3.JGT @ @ ++++++++++++++++83
The a(ove equation is the required reaction
R9at T! ?H5>+G3.JG_523 @ ! +GJH2G.?2@
7ut the relation (etween equili(rium constant and free energy change is given (y
R9! +KTln/e
$here /eis the equili(rium constant
Therefore
/e ! xp 8+R9AKT
! xp 8GJH2G.?2A>.3G_523
! xp 8H.G?
! "5
9) a) Define %aoult7s law !) *@plain the factors causing a real solution to de#iate from ideal !eha#iourAns:a) Raoults law:
The activity of a component of an ideal solution is equal to its mole fraction. That is in an ideal solution ofsolvent A and solute B,
Where
Activity of A
ole fraction of A!imilarly aB" #BWhere aB Activity of B#B ole fraction of B
( The deviations of a real solution from ideal (ehaviour are of two types they are
0ositive deviation and 2 Oegative deviation
The factors, which cause the deviations, are the interactions (etween similar and dissimilar atoms of the
components of the solution. %onsider a solution in which the solvent is BLC and solute is B7C and if the interactions
(etween similar atoms are FL+Land F7+7 and the interactions (etween dissimilar atoms are FL+7then
If FL+L! F7+7 ! FL+7 Then the solution (ecomes ideal
2 If FL+L F7+7 Then the solution under goes negative deviation
2 If FL+L>FL+7< F7+7 Then the solution under goes positive deviation
The deviations are shown (elow in the diagram
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= a) *@plain the terms: heat and wor with units State sign con#entions of heat and wor
!) *@plain a!out wor of e@pansion Deri#e equation for wor of e@pansionAns:
a) $eat and or:
$hen a change in the state of a system occurs, energy is transferred to or from the surroundings. This energy may (e
transferred as heat or mechanical wor'.
$eat: &eat is defined as the energy in transit
or: Mechanical wor' is defined as force N distance. 9r $ ! 0U
$here 0 # is pressure U # is the change in volume
8nits of $eat: In *.I units heat is measured in oules.
8nits of or:In *.I units $or' is measured in `oules.
Sign 2on#ention of $eat:
The sym(ol of &eat is ".
If the heat flows from the surroundings into the system to raise the energy of the systemP it is ta'en to (e positive,
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Ras at constant pressure
dl
Therefore Wor$ " %ressure # chan&e in 'olume
Qerivation of equation for wor' of expansion:
%onsider a gas contained in a cylinder fitted with a friction less piston. The constant pressure acting is 0. If the gas
expands at constant pressure, the piston moves through a distance let this distance is DdlE.
Then $or' ! force N distance
$ ! f N dl
7ut f ! 0 N L 8here L!area of cross section of the piston
Therefore $ ! 0 N L N dl
! 0 N U 8here U ! change in volume
1-) Deri#e the relation !etween & (&ressure) and (olume) in an adia!atic re#ersi!le process for an ideal gas
Ans:
Deri#ation of %elation !etween & > in an Adia!atic %e#ersi!le&rocess for an Ideal gas
In an adia(atic reversi(le process, the first law of thermodynamics ta'es the following form
" ! d) < 0dU according to first law
5 ! d) < 0dU since in an adia(atic process " ! 5
d) ! +0dU
Llso we have d) ! %vdT
%vdT ! +0dU
The state equation of an ideal gas is 0 U !
K T 8for mole of gas
Qifferentiating, 0dU < Ud0 ! KdT ++++++++++ 8(
*u(stituting 8a in 8(
0dU < Ud0 ! K 8+0dUA%v ++++++++++++++++++++++ 8c
7ut we have for I mole of ideal gas %p# %v ! K +++++++++++++ 8d
*u(stituting 8d in 8c
0dU < Ud0 ! 8%p# %v 8+0dUA%v
dT .+0dU
++++++++++ 8a%v
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Ans:
a) $ess7s law of constant heat summation
This law can (e stated as: If a chemical change can (e made to ta'e place in two or more different ways, whether in one or
several steps, the amount of total heat change is same no matter (y which method the change is (rought a(out.
