System: Any portion of the universe which is under study is called a system.
Surroundings: Rest of the universe other than the system is called surroundings. The system of surroundings is separated by a seal as an imaginary boundary.
Types of the system: There are three types of systems:
1. Open system: A system which can exchange energy as well as matter with the surroundings is called an open system.
e.g.: Hot water is taken in on the open beaker.
2. Closed system: A system which can exchange energy but not matter with the surroundings is called a closed system.
e. g: Hot water taken in a closed beaker.
3. Isolated system: A system which can exchange neither energy nor matter with the surroundings is called an isolated system.
e.g.: Boiling water is taken in “thermass flask” whose mouth is covered with a plate.
The macroscopic system of macroscopic properties
A system which contains matter in bulk i.e. large quantity of atoms, ions or molecular is called a macroscopic system.
The properties associated with the macroscopic system are called macroscopic properties.
e.g.: Pressure, volume, temp, density, suf. Index, surface tension, viscosity etc.
State of the system of state variables
State of the system means the condition of the system when the macroscopic properties have definite values. Any change in the macroscopic property changes the state of the system. Hence these properties are also called state variables.
e.g.: CO2 gas at 298k of 1atm. Pressure is one state of the system while CO2 gas at 400k & 1atm. Pressure is another state of the system.
If a physical quantity is said to be a state function, if it depends upon the state of existence of the system & is independent of the manner how that state is achieved.
e.g.: Internal energy, enthalpy, free energy, volume etc.
Alternatively, a physical quantity is said to be a state function if it changes in value depending upon the initial or final state & not upon the path followed.
These are the properties of the system which depend upon the quantity of matter open in the system.
e.g.: Mass, volume, energy, enthalpy of free energy etc.
These are the properties of the system which don’t depend upon the amount/quantity of matter but depend upon the nature of matter.
e.g.: Surface tension, viscosity & fusing point, boiling point temp., pressure, self-index, pm value, EMF of cell etc.
Force is an extensive property, but pressure is an intensive property because the pressure is force per unit also.
Extensive properties are additive property, but intensive properties are not.
The ratio of the two extensive properties becomes an intensive property.
e.g.: Mass of the volume is extensive property but mass per unit volume called or volume per unit mass called specific volume is intensive property.
Extensive property becomes an intensive property by specifying a unit amount of the substance. In fact, any mole property becomes intensive property as it is in the extensive property per mole of the substance.
A system is said to be in the equation where macroscopic properties like temp, pressure, density etc do not change with the passage of time.
When a system changes from one state to the other state, a thermodynamic process is said to have occurred. The various thermodynamic processes are:
Isothermal process: A process in which temp remains without
The isothermal process can be attained in an open vessel as a closed vessel having conducting walls.
Adiabatic process: A process in which no heat is allowed to enter or leave the system.
The process can be attained in an isolated system or adiabatic system.
Isobaric process: A process in which pressure remains constant. i. e
Isochoric process: It is a process in which volume remains constant. i.e.
Reversible & Irreversible process:
Reversible process: A process which is carried out infinitesimally slowly is that all the changes taking place is the direct process and be reversed of the system remains almost in a state of the equation at every stage of the reaction process.
In these processes, the driving & the opposing force differs only slightly.
Irreversible process: It is a process which is carried out rapidly, driving and opposing forces differ widely & equilibrium is attained only after the completion of the reaction.
Important Thermodynamic Functions (E, H, S, G)
The concept of internal energy & internal energy change
The energy possessed by a substance by virtue of its existence is called its internal energy or intrinsic energy. It is represented by U or E. it depends upon the nature of substance, pressure, volume composition.
Increase of molecules, internal energy is a combination of many factors such as: translational K.E [Et]], vibration energy [Ev], rotational energy [Er], chemcal bond energy [Ec], potential energy [Ep], electronic energy [Ee], nuclear energy [En] i.e.
Important characteristics of internal energy are:
Internal energy is a state function i.e. it depends upon the state of existence of sub & is independent of the moment how that state is achieved.
Internal energy is an extensive property. It depends upon the quantity /amount of current states.
The absolute value of internal energy can’t be calculated because it is difficult to measure the consequent energies accurately.
However internal energy changes (all can be measured accurately. It is equal to the difference of internal energies of products of reactants.
∆U = UP + UR
UP = internal energy of the product
UR = internal energy of the reactant
Modes of exchange of energy:
Two important mode of exchange of energy is heat & volume.
Heat: This mode of exchange of energy occurs only if the system & surroundings are of different temperatures.
If the system is at a higher temp the surroundings, then heat is transferred from system to surroundings though conducting wall i.e. heat is released by the system.
If the system is at a lower temp than surroundings, then heat is transferred from surroundings to the system. i. e heat is absorbed by the system.
According to sign convention
[ q is +ve if heat is absorbed (←)
q is -ve if heat is evolved ( →)]
The two important forms of heat are electrical work & mechanical work (pressure, volume, work)
Electrical work: this type of work is significant in an electro-chemical cell. The electrical energy produced, or electrical work is the product of the E.M.E of the cell & the quantity of electricity flowing through the cell.
