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CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
example-pressure, temperature, density, specific heat, surface
INTRODUCTION :---In a chemical reaction, the old bonds in the
tension, viscosity, refractive index, melting point, boiling point,
reactant species breaks & new bonds are formed
...
• Intensive properties are not additive in nature
...
If energy required for
Such thermodynamic properties which depends upon the quantity
breaking of bonds is greater than the energy released during bond
formation, the reaction will be endothermic but if energy released of matter, known as extensive properties e
...
mass, volume,
during bond formation is greater than energy required for breaking energy, enthalpy, work, heat capacity, internal energy, entropy,
Gibb’s energy etc
...

of old bonds, the reaction will be exothermic
...

original value of that property for the whole of the system is
Thermodynamics deals with energy in its various forms & their
called an Intensive property
...
Transformation of heat into work is only
values different from the values for whole of the system are called
possible by the use of a device, known asengine
...

Thermodynamics:--Branch of science which deals with the study
consider air in a room at temp of 300K, 1 atm pressure
...
Branch of science in room is divided by some boundary ( imaginary or real) into two
parts (equal or unequal) then in these two parts :
which we study the energy change during a chemical reaction,
*The temperature, pressure, density of the gas, concentration of
known as chemical thermodynamics
...
will have the same value as that of for
Laws of thermodynamics purely based on experience & have no
whole of the system
...
Thermodynamics does not explain rate of
*While the volume of two parts, mass of gas in two parts, total
reaction, mechanism of reaction, extent of reaction etc
...
will be different from the values of these properties as
•We can predict feasibility of the reaction that is the reaction will
for the whole of the system initially
...

•If Reaction take place then what are the energy changes involved • THERMODYNAMIC EQUILIBRIUM :
•When there is no change in any observable or measurable
during the reaction
...

be the equilibrium Concentrations of different reactants &
•Thermodynamic equilibrium consist of three types of
products, can be calculated with thermodynamics
...

LIMITATIONS OF THERMODYNAMICS :
(a)Mechanical equilibrium
• Laws of thermodynamics are applicable to matter in bulk or on
system as a whole, these can not be applied on individual particles (b)Thermal equilibrium
• thermodynamics cannot calculate time taken for completion of a (c)Chemical equilirbrium
STATE FUNCTIONS :-The variables which determines the state
reaction or for attainment of chemical equilibrium
...
g
...
Thermodynamics does not explain rate of
internal energy, enthalpy, entropy, free energy, no
...
It
reaction, mechanism of reaction, extent of reaction etc
...

thermodynamic system
...

•PATH FUNCTION : Quantities which are dependent on the
•Boundary : Anything which separates system & surroundings is
path/way the system has achieved a particular state
...
g
...
Boundary can be real or imaginary
...

* Boundary can be adiabatic (non-conducting) or diathermic
Thermodynamic process : Any method/process by which system
(conducting)
...
Combination of system & surrounding
to another state of thermodynamic equilibrium
...

Isothermal process:- Such type of thermodynamic process in
Type of system:-- A system is said to be open systemif it can
which temp
...

exchange the matter and energy both with the surrounding
...

in which no heat transfer occur i
...
total heat remain
A system is said to be isolated system if it can exchange neither
constant, known as adiabatic process
...
There is not even a perfectly
Isochoric process :-- Such type of thermodynamic
isolated system
...

process
...
e
...
A system is said to be heterogeneous if all
known as isobaric process
...

be cyclic process if the system undergoes a series of
THERMODYNAMIC PROPERTIES:-Such thermodynamic
changes and finally returns to its initial state
...
For
change is carried out so slowly that system and surrounding are
56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
always in near equilibrium, known as reversible process and if it
Let a system having energy E1& q amount of heat is supplied to it
...
An ideal reversible
Again w work is done on it so that its energy become E2 then,
process will take infinite time to get completed
...
All real process are irreversible
...
If the
work between the system & the surrounding
...
e
...
and
Heat : When the energy transfer takes place because of
hence ∆E =qv
temperature difference between system & surroundings
...
Energy that is transmitted from
is state function
...
e
...
It is a path
reduces to ∆E = w
...

internal energy will increase or if work is done by the system its
P=F/A ⇒ F=P
...

Work=Force ×Displacement =F
...
dl=P
...
dl=P
...
e
...

If Hr>Hp
∆H is Negative – exothermic reaction
W = PdV= (nRT/V)
...

= nRT lnV2/V1 =2
...
303 nRT log P1/P2 =2
...

W = P∆V = 2
...

= 2
...
Q=m c ∆t
W rev = -2
...
303nRT log10 P1/P2
as molar heat capacity
...
of unit mass of the
w = -pext V2 – V1)
substance by one degree centigrade
...

* IUPAC SIGN CONVENTION ABOUT HEAT AND WORK
C=dq/dt&dq=dE+PdV
Note:- heat given to system = positive
Cv=(dE +Pdv) /dT=dE/dT (as dv=0 at constant volume)
heat taken out from system = Negative
Cp=dE/dT + P dv/dT=dH/dT (because H=E+PV,
work done by system i
...
during expansion is (-) ve
dH =dE+PdV, dH/dT = dE/dT + P dV/dT )
work done on system i
...
during contraction is (+) ve
Also, PV=RT & H= E+ PV Hence, H=E + RT
SI unit of heat & work is joule
1000j =1 KJ
7
dH/dT= dE/dT +R, Cp=Cv+ R, Cp-Cv= R
1J =10 ergs
&
1 calorie = 4
...
atm = 101
...
206 cal
internal energy & pressure volume work done
...
Sum of different types of
It is a state function
...
It is not
measured as absolute value of E cannot be determined
...
The
∆H=Hp-Hr⇒ Thus, change in enthalpy is the heat energy
internal energy of a system is extensive property & a state
exchange between the system & surrounding at constant temp
...

∆E = Ep-ErIf Ep>Er⇒ ∆E Will be Positive
Again, ∆H=∆E+P∆V+ V∆P
&If EpIf ∆P=0 ie
...

Also, at constant pressure q=∆E+P∆V & hence ∆H=qp
ZEROTH law of thermodynamics :-- According to this law, two
objects at different temperature in thermal contact with each other Thus change in enthalpy is the sum of change in internal energy
and pressure volume work done
...

the amount of heat absorbed at constant pressure
...
If one of the two PV=nRT
PV1 = n1RT & PV2 = n2RT⇒ PV2-PV1= n2RT - n1RT
objects be thermometer, we can measure the temperature of other
⇒ P (V2-V1) = (n2-n1) RT⇒ P∆V =∆ngRT
object
...

also called law of conservation of energy& was given by
Helmholtz
...

enthalpy of reaction is the change in enthalpy during a chemical
”i
...
“total energy of the universe is constant
...

system that is isolated from its surrounding is constant
...
E
...
E
...

one allotropic form changes to another
...
3 Kcal
Enthalpy of Precipitation :
H2(g) + ½ O2(g) → H2O(g) → H2O(g) ; ∆H = -57
...

½ Cl2 (g) + aq→Cl- (aq) ∆Hro = ∆Hfo (Cl-, aq)
(d) condition of temp
...

By convention, the standard enthlpy of formation of H+(aq) is
Standard enthalpy of reaction is the enthalpy change for a
taken to be zero
...
)
∆Horesonance = ∆Hof,experimental - ∆Ho f, calculated
Heat of formation or Enthalpy of formation:-- Enthalpy of
=∆Hoc,calculated- ∆Hoc,expenmental
formation is the heat energy change or enthalpy change during the * Bond energy : The amount of energy required to break one
formation of 1 mole of a compound from its constituting elements mole of a particular type of a bond in the gaseous state is called
at a given temp
...
It is denoted by ∆Hf
...