The law also follows as a mere consequence of the first law of thermodynamics.
Set us suppose that a su(stance BLC can (e changed to BC directly
L ! < "
$here " is the heat evolved in the direct change. $hen the same change is (rought a(out through intermediate stages
L < 7 ! "2
7 < % ! "3
% < ! "G
The total evolution of heat ! "2
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Fe293< 3%9 ! 2Fe ! +2?.2 /@++++++++ 8>
qn 8> is the required eqn therefore N$-=9. /5 KC (Ans)
1) 2alculate the standard enthalpy and entropy changes at 2 for the reaction:
J2u0 + (4) . J 2u40
NFo
. /1==-/1' T log T 3 1"" T C
Ans:
2alculation ofstandard enthalpychange:
The enthalpy change in terms of Ro is given as
Therefore RoAT ! +J25AT+J.G log T < 23.3
! +J25AT+J.G ln TA2.353
! +J?J> @
! +J?.J> '@
Therefore the standard enthalpy change ! /1519 C
2alculation of standard entropy change:
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The enthalpy change in terms of Ro is given as
Therefore Ro! +J25+J.G T log T < 23.3 T
! +J25+?.2 T ln T < 23.3 T
! +?.2T_AT < ln T 1 < 23.3
! +?.2 < ln T1 < 23.3
! +?.2 < ln T1 < 23.3
7ut
so. ?.2 < ln T1+23.3
! ?.2 < ln 2>1 # 23.3
! ?.2
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2 2p.3 T / T/
$hereGG are constants%p +++ &eat capacity at constant pressure
T +++ Temperature in /elvin
( For a process at %onstant Uolume, there is no xpansion and %ontraction i.e. 8dU ! 5 *o $ ! 5
*o the equation " ! < $ will (ecome " U ! 6U&eat a(sor(ed at constant Uolume i.e., "Uis equal to the energy increase 6Uaccompanying the
0rocess
c Deri#ation of Fi!!s > $elmholtV equation
The Ri((s free energy for a closed system R ! &+T*
Qifferentiating the a(ove equation dR ! d& + Td* + *dT
9r dR ! d 8)
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+++++++++
++++ 8 dT T U
$e can ma'e two approximations, which will give us the %lausius+%lapeyron equation.
The first approximation is that the volume of a given amount of gas is much larger that the volume of an equivalent
amount of solid or liquid. There fore
U ! Ugas] Uliquidb Ugas
9r2 The second approximation is that the vapour produced (ehaves as an ideal gas therefore we can use the state equation
of the ideal gas and we can replace Ugas(y the following.
0U ! KT 8ideal gas equation if n!
That is, we set Ugas! KTAp.
$ith these two approximations quation 8 (ecomes
d0A0 ! 8 &A KT2 dT
d ln0 ! 8 &A KT2 dT
9r2
lnRTH
dTPd =
The a(ove equation is called the 2lausius/2lapeyron equation
The a(ove equation ta'es the following form if it is integrated (etween pressures 0 , 02and temperatures (etween T, T
assuming & is constant
2
,
2
,
2
:8lnT
dT
R
HPd
T
T
P
P
=
1,,
.ln2
22
,2
,
TTR
H
P
P
=
The a(ove equation is very useful for estimating vapor pressures at any temperature if the vapour pressure at another
temperature and the heat of vaporisation over that temperature range are 'nown.