WElectrical = EMF x Quantity of electricity flowing
WElectrical = Fcell x xF
X – mole
F – Faraday
Mechanical work: This type of work is significant for system ifses undergoing a change in volume under internal pressure.
Let us consider, 1 mole of gas in a cylinder fitted with a frictionless piston with an area of cross section “a”
Let the pressure on piston = P (Which is slightly less than the external pressure, so that the gas can expand)
Force acting on the piston = pressure x area = P × a
Work was done by the gas = force x distance = P × a × dl
(dl is the distance moved by the piston)
W = P × dv
If the gas expands from volume V1 to V2
W = ∫_V1^V2▒Pdv = P (V2-V1)
W = Pext (V2-V1)
However, if external pressure is more than the pressure of the gas, the work is done on the gas & the gas will contract.
Sign convention: if work is done on the system W is “+”ve.
If work is done by the system (expansion) W is “-“ve.
∴In above case:
Wexpansion = – Pexternal (V2-V1) ←
However, this solution can be applied to constriction work also as V2<V1 as W comes out to be “+”ve.
Hence the general expansion for an irreversible expansion against a constant external pressure is:
W = – Pexternal (V2-V1) ← irreversible
Work was done in the isothermal reversible expansion of an ideal gas:
The small amount of W.D “dw” when the gas expands through a small volume “dv” is given by the expression.
dw = – Pext dv
The pressure during reversible expansion
Pext = Pinternal – dP
=>dw = – (Pint – dP)dv
dw = – Pint x dv ————-(1) (dP x dv small, so neglected)
Now, for an ideal gas
PV = nRT
=>P = nRT/V put in equation (1)
dw = – nRT/V x dv
Thus the work doneis donen gas expands from volume V1 → V2, work done is given as:
W = ∫_V1^V2▒(-nRT)/V x dv
= (-nRT)/V ∫_V1^V2▒dv
= -nRT lnV2/V1
W = – 2.303xRT log_10〖V2/V1〗
At 0 constant temp.
P1V1 = P2V2
P1/P2 = V2/V1
=>W= -2.303 nRT log〖P1/P2〗
Question: Calculate the work done when a gas is compressed by an average pressure of 0.5 atm so as to decide its volume from 400cm3 to 200cm3.
Pext = 0.5 atm
∆V=v2 – v1 = 0.2 – 0.4 =- 0.2 L
W = – Pexternal (V2-V1)
= – 0.5 × -0.2 = + 0.1 L atm
W = 0.1 × 101.3 = 10.13 J [1 L atm = 101.3 J]
Question: 2.5 moles of an ideal gas at 2atm at 27o c is expanded isothermally to 2.5times of its original volume against an external pressure of 1 atm. Calculate
a) The work is done
b) If the gas expands isothermally in a reversible manner then what will be the work done?
W = – Pexternal (V2-V1)
= -1 (2.5 x 30.75 – 30.75) [ V1 = nRT/P=(2.5×0.082×300)/2=30.75L]
= -1 x 46.12 L atm = – 4.61 KJ
W= -2.303 nRT log〖V2/V1〗
=2.303 ×2.5×8.314×300 log_2.5〖y/x〗
1st law of thermodynamics:
This law can be stated in a number of ways:
Energy can neither be created nor be destroyed but one form of energy can be converted to the other.
Whenever energy in one form disappears, an equivalent amount of energy in another form must appear.
Some total energy of system & surroundings taken together remain constant.
The energy of the universe is conserved.
The validity of 1st law:
There is no direct & absolute proof in favor of this law as it is based on human experiences. It is rectified by the full observations:
It is impossible to construct a perpetual motion machine i.e. machine which produces work without the expenditure of energy.
Joule conducted a number of exp regarding the conversion of work into heat; he found that for energy 4.184 of worked one, 1 calorie of heat is produced irreversible of the manner how that work is done.
Moreover, the same amount of work done produces the same amount of energy, irrespective of the manner how that work is done.
Energy is consumed during chemical reaction also.
e.g.: during the electrolysis of 1 mol e of H2O, 286 KJ of electrical energy is consumed i.e.:
H2O (l) + 286.0 KJ mol-1 → H2 (g) + ½ O2 (g)
When 1 mole of liquid H2O is produced from gaseous hydrogen & gaseous oxygen, 286 Kj of heat energy is released.
H2 (g) + ½ O2 (g) → H2O (l) + 286.0 KJ (heat energy)
The mathematical statement of the 1st law of thermodynamics:
Let us consider a system having internal energy U1. Suppose q is the heat supplier to the system. Internal energy increase & becomes U1+q. Now suppose ‘w’ is work done on the system.
Internal energy further increase & becomes
Final internal energy, U2 = U1+q+w
Or, U2-U1 = q+ w
if ∆U is internal energy change:
[Internal energy change] = [Heat absorved] + [Word done on the system] (→expansion work)
Relation (1) & (2), mathematical statements of 1st the law of thermodynamics.