Standard molar enthalpy of formation is the standard enthalpy
It is generally expressed in terms of kJ/mol
...

matter whether it takes place in one or more ways involving one
∆Hf0 = ∑Hf0 product - ∑Hf0 reactants
or several steps Suppose a substance A is changed into D (i)
•"The standard enthalpy of formation of every element in its stable directly and (ii) indirectly
...

B → C + q2Kj& C → D + q3 kJ
A few exmaples are
∆Hfo-(O2,g) = 0
Thus total heat evolved = q1 + q2 + q3 = Q2 kJ (say)
...

0
Applications of Hess’s law : Hess’s law is used for the
Enthalpy of combustion:--- Change in heat energy or enthalpy
calculation of heat changes for processes for which direct
during the combustion of 1 mole of a substance in excess of air or experimental measurements are not possible
...
It is always
Levoisier and Laplace Law: According to this ‘‘enthalpy’’ of
exothermic & hence ∆Hc will always be negative
...
e
...
, C(S) + O2(g)
defined as the net amount of energy released during the
→ CO2(g); ∆H=-94
...
3 Kcal
oxygen or air
...
/gram or KJ/gram (S
...
unit)
...

∆E = ∆Cv (T2 –T1) or (∆E2 - ∆E1/T2 -T1) = ∆Cv
Enthalpy of solution: - The change in enthalpy or heat energy
∆H = ∆CP (T2–T1) or (∆H2 -∆H1/ T2 – T1) = ∆CP
when 1 mole of the substance is dissolved in such a large excess
where ∆Cp = (Cp of products - Cp of reactants) ∆Cv = (Cv of
of the solvent at a given temp
...
If water is used as solvent then aq
...

∆E2, ∆E are change in heat internal energy at temperature T2& T1
Enthalpy of hydration:-- Net change in enthalpy or heat energy
respectively
when one mole of anhydrous salt changes to hydrated salt by
* If molar heat capacities of products and reactants are same then
combining with specified number of moles of water, known as
heat of reaction is independent of temperature
...

reactions involving solids
...
aq
...
Enthalpy of neutralization of a strong acid vs
Joule - thomson effect
...
-57
...
It is always
allowed to expand at room remperature because their inversion
exothermic
...
1 KJ/gram
expansion at room temperature because its inversion temp is –
equivalent due to less tendency of ionization
...
There are two types of Calorimeters
into ions
...
8 Kcal
enthalpies of combustion known as the bomb calorimeter is shown
Enthalpy of Transition :
in figure
...
A
56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
modified form of the apparatus shown in figure consists of a
a decrease in potential energy e
...
rolling down a slope of a rock,
sealed combustion chamber, called a bomb, containing a weighed falling down of an object, burning of octane etc
...
The Bomb calorimeter is used to measure the
2N2O5 (g) → 4NO2 (g) + O2(g) ∆H =219KJ/ mole
heat exchanged at constant volume
...
8 H2O (S) + NH4SCN (s) →Ba(SCN)2(aq) +2NH3(aq) +10
H2O (l) ∆ H= +ve
Expansion of an ideal gas is an example of a spontaneous process
in which there is no change of energy at all
...
To achieve stability they
having tendency to decrease their energy and hence all exothermic
process should be spontaneous
...
(iv) All exothermic reaction should be spontaneous as
heat content of product decreases
...
H2 (g) + ½ O2 → H2O (l) ∆ H= - 286
...
4 KJ / mole ;
the substance taken, S is specfic heat
...
endothermic in nature ie
...
In other word spontaneous eg
...
, decomposition of CaCO3 on
heating to produce CaO& CO2, decomposition of mercuric oxide
to take place is called spontaneous process
...

A process which can neither take place either by itself or by
Similarly in certain cases ∆ H =0 but spontaneous eg
...

Esterification, In some cases ∆ H is –ve but are non spontaneous
Example of spontaneous process which occur by itself :--- Flow
of heat from hot body to cold body, Dissolution of salt or sugar in in nature
...
Thus based on
I2 to form HI, Combination of an acid & base to form salt & water energy factors, not even a reversible reaction should be
spontaneous but it is not so
...

explain about the feasibility of a reaction
...
Example- Spreading of a drop of ink in a beaker filled
an electric spark , Reaction of CH4& O2 to form CO2& H2O by
with water, evaporation of water, dissolution of NH4Cl, KCl etc
...

melting of ice, decomposition of solid CaCO3&HgO etc
...
Randomness factor but it depend upon resultant value of
water, Diffusion of gas from low pressure to high pressure
...

Higher
the
randomness
higher
will be entropy
...
”
st
It
is
represented
by
S
...

Value of
But 1 law fail to explain , “why do transformations take place
entropy
is
least
for
solid
&
highest
for
gasses
...

the energy is conserved in both the cases
...
g
...

amount of heat (q) absorbed isothermally and reversibly divided
Spreading of ink in water through the beaker making the whole
by absolute temp
...
Unit of ∆S is cal
solution coloured is spontaneous process but reverse of it is not
K-1 mol-1 or JK-1 mol-1
...
Similarly mixing of two gases at same pressure is
spontaneous but separation of gaseous mixture is non spontaneous The entropy of fusion is the change in entropy when one mole of
the solid substance changes into liquid form at the melting temp
...
Thus 1st law of
∆S fusion =S liquid –S solid = ∆H fusion /Tm
thermodynamics fails to explain the spontaneity of a process
...

Spontaneous process involving macroscopic objects proceed with Entropy of vaporization is the entropy change when 1mol of a
liquid changes into vapours at its boiling temperature
...

1 All spontaneous process are thermodynamically irreversible i
...

∆S sublimation = S vap – S solid = ∆H sublimation/ Tc
without help of external agency, a spontaneous process cannot be
Entropy change is also positive in such cases for which
reversed
...
of moles of products is higher (more) than no
...

without leaving some effect elsewhere
...
e
...
For
+ ∆S surrounding) must be positive
...

∆S total = ∆S system + ∆S surrounding> 0 i
...
positive
...

In all spontaneous process entropy increases until equilibrium
Effect of temp
...
Thus at equilibrium, the entropy
For exothermic reaction — +
— any temp
...
spontaneous
∆S for water → ice at 272K, 273K & 274K are (+ve),
— —
+ high temp
...
These show that freezing
For endothermic reaction + —
+ any temp
...
non spontaneous
are in equilibrium state and at 274K freezing of water
+
+
— high temp
...

Thus
an
exothermic
reaction
which
is non spontaneous at low
Entropy change in ideal gas
temp
...

Similarly an exothermic
For an ideal gas changing its state from (P1, V1,T1) to (P2, V2,T2)
process
which
is
nonspontaneous
at
high
temp
...

= nCP InT2/T1-nRInP2/P1
STANDARAD FREE ENERGY CHANGE
1
...
For an isobaric process, ∆S = nCP In T2/T1
state
(1atm
...
For an isochoric process, ∆S = - nCvInT2/T1
state
...
For an adibatic process, Sincedq = 0, ∆S = 0
∆Go = Sum of standard free energy of formation of product
Such a process is called isoentropic process
...
303 RT log k
represented by equation G= H-TS
Standard free energy of formation of a compound is defined as the
for isothermal process, T1 =T2
free energy change which takes place when one mole of the
& hence G1 = H1 -TS1 &
G2 = H2 -TS2
compound is formed from its elements taken in their standard
∆G = (H2-TS2) – (H1-TS1) = H2- H1- T (S2- S1) = ∆H -T∆S
state
...

st
THIRD LAW OF THERMODYNAMICS
We have also, q = ∆E+W from 1 law of thermodynamics
...
and
= ∆E + W expansion + W non - expansion
vice
versa
...
(because, ∆S =qrev/T & ∆H =∆E+P∆V)
solids
may
be
taken
as
zero
...
= ∆H - T∆S thus W non expansion = -∆G
chemist
waltherNernst
...
It’s unit is J
...

process is equal to the work done by the system
...
is known as standard molar entropy and
is free for conversion to useful work and is therefore known as
represented by Som
...

absolute entropy
...
i
...
free energy change is maxm
...