1) Deri#e the ant $off7s *quation for Isotherm for the reaction of the type aA 3 !B . c2 3 dD
Ans:
%onsider the given reaction at a temperature 8T and a pressure 80
aL < (7 ;++4 c% < dQ
The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a,( are the no. of
moles of reactants and c,d are the no. of moles of products
The free energy change for the a(ove reaction is written as
6R ! R products# R reactants++++++++++++++++++ 8
R products! cR% < dRQ++++++++++++++++++++++++++ 82
R reactants! aRL < (R7+++++++++++++++++++++++++++ 83
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7ut R ! R5< KT lna 8BaC is activity
Lpplying the a(ove equation to 82 and 83
cR%! cR5< cKT lna% 8 BacCis the activity of B%C
*imilarly
dRQ! dR5< dKT lnaQ +++++++++++++++++++++++ 8G
aRL! aR5< aKT lnaL++++++++++++++++++++++++ 8H
(R7! (R5< (KT lna7++++++++++++++++++++++++ 8J
*u(stituting the a(ove 8G, 8H and 8J in 8
6R ! 8cR%
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$hen L+7 interactions are wea'er than L+L or 7+7 interactionsP then is 4 8positive deviation from ideality,
system can then tend towards phase separation and mixing is endothermic.
xample: Fe+%u liquid alloy
;actors causing de#iations from ideal !eha#iour
( The deviations of a real solution from ideal (ehaviour are of two types they are 0ositive deviation and 2 Oegative deviation
The factors, which cause the deviations, are the interactions (etween similar and dissimilar atoms of the
components of the solution. %onsider a solution in which the solvent is BLC and solute is B7C and if the interactions
(etween similar atoms are FL+Land F7+7 and the interactions (etween dissimilar atoms are FL+7then
If FL+L! F7+7 ! FL+7 Then the solution (ecomes ideal
2 If FL+L; FL+74 F7+7 Then the solution under goes negative deviation
2 If FL+L4 FL+7; F7+7 Then the solution under goes positive deviation
15 a) rite any fi#e differences !etween re#ersi!le and Irre#ersi!le processes !) Define Thermodynamic
equili!rium c) e@plain the criteria for the esta!lishment of thermodynamic equili!rium
Ans:
Keversi(le process Irreversi(le process
It ta'es place with infinitesimally small change It ta'es place with a finite change
2 These are ideali=ed and true in principle only 2 Lll actual processes which occur are Irreversi(le
3 *ystem is at equili(rium at all stages of the
process
3 *ystem is at equili(rium only at the initial and
final stages of the process
G $or' done is maximum G $or' done is less than that of reversi(le process
H fficiency of reversi(le process is maximum H fficiency is less
J This process can (e retraced (ac' in the opposite
direction
J this process cannot (e retraced (ac' in the
opposite direction
? xample: Ice can (e formed (y withdrawing
heat from water at 55% under normal pressure.
Further if heat is supplied at the same rate to ice at
55% under normal pressure water is formed at the
same rate.
? xample: when two gases are allowed to mix up
into a given volume at a constant temperature, the
pressure in the container increases. 7ut the two
gases cannot (e separated (y `ust with drawing the
pressure from the mixture.
( L *ystem is said to (e under Thermodynamic equili(rium if the system under goes infinitesimal change so that we
L(le to determine the values of its state properties such as 0ressure, Temperature and Uolume.
c For the esta(lishment of Thermodynamic equili(rium the system should (e in
Thermal equili(rium
Mechanical quili(rium and
%hemical equili(rium
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Thermal *quili!riummeans the temperature of the system must (e uniform throughout the system I.e. dT ! 5 $here
BdTC represents change in temperature
Mechanical *quili!riummeans the 0ressure of the system must (e uniform throughout the system I.e. d0 ! 5 $here
Bd0C represents change in pressure.
2hemical *quili!riumthe %hemical %omposition of the system must (e uniform throughout the system I.e. d% ! 5
$here Bd%C represents change in temperature
19 a) Define heat capacity at constant pressure (2p) and heat capacity at constant #olume(2#)
!) Deri#e the relation !etween 2pand 2#for an ideal gas from first principles
Ans:
a)
$eat capacity:The heat capacity is the amount of heat required to raise the temperature of the system (y one degree%elsius.