For electrical work done, - ∆G = W max = nFE
∆S = S-So =
CP dT / T
and -∆Go = nFEo
but according to third law, S0 =0 at 0 K for solid & hence,
Entropy change for a system which is not isolated,
S=
CP dT / T =
CP(dT / T) =
CP d ln T
∆S total = ∆S syst+ ∆S surrounding
Let a process occur at const
...
& pressure in which heat lost
= CPln T = 2
...
dT/T + ∆Hf / Tf +
∆S surrounding = - q p / T and also,
at constant pressure, ∆H =q p& T∆S surrouding = - ∆H
CP(l)
...
dT / T
∆S total = ∆S system - ∆H / T,
T∆S total = T∆S system -∆H
Standard entropy of formation of a compound is defined as the
T∆S total = - (∆H - T∆S) = -∆G ⇒∆G = - T∆S total
entropy change that takes place when one mole of the compound
For equilibrium stage ∆S = 0 & ∆G= 0
in the standard state is formed from it’s elements also taken in the
For spontaneous process, ∆S is positive = ∆G is negative
...
It’s value in never zero
...

COUPLED REACTION
56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
In certain reactions ∆G is not negative and hence process in non
20
...
on entropy & define entropy
...
These non-spontaneous process may be converted
entropy of universe constant
...

having a very large negative free energy of reaction e
...

21
...

[2CO (g) + O2 (g)→2CO2(g) ∆G0= -514
...
What was requirement to introduce Gibb’s energy change
when enthalpy change & entropy was capable to explain the
coupled reaction,2Fe2O3 (s)+ 6 CO (g)→ 4 Fe(s) + 6CO2 (g)
0
feasibility of a process
...

∆G = - 56
...
Define Gibb’s free energy & what does decrease in free
* Discuss the feasibility of a reaction,
energy indicate
...
Derive Gibb's Helmholtz equation
...
(a) For the reaction both ΔH & ΔS are negative
...
4 – 4x 27
...

= - 549
...
Under what
conditions the reactions occurs spontaneously
...
What is effect of temperature on the spontaneity of the
∆ S surrounding = - ∆H / T = q /T
-1
-1
-1
-1
reaction
...

= 1648000 /298 JK mol = 5530 JK mol
26 How TΔS determines the feasibility of a process
...
Write physical significance of free energy, why it is so called,
= - 549
...
6 JK mol
reaction
...
What will be the sign of ΔG for melting of ice at (i) 267K (ii)
feasible (i
...
spontaneous)
274K
...
At a certain temp
...
Explain open, closed & isolated system
...
Find sign of Δs for reaction A →B & ΔG for
is not possible
...
From thermodynamic point of view, to which system animal & reverse reaction B→A
30
...

31
...

3
...

32
...
Explain the state variable & state function
...
Derive ΔG system = _ TΔS total
...
Explain how and under what conditions do you arrive at
temperature, viscosity, heat capacity, vapour pressure, surface
above relationship
...

34
...
P = useful work
...
Differentiate between the following:
35
...

(A) Isothermal and adiabatic processes
...

(B) Exothermic and endothermic reactions
...
Explain second law of thermodynamics for isolated and non(C) Reversible and irreversible processes
...

6
...
What are internal energy
37
...

change and enthalpy change? How are these related & under what
38
...

39
...
How it is useful to
7
...

how the value of ΔH change?
40
...

8
...
What are spontaneous and non-spontaneous processes
...

9
...
What is the value of entropy of perfectly crystalline solid at
changed
...

10
...

43
...
If q p is always greater than q v or not
...

H2O & solution becomes colder
...
Deduce relation between qp & q v
...
How are internal energy at constant pressure P, ΔG & ΔS are
q p =q v
interrelated to each other
...
Write mathematical representation of 1 law of
Q45
...
Also explain its limitations
...
Show with the help of diagram how Gibbs energy
14
...
Why
changes with the extent of a reaction
...
What is phase transition & how entropy can be calculated for
15
...

feasible process? How temp
...
Define efficiency of a process & Write formulae to find
16
...

centric cubic close packed structure
...
How is free energy change of an electrochemical cell related
17
...

process ΔS universe >0
...
Write relation between standard free energy change &
18
...

equilibrium constant
...
How enthalpy & entropy change explain feasibility of a process
...

gas from volume V1 to V2 , the work done by the system is equal
50
...

viii) If heat of formation of CCl4 is 316 kcal mol-1 , the bond
51
...
energy of C–Cl is -------52
...
Explain with example
...
Define the following: -- (a)Standard enthalpy of formation (b) xi) The efficiency of a carnot engine can be increased by ------Bond enthalpy (c) Hess’s law of constant (d) heat summation (e)
sink temperature when the source temperature is held constant
...

formation
...

54
...

xvi) if heat content of X is smaller than that of Y then the process
55
...
Give reasons for the following:
xvii) C (Diamond) → C (Graphite) + x KJ
...

xviii) As per reaction, N2(g) + 202(g) →2N 02(g) –66 KJ the value
(b) Enthalpy of neutralization of strong acid with strong
of ∆Hf of NO2 is --------base is always constant
...

xix) The heat of combustion of graphite and carbon monoxide
(c) The heat energy exchanged in chemical reactions is expressed respectively are –393
...
Thus, heat of
as ΔH and not as ΔE
...

xx) At 298 K, the bond energies of C–H, C–C, C = C and H–H
(e) Internal energy is a state function but work is not
...

mol-1
...

+ H2(g)→ H3C–CH3 (g) at 298 K will be -------(h) Enthalpy of fusion of NaCl is greater than solid O2?
xxi) For the reaction, C3H8 (g) + 5O2(g) →3CO2 (g) + 4H2O (l)
(i) Endothermic reaction become more feasible at increased
at constant temperature, ∆H – ∆U is -------temperature
...

vii) - RT ln (V2/V1) viii) 79 K cal
ix) Carbon, hydrogen
(k) Neither q nor w is a state function, but q+w is a state function
...
xi) Decreasing
...
xiii) Decrease
...

xiv) Positive
...

transition xviii) 5 + 33 kJ mol-1 xix) -110
...
0 Kj xxi) (n) Standard entropy of an elementary substance is not zero
3RT
whereas standard enthalpy of formation is taken as zero
...

SECTION (A) : Basic definition, Ist Law thermodynamics
(P) Entropy is a state function
...
If 500 calorie of heat energy is added to a system and the
(Q) Entropy of a solution is higher than that of pure liquid
system does 350 calorie of work on the surroundings
...

energy change of the system
...

Work done by the system, w = -350 cal
(T) Why does Entropy of solid increases on fusion
...

∆E = q + w = 500 + (-350) = 150 calorie
(v) Evaporation of water is endothermic but spontaneous
...

EX 2
...
Comment on
5 litre against a constant pressure of 3 atm
...

during expansion be used to heat 10 mole of water initially present
0
b) K for a reaction is 1or more if ΔG for it is less than zero
...

ANSWER:- Work done in expansion = P x ∆ V=6
Fill in the blanks the following questions
We have 1 litre atm = 101
...
3 J = 607
...

(iv) A reaction will proceed spontaneously if the reactant have ---- Therefore, P∆V = M x S x ∆T
607
...
184x ∆T ⇒ ∆T = 0
...

Tf = T1 + ∆T = 290
...
Calculate the work performed when 2 moles of hydrogen
system and w is the work done on the ----expand isothermally and reversibly at 25°C form 15 to 50 litres
...

Sol
...
303 n RT logV2/V1
56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
= -2
...

EX
...
0 g of aluminium from 35°C to 55°C
...

273 K
...

the heat capacity required to raise the temperature of 27g Al
Sol
...
303 nRT log P1/P2
through 1 K is 24 Joules
n = number of moles of hydrogen = 5 moles
...
066kJ
Thus W = -2
...

EX
...
If heat of combustion of ethylene is 1411 kj, the
therefore, at constant temperature, internal energy will not change volume of O2 (at NTP) that entered in the reaction is
i
...
,ΔE=0
...
5 Ml (B) 296
...
4 litre (D) 22
...

Sol
...
4/1411= 296
...
external pressure of the
EX: Assume that for a domestic hot water supply 150 kg of water
liquid is suddenly increased to 100 atm and the liquid gets
per day must be heated from 10°C to 65°C and gaseous fuel
compressed by 1 L against this pressure then find (i) work (ii) Δu
propane C3H8 is used for this purpose
...
∆H for combustion of propane is - 2050 kJ mol1
Δq=0 Δw = Δu ⇒100 = Δu
& specific heat of water is
...
184 x 10-3 kJ/g
...

= 150 x 103 x 4
...
For Ag,
(JK-1 mol-1) is given by 24 + 0
...
Calculate
34158 kJ heat is provided by = 34518/2050
∆H if 3 mol of silver are raised from 27° C to its melting point
=16
...

Volume of C3H8 at NTP = 16
...
4 litre =3
...
A thermally isolated vessel contains 100 g of water at 0°C
...

freezes and some evaporates at 0°C itself
...
006 (T22 – T12 ) J mol-1
the ice formed such that no water is left in the vessel
...
10 x 106 J/kg and latent of
EX
...
36 x 105 J/kg
...
Total mass of the water = M = 100 g
1 atm ? Assume that steam obeys perfect gas laws
...
0 x 105 J/Kg
molar enthalpy of vaporization is 9
...
36 x 105 J/Kg
change of internal energy in the above process ?
Suppose, the mass of the ice formed = m
(A) 1294
...
4 cals, 11074 cals
Then the mass of water evaporated = M — m
(C) 1029
...
6 cals
(D) 1129
...
Mole of H2O = 1
...
39 x 0
...
80
...
1 mole of ice at 0°C and 4
...
dv = 1 x [42
...
= - 42
...
325 J = water vapour at a constant temperature and pressure
...
80 x 101
...
2cal = 1024
...

∆E if the latent heat of fusion of ice is 80 cal/g and latent heat of
ΔH = ΔE + [PΔv]
...
39+ 1024
...
6 cal/s
...
There is 1 mol liquid (molar volume 100 ml) in an adiabatic
ice in comparison to that of water (vapour) is neglected
...
No
...

increased to 100 bar, and the volume decreased by 1 ml under
Pv = nRT ⇒ (4
...
82 × 273
constant pressure of 100 bar
...
[Given 1 bar = v = 3699 lit= [3700 lit]
105 N/m2]
[ JEE 2004,2]
latent heat of fusion = 80 cal/gram
(A) ∆E = 0 J, ∆H ≠ 0 J (B) ∆H = 0 J, ∆E = 10 J
Latent heat of vaporisation = 596 cal/gram
(C) ∆E = 20 J, ∆H = 890 J (D) ∆E = 0 J, ∆H = 10 J
∆H = 80 × 18 + 596 × 18 = [80 + 596] × 18 = 12168 cal
sol
...
6/760) [3700 –1] × (101
...
2)
qp=0, ∆H = qp=0
∆E = 12168 – 540
...
28 cal
...
Enthalpy of neutralization of HCl by NaOH is -57
...
1 kJ/mol
...
If water vapour is assumed to be a perfect gas, molar enthalpy dissociation of NH4OH
...
Given that
41 kJ mol-1
...
1 kJ/mole
water is vapourised at 1 bar pressure and 100°C
We may consider neutralization in two steps
...
3 x 10 x 373
(ii) Neutralization,
∆E = 37
...
Jmol-1
H+(aq) + OH-(aq)→H2O(ℓ) ∆H2 = - 57
...
1 kJ/mol + 57
...
0 kJ/mol
Enthalpy of neutralization = - 5706
...
4j
EX
...
What is the enthalpy of formation of OH- ions ?
EX: Compute the resonance energy of gaseous benzene from the
Sol
...

∆neutH = ∆fH(H2O, ℓ)-∆fH(OH-,aq)
B
...
3 kJ /mol , B
...
4 kJ/ mol
Hence ∆fH (OH-, aq) = ∆fH(H2O, ℓ) - ∆neut H
B
...
1 kJ/ mol, ∆H0sub [C, (graphite)] = 7184 kj/mol
-1
-1
= [-285 - (-55 )] kJ mol = -230 kJ mol
∆H0sub [H2(g)] = 82
...
9 kJ/mol
+
EX
...
1KJ
...
2 M
of ∆H0f of (benzene, g) with the given one
...
1 M KOH solution will be
To calculate ∆H0f (benzene, g), we add the following reactions
...
426KJ
(B) 13
...
2KJ (D) 55 KJ
6C(g) + 6H(g) → C6H6 (g)
SOL
...
2x2x400/ 1000 =0
...
B
...
EC=C + 6B
...
1/1000 = 0
...
4 KJ/ mol
Number of equivalents of acid and bases which neutralised each
Add 3 H2 (g) →6H(g) ∆H0 = 3 x 435
...
06
...
06 x 57
...
426 kj
The corresponding enthalpy change is
EX
...
35kJ mol-1 using
∆H0f = - (3B
...
E C=C+ 6B
...
4 + 3 x 455
...
Hence ∆H of H2C2O4⇌ C2O42- + 2H+
kJ/mol
(A) 5
...
88 KJ (C) - 13
...
9KJ
= [-(3 x 331
...
1 + 6 x 416
...
4 + 3 x 435
...
By the definition of heat of neutralisation, we have
KJ/ mol= 352
...
35 kj
The given ∆H0f (benzene, g) = 82
...
35 kj ---(i)
This means benzene becomes more stable by ( 352
...
9)
H++ OH- → H2O
∆H = -57
...
e
...
7kJ/mol
...

Subtracting equation (ii) from (i), we get
EX
...
95
(l) and benzene (l) at 25° C are -156 and + 49 KJ mol-1
2+
H2C2O4→ C2O4 + 2H ∆H= +7
...
standard enthalpy of hydrogenation of cyclohexene
EX
...
0 mL of 0
...
0 mL of 0
...
Use these data to estimate the
NaOH
...
0°C
...

enthalpy of neutralization per mole of HCl
...
Enthalpy of formation of 3 carbon-carbon double bonds
(A) -2
...
3×102 kJ/mole
= -156 - (+ 49) kJ = -205 kJ
...
4×101 kJ/mole (D) -6
...
Assuming density of solution is 1g/cc,
Given that,
and specipic heat is 4
...
2 ×3
Theoretical enthalpy of formation of 3 double bonds in benzene
millimoles of acid neutralized = 5
ring =3 × (-119)kJ = - 357 kJ
...
2 × 3 × (1000/5) = - 2
...