The relationship (etween heat and temperature change is usually expressed in the form shown (elow
$here % is the heat capacity
$eat capacity at constant pressure82&:
$here"p ! &eat content of the system at constant 0ressure
$eat capacity at constant #olume82#:
$here
E#! &eat content of the system at %onstant Uolume.
!) Deri#ation of the relation !etween 2pand 2#from first principles
Lccording to the definition of heat capacity at constant 0ressure 8%p, %p is given (y the following equation
2 . E
,T
2p .Ep
,T
2p .E#
,T
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2p !
Ep//////////////(1)
NT
9r
2p !R$
//////////////()NT
$here Ep or R$is heat content of the system at constant pressure Tis change in temperature
Lccording to the definition of heat capacity at constant 0ressure 8%v, %v is given (y the following equation
2# !E#
///////////////(")NT
4r
2# !R8
///////////////(')NT
$hereE or R8 is heat content of the system at constant volume. Tis change in temperature
Lccording to first law of thermodynamics: E . d8 3
That is E . d8 3 &d
Lt constant pressure: Ep. d8 3 &d
Qividing the a(ove equation with dT 8change in temperature)
*u(stituting (1)and (')in () we get
For one mole of an Ideal gas, the state equation is & . %T
Qifferentiating on (oth sides: &d 3 d& . %dT
Lt constant pressure: d& . -
Therefore &d 3 - . %dT
That is &d . %dT
d.
%dT &
*u(stituting this in eqn ()
Ep.
d83
&d////////////////()
dT dT dT
2p . 2# 3 p d ///////////////()dT
2p . 2# 3& %
&
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Therefore
9r
(1=) ;e4"G &!4 and 2u4 present in a dead roasted tin ore react according to the following equationsG during
leaching with 111 M $2l at -2
;e4"3 $2l (- $4) . ;e2l"(1- $4) 3 "$4
&!4 3 $2l (- $4) . &!2l(1-- $4) 3 $4
2u4 3 $2l (- $4) . 2u2l (- $4) 3 $4
2alculate the heat of leaching process at -2 per 1-- Kg of oreG assuming that the Si4and Sn4present in the
ore are unaffected during leaching
Fi#en:
(i) 2omposition of dead roasted tin ore is 1-? ;e4"G ? &!4G ? 2u4G 1? Si4and rest Sn4
(ii) Standard heats of formation of ;e4"G 2u4G &!4G ;e2l"(1- $4)G 2u2l (- $4)G &!2l(1-- $4)G $4G $2l
(- $4) at -2 are /1="G /'--G /'G /=-G /"-=G /9"G /9" and /15' cal3.2
&29 +J>.32
&%l 8H5 &29 +?.G
Fe293< J &%l 8H5 &29 ! 2Fe%l38H5 &29 < 3&29 +++++++++++ 8
&eat of leaching due to mole of Fe293is ! 2&f Fe%l3 8H5 &29< 3&f &291+&f Fe293< J &f&%l 8H5 &291
! 2_8+2.5 < 38+J>.321 # +J.3>.2J 'cal
That is heat of leaching due to J5 g 8since mole of Fe293 ! J5 g Fe293 ! +>>.2J ' cal
2p . 2# 3 %
2p / 2# . %
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Lmount of Fe293present in 55 'g of ore ! 5 of 55 'g ! 5 'g
Therefore heat of leaching due to 5 'g or 5555 g of Fe293! 8+>>.2JAJ51_5555 ! +HHJ.2H ' cal
0(9 < 2 &%l 8H5 &29 ! 0(%l2855 &29 < &29+++++++++ 82
&eat of leaching due to mole of 0(9 is ! &f 0(%l2 855 &29< &f &291+&f 0(9< 2 &f &%l 8H5 &291
! 8+>3.2 < 8+J>.321 # 8+H2.G < 28+?.G1
! +JG.32 'cal
That is heat of leaching due to 223.2 g 8since mole of 0(9 !223.2 g 0(9 !+JG.32 ' cal
Lmount of 0(9 present in 55 'g of ore ! H of 55 'g ! H 'g
Therefore heat of leaching due to H 'g or H555 g of 0(9 ! 8+JG.32A223.21_H555 ! +GG5. ' cal