SECTION (G) ENTHALPY OF FORMATION
EX
...
What is the heat of formation of CS2, if heat of combustion of
(HA) in 1M solution with a strong base is -55
...
If the
C, S and CS2 are "x, y and z" kJ mol-1?
unionized acid is required 1
...
3 kJ/mol
...
C
EX
...
N2(g) + 3H2(g)→ 2NH3(g) ; ∆rHo = –92
...
57%
(C) 35
...
7 (B) -46
...
5 (D) - 52
...
(B) HA → H + A ; ∆rH = 1
...
B
∆Hneutralization = ∆Hionization + ∆rH (H + OH → H2O)
EX
...
95 = ∆Hionization- 57
...
Then enthalpy of formation of NH3 is
∆Hionization for 1 M HA = 1
...
B
= (1
...
4) × 100 = 96
...
43 = 3
...

Fe2O3(s) + 3C(graphite)→ 2Fe(s) + 3CO(g), ∆H0= 492
...
A constant pressure calorimeter consists of an insulated
FeO(s) + C(graphite)→ CO2(g), ∆H0= 155
...
75
C(graphite) + O2(g)→ CO2(g), ∆H0 = -282
...
The beaker contains 100 mL of 1 M HCl of 22
...

which 100 mL 1 M NaOH at 23
...
The final
Sol: Fe(s) + CO(g)→FeO(s) + C(graphite) ∆H0= 155
...
3°C
...
5 KJ/mol
per mole for this neutralization reaction ? Assume that the heat
CO2(g) →CO(g) + 1/2 O2(g), ∆H0 = -282
...

On adding Fe(s) + 1/2 O2(g)→FeO(s)
∆H = -393
...
Initial average temperature of the acid and base = (22
...

23
...
0°C
EX
...
3 - 23
...
3°C
½ H2(g) + ½ Cl2(g) →HCl(g); ∆Hf0 = -92
...
75 + 200 x 4
...
3
HCl (g) + nH2O → H+(aq) + Cl- (aq) ; ∆Hf0 = - 74
...
8) x 6
...
54 J
56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
SECTION (J) BOND DISSOCIATION ENERGY
∆Hf0H+Aq = 0
...
The heat of combustion of acetylene is 312 kcal
...
Given,1/2 H2 (g) + aq
...
(i)
formation of CO2 and H2O are -94
...
38 kcal
1/2H2(g) + 1/2C12(g) → HCl(g); ∆HO = –92
...
(ii)
respectively, calculate C ≡ C bond energy
...
8 kj (iii)
atomisation of C and H are 150
...
5 kcal respectively and C
Add
...
(ii) and (iii)
- H bond energy is 93
...

½ H2(g)+ 1/2 C12 (g) + nH20 →H + (aq) + Cl- (aq); ∆HO = –167
...
(iv)
C + O2 →CO2 ;∆H = -94
...
(i) from (iv)
H2 + ½ O2 →H2O ;∆H = -68
...
+ e →Cl-(aq); ∆HO = –167
...
2 kJ
2C+ H2 + 5/2 O2 → CO2 + H2O; ∆H = -257
...
(iv)
SECTION (H) ENTHALPY OF REACTION
0
EX: The ∆Hf for CO2(g), CO(g), and H2O (g) are - 393
...
5 (iv) - (i), 2C + H2→ C2H2; ∆H = 54
...
8 kJ mol-1 respectively
...
5 KJ
...
5 kJ ----------- (ii)
54
...
64]
H2 + ( 1 /2)O2→ H2O; ∆H= - 241
...
86kcal
EX
...
2
F(g) are : -1100, 275 and 80kJmol-1 respectively
...
For the given reaction
average S - F bond energy in SF6 is
H2(g)+F2(g) → 2HF(g) ∆H° = –124 kcal
(A) 52
...
8 kcal then the value of ∆H for
Solution: S(g) + 6F(g) → SF6(g) ∆Hf = -1100kJmo1-1
H(g) + F(g) → HF(g) is
S(s) → S(g)
∆H = +275 KJmol-1
(A) 142 kcal (B) – 132
...
(B) H2(g)+F2(g) → 2H F(g); ∆H° = -124 kcal
Therefore heat of formation = Bond energy of reactants - Bond
∆H° =ΣB
...
(reactants) - ΣB
...
of (products)
energy of product ⟹ -1000 = [275 + 6 x 80] - [6 x S - F]
or -124 =∆HH-H + ∆HF-F - 2∆HH-F =104 + 37
...
0 kJ
2∆H H-F =104+37
...
8 kcal
mol-1 and 941
...
8/2=132
...
What is enthalpy of atomization of NH3 (g) ? What is
∆H° for the given reaction= - 132
...
Calculate enthalpy change of the following reaction:
Solution: N2(g) + 3H2(g) → 2NH3(g); ∆H = -2 x 46 KJ/mol
H2C = CH2(g) + H2(g) → H3C - CH3(g)
∆H = Σ(B
...
E)p = (941
...
E
...

x
=
390
...
∆Hreaction = Bond energy data for the formation of bond +
NH3→ N + 3(H)
Bond energy data for the dissociation of bond
Heat of atomization 3 x 390
...
9 KJ/mol
= -[1(C - C) + 6(C - H)] + [1 (C = C) + 4(C - H) + 1 (H - H)]
Problem 9: Using the data ( all values in k cal mol-1 at 25° C)
= - 347-2 x 414+615+435 = -125 kJ
given below, calculate bond energy of C- C& C- H bonds
...
0 kcal
as ∆H at 298
...
325 kPa for the process below:
C(s) +O2→CO2; ∆H = -94
...
2 Kj for C2H6 and - 87
...
2 kJ (a)
∆H2 =- [1(C-C) +6(C-H)]+[2CS→g + 3(H-H)] ------ (ii)
C3H8 +2H2→ 3CH4 ∆H= - 87
...
0 kcal -----(v)
CH4(g) + C3H8(g)→ 2C2H6(g) ∆H = ? (c)
H
+1/2O
→H
O(1)
∆H
=
-68
...
4 + 130
...
At 300 K, the standard enthalpies of formation of
C3H8+5O2→3CO2+4H2O; ∆H=-530kcal------(viii)
C6H5COOH(s), CO2(g) and H20(l) are -408, -393 and -286 kJ/ mol By inspection method : 2 x (v) + 3 x (vi)- (vii) gives
respectively
...

(ix)
(i) Constant pressure (ii) constant volume
and 3 x (v) + 4x (vi) - (viii) gives
SOL
...
Given that
C6H5COOH(s) + 15 02(g)→7CO2(g) + 3H20(l)
2H2(g) + O2(g) →2H2O(g), ∆H = -115
...
5) x 8
...
2471 = -3199
...
E
...
E
...

For the reaction, 2H - H(g) + O = O(g) →2H - O- H(g) ∆H = EX
...

115
...
E
...
E
...
5 kJ/mol; ∆rCP = 0
Since one H2O molecule contains two O- H bonds
SoN2(g) = 192 JK-1 mol-1; SoH2(g) = 130
...
4 = (2 x 104) + 119 - 4 (O - H) bond energy
SoCl2(g) = 233 JK-1 mol-1; SoNH4Cl(s) = 99
...
4
All given data a 300 K
...
, O- H bond energy = (2x104)+119+115
...
6 kcal mol-1
(A) -198
...
7 kJ/mol (C) -202
...
The heats of combustion of yellow phosphorus and red
= SoNH4Cl(s)–[(1/2)SoN2 + 2SoH2 + (1/2)SoCl2] = -374 JK-1 mol-1
phosphorous are - 9
...
78 kJ respectively, then heat of
∆fCP = 0
transition of yellow phosphorous to red phosphorous is
∴∆fSo310 = ∆fSo300 = -374 JK-1 mol-1
Answer: (i) P4(yellow) + 502(g) → P4O1O + 9
...
5
(ii) P4(red)+5O2(g) → P4O10 + 8
...
5- [310(-374)]/1000
substracting, P4(yellow) - P4(red) =1
...
56 kJ/mol
so, heat of transition of yellow of red phosphorous is -1
...

EX
...

hydrogen, cyclohexene (C6H10) and cyclohexane (C6H12) are -241, ∆rCP is 20 JK-1 mol-1 then ∆rHo at 400 K is :
- 3800 and - 3920 kJ/mole respectively
...
98 kJ/mol (C) 28 kJ/mol (D) None of these
hydrogenation of cyclohexane
...

C6H12 + 9O2→ 6CO2 + 6H2O, ∆H4 = - 3920 KJ/mole ------- (4)
SECTION (O) EQUILIBRIUM CONSTANT
C6H10 + H2→ C6H12
Ex
...
754 × 10-5
...
633 × 10-5 What are ∆H° and ∆S° for the ionisation of
= (-241 - 3800) - (-3920) = -4041 + 3920= -121 kJ/mole
CH3COOH?
SECTION (M) ENTROPY /ENTHALPY/ MP/BP
Sol
...
303RT log K = -2
...
314 × 298 ×1og
EX
...
754 ×10-5) = 27194 J
...
303 × 8
...
633 × 10-5) = 29605 J
...
at 300 K
...
O2 (g) and H2O( l) ∆GO = ∆Ho – T∆So 27194 = ∆HO – 298 ∆So
are 126
...
20 and 68
...
4 JK–1 mol–1 (B) 318
...
55 kJ/mol & ∆S° = – 96
...
K
(C) 31
...
2H 2 (g) + O2 (g)→ 2H2O(l)
Ex: Show that the reaction, CO(g)+(1/2)O2(g)→ CO2 (g)
ΔS = [2∑H2O-(2 ∑N2 +∑O2 )] = [2 × 68 – (2 × 126
...
20)]
at 300 K is spontaneous and exothermic, when the standard
= [138 – (253
...
20)] = – 318
...
094 kJ mol-1 K-1
...
What is the entropy change for the conversion of one gram of energies of formation of CO and CO are - 394
...
2 KJ
2
ice to water at 167 K and one atmospheric pressure?
mol-1 respectively
...
025 kJ mol
Sol: The given reaction is, CO(g)+(1/2)O2(g)→ CO2 (g)
Sol: ∆Hfusion=6
...
72 Jg-1;
∆G0 (for reaction) = GCO2 -GCO –[1/2]GO2 = -394
...
2) -0 = ∆Sfusion= 334
...
2 kJmol-1
EX
...

∆G0 =∆H0 - T∆S0 ⟹ - 257
...
094)
When 117 g benzene vaporizes at it's normal boiling point, the
Or ∆H0 = -288
...
5 JK-1 (C) 85 x 1
...

42
...
5
SECTION (Q) TEMPERATURE FOR FEASIBILITY
∆Ssystem = 1
...
5 × 85 J/K
Ex
...
4 KJ
SECTION (N) GIBB’S FREE ENERGY
and ∆ S = -198
...
Calculate the maximum temperature at
EX
...

O2(g)⇌ 2O3(g) at 298 K
...
47 x 10-29 Sol: ∆G = ∆H - T∆S, For a reaction to be spontaneous, ∆G = -ve
∆H- T∆S = -ve or ∆H > T∆S
or
∆H/∆S> T
(A) 152 KJ mol-1 (B) 163 KJ mol-1
or - 95
...
3 >T or
481
...
303RT log kP
would be spontaneous
...

= 2
...
3l4x298xlog 2
...
228Kj/mol
An increase in temperature above 481 k will develop non-spontaneity
EX
...

dissolved in water at 298 K
...
Given the following data :
778 kj mol
-1
Substance
∆Ho (kJ/mol) So(J/mol K) ∆Go (kJ/mol)
(b) Hydration energy of NaCl = -775 kj mol
FeO(s)
-266
...
49
-245
...
74
0
Sol: ∆Hdissolution= ∆H (ionisation) + ∆H(hydration)
(A) 110
...
6 kcal (C) 105 kcal / mol (D) -105 kcal

56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
13
...
The heat released was 743
kJ mol-1
...
5 W mol-1
...
75 kJ mol-1
(C) -743
...
25 kJ mol-1
14
...
43 (B) + 3
...
72 (D) +7
...
The reaction of cyanamide, NH2CN (s), with dioxygen was
carried out in a bomb calorimeter, and ∆U was found to be -742
...
Calculate the enthalpy change for the reaction
1
...
NH2CN (g) + 3/2 O2 (g) →N2(g) + CO 2(g) + H20(l)
work is done by the system
...
5 (B) -741
...
The heat of formation of methane C(s) + 2H2(g)→ CH4 (g) at
SECTION (B) : Work calculation in different type of processes
constant pressure is 18500 cal at 25°C
...
If a gas at a pressure of 10 atm at 300 k expands against a
constant volume would be
constant external pressure of 2 atm from a vol
...
(∆H–∆U) for the formation of carbon monoxide (CO) from its
3
...
314 J K-1 mol-1) [ AIEEE 2006]
of an ideal gas from an initial pressure 1 bar to a final pressure 0
...
78 J mol-1 (B) –2477
...

(C) 2477
...
78 J mol-1
4
...
It is provided 58
...
1 mole of napthalene (C10H8) was burnt in oxygen gas at 25°C
that its volume becomes 2
...
Calculate change in its
at constant volume
...
8 KJ
...

Calculate the heat of reaction at constant pressure
...
314 JK-1
5
...

at 300 K and occupying a volume of 5 dm3 isothermally until the
19
...
The valve on a cylinder containing initially 10 liters of an ideal 2CO + O2→2CO2
2moles
1
mole
gas at 25 atm and 25°C is opened to the atmosphere, where the
is carried and in one litre container, if the pressure in the container
pressure is 760 torr and the temperature is 25°C
...

the process is isothermal, how much work (in L
...
[1 L atm = 0
...
An ideal gas expands in volume from 1 X 10 m to 1 X 10 m at 20
...
0
atm, 3
...
The work is :
energy, ∆E = 30
...
The change in enthalpy, ∆H, of the
[AIEEE 2004]
process in L atm is
(A) - 900 J(B) - 900 kJ(C) 270 kJ (D) + 900 kJ
(A) 40
...
3
8
...
0(D) not defined, because pressure is not constant
from 1 litre to 10 litre
...

21
...
7 (C) -11
...
6 kJ(B) - 1
...
15 kJ
(D) + 11
...
2 mole of ideal gas expands isothermally and reversibly from 1
SECTION (D) MOLAR HEAT CAPACITY
L to 10 L at 300 K
...
Work done in expansion of an ideal gas from 4 litre to 6 litre
(A) 4
...
47 KJ
(C)- 11
...
The molar enthalpy of vaporization of benzene at its boiling point against a constant external pressure of 2
...
If specific heat of water is 4
...
84 kJmol-1 What is the molar internal energy change?
what is the final temperature of water ?
For how long would a 12 V source need to supply a 0
...
8 g of the sample at its boiling point ?
23
...
36g of glucose was burned in a bomb calorimeter
11
...
Calculate
expanded reversibly and isothermally to a final volume of 5L
the standard molar enthalpy of combustion
...
Heat of neutralization between HCl and NaOH is –13
...
cal
...

If heat of neutralization between CH3COOH and NaOH is –11
...
cal
...

using an external constant pressure of 5 atm
...
The heats of neutralisation of four acids A,B,C,D are - 13
...
4, - 11
...
4 kcal respectively when they are neutralised
(c) In the above two processes what is the net heat gained by
by a common base
...
A sample of 4
...
The amount of heat released when 100 ml of N/10 H2SO4 is
and then undergoes adiabatic expansion against a constant
pressure of 600 Torr until the volume has increased by a factor of mixed with 150 ml of N/10 NaOH solution is
3
...
Calculate q, w, ∆T, ∆U and ∆H
...
Enthalpy of neutralisation of hydrocyanic acid by potassium
hydroxide is – 12
...
Calculate the heat of ionisation of
is not necessarily 600 Torr)
...