%u29 < 2 &%l 8H5 &29 ! 2%u%l 8H5 &29 < &29+++++++++ 83
&eat of leaching due to mole of %u29 is ! 2 &f %u%l 8H5 &29< &f &291+&f %u29< 2 &f &%l 8H5 &291
! 28+35. < 8+J>.321 # 8+G5.5 < 28+?.G1
! +HH.32 'cal
That is heat of leaching due to G3. g 8since mole of %u29! G3. g %u29! +HH.32 ' cal
Lmount of %u29present in 55 'g of ore ! H of 55 'g ! H 'g
Therefore heat of leaching due to H 'g or H555 g of %u29! 8+HH.32AG3.1_H555 ! +33 ' cal
Therefore the heat of leaching process at 2H5% per 55 /g of ore
! +HHJ.2H ' cal < +GG5. ' cal < +33 ' cal
! /99=-1 cal (Ans)
-) a) State the concept of ;ree energy and gi#e its mathematical equation !) 2hromium and 2ar!on present in
stainless steel form 2hromium car!ide at ---2 Show !y thermodynamic calculations which of the metals among
SiG TiG and should !e alloyed to stainless steelG so as to pre#ent the formation of 2hromium car!ide
Fi#en:
" 2r 3 2 . 2r"2O NF-. /-9--/'- T cal
Si 3 2 . Si2O NF-. /155- 3 1 T cal
3 2 . 2O NF-. /---- 3 1 T cal
Ti 3 2 . Ti2O NF-. /'--- 3 9 T cal
Ans:
a The energy freely availa(le in the system for doing useful wor' is called free energy. It is represented (y BRC. This is
also 'nown as thermodynamic potential or Ri((sCs free energy. The mathematical definition of free energy is given (y
F . $ > TS
$here &+enthalpy of the system
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T+a(solute temperature
*+entropy of the system
The a(ove equation infers that the total energy availa(le in the system is B&C Lnd the energy which is not used for doing
wor' is BT*C this is also called the (ound energy, the remaining energy that is BRC is only availa(le in the system for doing
useful wor' which is the free energy.
!) Solution:
First find the numerical values of the R 5at T ! J55 < 2?3 ! >?3 / of the given reactions
3 %r < 2% ! %r3%2P R5! +25>55+G.5 T cal
R5of %r3%2 at >?3 / ! +25>55+G.5_>?3 ! +2G22 cal
*i < % ! *i%P R5! +2??5 < .JJ T cal
R5of *i% at >?3 / ! +2??5 < .JJ_>?3 ! +325.> cal
U < % ! U%P R5! +25555 < .J T cal
R5
of U% at >?3 / ! +25555 < .J_>?3 ! +>J53.2 calTi < % ! Ti%P R5! +GH555 < 2.> T cal
R5of Ti%at >?3 / ! +GH555 < 2.>_>?3 ! +G2HHH.J cal
Oow %ompare the Ualues of R5of the a(ove compounds I,e 2r"2O Si2O 2O and Ti2O
9n comparison, it can (e concluded that R5of Ti% is more negativeP this means the compound Ti% is more sta(le than
%r3%2
Therefore, to prevent the formation of %hromium car(ide Tishould (e added
1) a) *@plain a!out the internal energy of a thermodynamic system !) Deri#e Fi!!s > $elmholtV equation
Ans:
a) Internal energy:
The total of all the possi(le 'inds of energy of a system is called its internal energy. Internal energy of a system is
made up of (y
Internal potential energy
Internal 'inetic energy
Internal potential energy :
This consists of the energy
Que to electrostatic force of attraction (etween nucleus and electrons of the atom
Que to inter molecular attractions
Que to chemical (onds
Internal 'inetic energy:
This consists of the energy
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Que to translational, vi(rational and rotational motions of molecules, atoms and electrons etc of the
system
Internal energy is a state property and extensive in nature.