SECTION (C) : RELATION BETWEEN ∆H & ∆U
Fe(s)
0
27
...
5
197
...
15
Determine at what temperature following reaction is spontaneous ?
FeO(s) + C (Graphite) → Fe(s) + CO(g)
(A) 298 K
(B) 668 K
(C) 966 K
(D) ∆Go is +ve, hence the reaction will never be spontaneous
Sol: FeO (s) + C (graphite) → Fe (s) + CO (g)
(1) ∆H = 0 + [-110
...
3) = 110
...
3 = + 155
...
28 + 197
...
49 - 5
...
65 J/mole
(3) ∆G = ∆H - T∆S = 0 = 155
...
65 × 10-3
T × 161
...
8 ⇒ T = 963
...
(C)
PRACTICE PROBLEMS SECTION I LEVEL I
SECTION (A) : Basic definition, Ist Law thermodynamics

56C/2/1 ASHOK MOHALLA, BHUTONWALI GALI NO – 1, NANGLOI, DELHI-41 # 9891351471#

CHEMICAL THERMODYNAMICS NOTES BY PAWAN SIR
28
...
7 kcals?
(C) 172 kJ/mol (D) - 172 kJ/mol
(1) HCI, NH4OH (2) HNO3, KOH
SECTION (G) ENTHALPY OF FORMATION
(3) NaOH, CH3COOH (4) H2SO4 , NH4OH
41
...
Heat of neutralization of a strong acid by a strong base is
equations
...
0 kcal
(A) H+ + OH-→ H2O (B) H2O + H+ → H3O+
H2(g) + ½ O2(g) → H2O(1); ∆H = -68
...
0 kcal
30
...
Calculate the standard enthalpy of formation of CH3OH(l)
with two mole of NaOH is
from the following data :
(A) 13
...
7 kcal
CH3OH(l) + 3/2 O2(g)→ CO2(g) + 2H2O(l); ∆rHo = -726 kJmol-1
(C) More than 13
...
A solution of 200 mL of 1MKOH is added to 200 mL of 1M
H2 (g) + 1/2O2 (g)→ H2O(l); ∆HOf = -286 kj mol-1
HCl and the mixture is well shaken
...
The experiment is repeated by using 100 mL of each
(C) +57 kj mol-1(D) -55 kJ mol-1
solution and increase in temperature T2 is again noted
...
If
...
2 Kj -------- (i)
the following is correct?
SO2+ ½ O2 → SO3 ∆H = -98
...
2 Kj -------(iii)
(3) T1 is twice as large as T2 (4) T1 is four times as large as T2
...
3 KJ ------ (iv)
32
...
4 KJ (B)+320
...
3 KJ (D) – 433
...
7 kcal
(2) 57 KJ (3) 5
...
H2(g) + (1/2)O2(g) = H2O(l); ∆H298K = -68
...
Enthalpy neutralization of H3PO3 is –106
...
Heat of vaporisation of water at 1 atm and 25°C is 10
...

If enthalpy of neutralization of HCl by NaOH is –55
...
∆H
The standard heat of formation (in k cal) of I mole of water
ionization of H3PO3 into its ions is x kJ/mol
...

vapour at 25°C is
34
...
84 (B) 78
...
80 (D) -57
...
5 M H2SO4 solutions
...
Calculate the heat of formation of methane, given that
volume of 100 mL produce the highest rise in temperature
Heat of formation of water=-286 kJ mol-1
(A) 67 : 33
(B) 33 : 67 (C) 40 : 60 (D) 50 : 50
Heat of combustion of methane=-890 kJ mol-1
35
...
5 kJ mol-1
(i) CHCℓ2- COOH by NaOH is 12830 cals ;
SECTION (H) ENTHALPY OF REACTION
(ii) HCℓ by NaOH is 13680 calories,
46: The standard enthalpies of formation at 298 K for CCl4(g),
(iii) NH4OH by HCℓ is 12270 calories
...
7, - 241
...
7 and -92
...
Calculate ∆Ho298K for the reaction
NH4OH ? Calculate also the heats of ionization of dichloro acetic
CCl4(g) + 2H2O(g) → CO2(g) + 4HCl(g)
acid & NH4OH
...
150 m L of 0
...
35°C were mixed with
Cgraphite+ 1/2O2→ CO; ∆H = -110
...
5 N NaOH solution at the same temperature in a
CO + 1/2O2→ CO2
∆H = - 283
...
The final temperature was raised to 28
...

∆H for the reaction, Cgraphite+ O2→ CO2 is
Calculate the heat of neutralisation of HCl and NaOH
...
7 KJ (B) +393
...
7 KJ (D)+172
...
184
48:Calculate the standard internal energy change for the reaction
Jg-1 K-1
...
Given standard
37
...
1 kJmol-1
...
J
of the acid is 1
...
Determine enthalpy change of reaction, C3H8(g) + H2(g) →
strong acid with a strong base is - 57
...
At 250C, using heat of combustion value under
ionization of the weak acid in molar solution (assume the acid to
standard conditions
...
8 -890
-1560
-393
...
Calculate the heat of neutralisation of (aq
...
8
KJ/mol
3 8
aqueous hydrochloric acid from the following data:
50
...
4 kcal
(aq) → H2O (l)
HCl(g) + aq=HCl(aq); ∆H=-17
...
2 kcal
51
...
The enthalpies of formation
39
...
5, -285
...
42
-1
-1
∆HC=O = 339kJ mol , ∆HO=O = 498kJ mol
kJ mol-1 respectively
...
0 kJ mol-1
...
For the reaction, C7H8(1) + 9O2 (g) → 7CO2 (g) + 4H2O(l),
the calculated heat of reaction is 232 kJ/mol and observed heat of
reaction is 50
...
2 kJ/mol (B) + 182
...
1, – 94
...
3 kcal respectively
...


53
...
What is the free energy change (∆G) when 1
...
Diborane is a potential rocket fuel which undergoes
1 atm pressure ?
combustion according to the reaction,
(A) 80 cal
(B) 540 cal
(C) 620 cal
(D) Zero
B2H6 + 3O2(g)→B2O3 (s) + 3H2O(g)
65
...
7 J mol-1 for the transition
...

surroundings to be an ice-water both at 0°C :
ii) H2(g) + (1/2) O2(g)→H2O(l); ∆H = -286 kJ mol-1
...
09 JK-1 (6)1
...
38 JK-1(D) None of these
-1
iii) H2O(l) →H2O(g); ∆H = 44 kJ mol
...
Calculate entropy change involved in conversion of 1 mole
iv) 2B(s) + 3H2(g)→B2H6(g); ∆H = 36 kJ mol-1
...
Latent heat of vaporization
55
...
257 KJ/gm
...
At 373K, the entropy change for the transition of liquid water
498, & 464 kj/mol respectively
...
Find ∆H for this process
...
∆Hvap for water is 40
...
The enthalpy of combustion of H2(g) at 298 K to give H2O(g)
The temperautre at which water is in equilibrium with water
is - 249 kJ mol-1 and bond enthalpies of H- H and O=O are 433 kJ vapours is
mol-1 and 492 kJ mol-1, respectively
...
67 0C (B) 260
...
69 K (D) 460 K
(A) 464 kJ mol-1 (B) - 464 kJ mol-1
69
...
Find boiling temperature
(C) 232 kJ mol-1 (D) -232 kJ mol-1
at 1 atm
...
The bond energy enthalpies of H - H and C1- Cl are 430 and
(A) 400 K (B) 300 K (C) 150 K(D) 425 K
242 kJ mol-1
...
Find ∆So for a reaction for which ∆Ho = 28
...
8×10-7 at 298K
...
Find ∆Go and ∆H° for that the reaction CO(g) + ½ O2 (g) →
58
...
The
change is – 0
...
The standard Gibbs free energies
C - C and C - H bond energies are respectively
of formation for CO2 and CO are - 394
...
2 kJ mol-1,
-1
-1
(A) 80 and 60 KJ mol (B) 80 and 90 KJ mol
respectively
...
2 kJ/mol, ∆H° = 285
...
4 kJ/mol, ∆H° = -570
...
The heat of transition for CDiamond → cAmorphus from the
(C) ∆Go = +514
...
8 kJ/mol
following is
(D) ∆Go = - 257
...
4 kJ/mol
Cdiamond + O2 (g) → CO2 (g)
∆H = -94
...
For the reaction, 2A(g) + B(g) → 2D(g)
Camorphous + O2 (g) → CO2 (g)
∆H = -97
...
5kj and ∆S0 = -44
...
3 kcal / mol
Calculate ∆GO for the reaction at 300 K
...
3 kJ mol (D) – 3
...
164 kJ (B)0
...
017 kj (D) 0
...
If ΔG = - 177 K cal for (1) 2 Fe(s) + 3/2O2 (g) → Fe2O3(S)
60
...
find the standard
What is the Gibbs free energy of formation of Fe3O4 ?
enthalpy change when acetylene is hydrogenated to ethane
...
6
kcal/ mol (B) - 242
...
6 kcal/mol
SECTION (M) ENTROPY /ENTHALPY/ MP/BP
74
...
What will be the
61
...
314 JK-1 mol-1, T = 300 K)
Fe2O3 (s) + 3H2(g)→2Fe(s)+3H2O(l)
(A) -1
...
744 kJ/mol
Given : S°m (Fe2O3, S) = 87
...
3, S°m (H2, g) =
(C) +1
...
27 kJ /mol
130
...
9 JK-1 mol-1
...
For the auto-ionization of water at 25°C, H2O (l)⇌H+ (aq) +
(A) –212
...
2 JK-1 mol-1 (C) –120
...
What is ΔG0 for the process ?
mol-1 (D) None of these
(A) ≈8 x 104 J (B) ≈ 3
...
Calculate the entropy change (J/mol K) of the given reaction
...
For a reaction A (g) ⇌ B(g) at equilibrium
...

substance
...
2] + 3O2 (g) [205
...
5] + 2SO2 (g)
(A) RT ln 4(B) - RT ln 4(C) RT log 4
(D) - RT log 4
[248
...
Calculate ∆G when 1 mole of NaCl is dissolved in water at
(A) -113
...
3
(C) +72
...
2
298 K
...
8 KJ/mol, hydration energy
63
...
1KJ/mol, ∆S=0
...
(Guide- ∆H =
out by the following shown path:
hydration energy – lattice energy)
[ JEE 2006]
∆S(A → C) = 50 ; ∆S(C→ D) = 30; ∆S(B→ D) = 20
The entropy change for the process A →B is
(A) 100 (B) - 60 (C) -100 (D) + 60

78
...
28V
...

SECTION (O) EQUILIBRIUM CONSTANT

79
...
W = –2
...
Calculate the value of ∆G for the reaction
...
303 x 1 x 8
...
0 /0
...
2 J
At constant temperature, for expansion ∆T = 0 ⇒ ∆E= 0
= 50 kJ / mol, ∆Gf0 (N2O4) = 100 kJ / mol and T = 298 K
...
2 J
to attain equilibrium
4
...
5 – 2
...
303 RT = 5
...
]
[ JEE 2004,2]
-1
= - 0
...
6
...
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK
-1
Work is carried out at constant P
...
For the reaction,
q=∆U–W ⇒ 58
...
63 ⇒ ∆U = 8 joule
½ X2 + 3/2 Y2→ XY3∆H = - 30 kJ, to be at equilibrium the
5
...
w = –2
...

temperature will be
[AIEEE AIEEE 2008]
6
...
atm, 7
...
(A), 9
...
27
...
, V=W/Q=W/It
81
...
(a) + 815
...

12
...
2kJ, q = 0, ∆T = -38K, ∆U = -3
...
5kJ
o
5
3CH ≡ CH(g) ⇌ C6H6(g) at 298K, ΔG f of C2H2 (g) = 2
...
A, 14
...
43kJ
o
5
J/mol, ΔG f of C6H6 (g) = 1
...
314J/K/mol
15
...
A, 17
...
5143
...
Calculate ΔGo for the reaction at 25oC
...
563, 20
...
3KJ/mol, ΔGo 21
...
299 K, 23
...
2 MJmol-1, 24
...
Sol: Q=13
...
7=2
HF(g)= 0
...
Also calculate equilibrium constant at 25oC 25: (B) Lower is heat of neutralisation , more is dissociation energy,
for the reaction
...
Find ∆G, ∆H, ∆S & Kp at 300K for the reaction,
N2(g)+3H2(g)→2NH3(g) ∆Ho for NH3 is 46
...
6, 191
...
5 JK1 mol-1 respectively
...
What is equilibrium constant K for the following at 400K,
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
...
2 KJ/mol, So = 111JK-1mol-1
85
...
46V (T=298K,F =
96500C/mol, R=8
...
A reaction has a value of ∆S = -200 KJ/mol-1 at 300 K
...
it is not
...

87
...
5Kcal,
∆S= -10
...
Find ∆G & explain its
feasibility
...
Find ∆G & explain spontaneity if 1mol N2O4 (g) forms 2
moles of N2O (g) at 298K
...
8 JK-1 mol-1 and
∆Ho for NO2(g) & N2O4(g) be 33
...
2 KJ mol-1 respectively
...
For the reaction Ag2O → 2Ag(S) + ½ O2(g), ΔH =30
...
066KJ/mol/K
...
Also predict the direction of reaction at (i) this
temperature (ii) below this temperature(iii) Above this temp
...
Find out whether it is possible to reduce MgO using C at
298K
...
it becomes spontaneous
...
18 KJ/mol
& ∆rSo = 197
...


26
...
eq of H2SO4 = 100x 1/10 = 10 meq
no of gm
...
3 = 0
...
- 45
...
2, 29
...
(C) , 31
...
4, 33
...
D
35
...
, 850 cal
...

36
...
237 Kj, 37
...
-12
...
B,
40
...
4 - 232
...
2 kJ mol-1
41
...
4 kcal, 42
...
(C), 44
...
- 75
...
4 kJ mol-1 , 47: (A), 48: 312
...
(55
...
- 2091
...
0 Kcal, 53
...
- 2035 kJ mol-1
56
...
(D), 58
...
(B) , 60
...
(B), 62
...
D, 64
...
C
68
...
D, 72
...
D
74
...
303RT log k, 75
...
A
79
...
B

SECTION (R) COUPLED REACTION

91
...
On the basis of ∆Go at 1073 k, show that roasting of ZnS to
form Zinc oxide is spontaneous – S2(s) +2O2 (g)→2SO2(g),
∆Go = -544 KJ/mol ; 2Zn(s)+O2(g)→ 2ZnO(s) ∆Go = -480 KJ/mol
; 2Zn(s)+S2(s)→2ZnS(s) ∆Go = - 293 KJ/mol
...
Gibb's free energy of formation of Al2O3, CO & CO2 are 11250, -250 & -380 KJ mol-1
...

ANSWERS (PRACTICE PROBLEMS) SECTION I LEVEL I
1
...
Sol Process is irreversible
Work done in expansion = -P x ∆ V=-2X 10X 101

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