!) Deri#ation of Fi!!s > $elmholtV equation
The Ri((s free energy for a closed system R ! &+T*
Qifferentiating the a(ove equation dR ! d& + Td* + *dT
9r dR ! d 8)
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Lccording to this equation, the rate at which the natural logarithm of the vapor pressure of a liquid changes with
temperature is determined (y the molar enthalpy ofvapori=ation of the liquid, the ideal gas constant, and thetemperature of the system.
If we assume that Hvdoes not depend on the temperature of the system, the %lausius+%lapeyron equation can
(e written in the following integrated form where Cis a constant.
CRTHP V +=ln
The plot of vapor pressure versus temperature is shown (elow
(
For solid \ liquid, equili(rium
d0 ! &fusiondT T U
$here d0Xchange in pressure &fusionXSatent heat of fusion
TXmelting point dTXchange in melting point
U X change in volume of liquid and volume of solid 8i.e., Ul# Us
2 For liquid \ vapour, equili(rium
d0!
&vap
dT T U
$here d0Xchange in pressure &vapXSatent heat of vaporisation
TX(oiling point dTXchange in (oiling point
U X change in volume of Uapour and volume of liquid 8 i.e., Uv# Ul
3 For solid \ Uapour, equili(rium
d0!
&su(limation
dT T U
$here d0Xchange in pressure &su(limationXSatent heat of su(limation
TXsu(limation point dTXchange in su(limation point
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U X change in volume of Uapour and volume of solid 8i.e., Uv# Us
") Deri#e the ant $off7s *quation for Isotherm for the reaction of the type aA 3 !B . c2 3 dD
Ans:
%onsider the given reaction at a temperature 8T and a pressure 80
aL < (7 ;++4 c% < dQ
The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a,( are the no. of
moles of reactants and c,d are the no. of moles of products
The free energy change for the a(ove reaction is written as
6R ! R products# R reactants++++++++++++++++++ 8
R products! cR% < dRQ++++++++++++++++++++++++++ 82
R reactants! aRL < (R7+++++++++++++++++++++++++++ 83
7ut R ! R5< KT lna 8BaC is activity
Lpplying the a(ove equation to 82 and 83
cR%! cR5< cKT lna% 8 BacCis the activity of B%C
*imilarly
dRQ! dR5< dKT lnaQ +++++++++++++++++++++++ 8G
aRL! aR5< aKT lnaL++++++++++++++++++++++++ 8H
(R7! (R5< (KT lna7++++++++++++++++++++++++ 8J
*u(stituting the a(ove 8G, 8H and 8J in 8
6R ! 8cR%
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( Qepending on the (ehaviuor of solutions they are classified as Ideal solutions 2 Keal solutions and 3 Kegular
solutions
1) Ideal solutions: L *olution is called ideal solution if the interactions (etween similar and dissimilar atoms is the same
that is in an ideal solution of L and 7, the L+L, 7+7 and L+7 interactions are identical
9r FL+L! F7+7! F%+% $here F indicates the interaction
xample: %u+Oi alloys
For an ideal solution &mix,ideal ! 5
*mix,ideal 5
Umix,ideal ! 5
) %eal solution:L *olution is called real solution if the interactions (etween similar and dissimilar atoms is different
that is in a real solution of L and 7, the L+L, 7+7 and L+7 interactions are different
9r FL+L F7+7 F%+% $here F indicates the interaction
For a real solution &mix,real 5
*mix,real 5
Umix,real 5
xample: Fe+Oi, Fe+%u
") %egular solution:L special type of a non+ideal or real solution whose (ehaviour is in (etween the ideal and rea
solutions for which the enthalpy of mixing is non+=ero (ut the entropy of mixing is identical to that of ideal solutions
For a regular solution &mix,reg 5
*mix,reg ! *mix,ideal
Umix,reg 5