Special cases of the 1st law of thermodynamics:
For an isothermal irreversible change,
for isothermal, ∆U=0 (T constant)
=>q= – Pext ∆V
i.e. heat applied to the system is useful work done by the system.
For isothermal, reversible change.
q = – Wrev = -2.303 nRT log〖V2/V1〗 = -2.303 nRT log〖P1/P2〗
For an adiabatic process,
q = 0
=>∆U=0 + Wadiabatic
When work is done on system internal energy increase & if done by system internal energy decrease.
Thus, in an adiabatic process, work is done at the expand of internal energy.
For an isochoric process:
∆U=0 (W = p∆u)
i.e. internal energy change is a measure of heat change during a process of constant volume. i.e. (∆u=qu)
The concept of enthalpy and enthalpy change:
Enthalpy: Enthalpy is the total energy associated with the system which involves its internal energy & energy due to environmental factors such as pressure, volume, conditions. It is also called the heat content. It is represented by H and is given by the solution:
H = U + PV
Important characteristics of enthalpy are:
It is a state function i.e. it depends upon the state of existence & is independent of the manner how that state is achieved.
It is an extensive property i.e. it depends upon the quantity of matter.
The absolute value of H cannot be determined however enthalpy change during a process can be determined accurately.
It is the difference between the enthalpies of products & reactions i.e. ∆H= HP – HR
Significance of ∆H:
According to the 1st law of thermodynamics
q= [u2 – u1] + P [v2-v1]
q= [u2 + P v2] – [u1 + Pv1]
q=H2 – H1 = ∆H
Thus, enthalpy change is the measure of heat change during a process at constant pressure i.e.
Exothermic and endothermic reactions:
Exothermic reactions: Chemical reactions which proceed with the evolution of heat are called exothermic reaction.
# When a reaction is carried out at constant volume, heat changes are expressed in term of ∆U.
# When a reaction is carried out at constant pressure, heat changes are expressed in terms of ∆H.
e.g→ C(s) + O2 (g) → CO2 (g) + 393.5 KJ
C(s) + O2 (g) → CO2 (g) ∆U= – 393.5 KJ (equation at constant volume)
C(s) + O2 (g) → CO2 (g) ∆H= – 393.5 KJ (at constant pressure)
H2 (g) + ½ O2 (g) → H2O (l) + 286.0 KJ
H2 (g) + ½ O2 (g) → H2O (l) ∆H = – 286.0 KJ
Question: Why ∆H is –ve for an exothermic reaction?
Answer: Exothermic reaction proceed with the evolution of heat. Thus, enthalpy of reactant is more than the enthalpy of product.
i.e. HR > HP
but, ∆H = HP – HR
hence, ∆H is –ve.
Chemical reactions which proceed with absorption of heat are called endothermic reaction.
e.g N2 (g) + O2 (g) → 2NO (g) – 180.5 Kj
N2 (g) + O2 (g) →2NO (g) ∆H = + 180.5 Kj
2HgO (s) → 2 Hg (l) + O2 (g) – 180.4 Kj
2HgO (s) → 2 Hg (l) + O2 (g) ∆H= + 180.4 Kj
Question: Why ∆H is +ve for endothermic reactions?
Answer: Endothermic reaction proceed with the absorption of heat thus enthalpy of reactant is less than that of product. i.e. HP>HR
Also, ∆H = HP – HR Similarly for internal energy also, ∆U = UP – UR
=>∆H is +ve
Question: what are thermochemical reactions?
Answer: A chemical reaction which besides fulfilling all the other requirements also indicate the heat change is called a thermo-chemical reaction.
The heat change during a thermo-chemical reaction can be written in two ways:
As a term along with products
As using ∆H rotation.
e.g C (s) + O2 (g) → CO2 (g) + 393.5 Kj
C (s) + O2 (g) → CO2 (g) ∆H = – 393.5 Kj
# It should be noted that while writing a thermo-chemical reaction, the physical state of reactants & products must be motioned because of ∆H changes with a change in physical states of reactants & products.
e.g H2 (g) + ½ O2 (g) → H2O (l) ∆H = – 286.0 KJ
H2 (g) + ½ O2 (g) → H2O (g) ∆H = – 249 KJ
Similarly, C (graphite) + O2 (g) → CO2 (g) ∆H = – 393.5 KJ
C (diamond) + O2 (g) → CO2 (g) ∆H = – 395.4 KJ (=+1.9KJ)
Graphite & diamond are not different because of ∆H as its value is very small but they difference being of difference process.
The relationship between ∆H & ∆U:
Let us consider a gaseous isothermal process at constant pressure.
VR = Volume of reactant
VP = Volume of products
nR = no. of moles of reactants
nP = no. of moles of products
for an ideal gas, PV = nRT
∴PVP = nPRT ( T constant)
PVR = nRRT (P constant)
=> PVP – PVR = nPRT – nRRT = (nP – nR) RT
=>P∆V = ∆ng RT ———————-(1)
Also, by definition H = U + PV
=> Change in enthalpy at constant pressure, ∆H = ∆U + P∆V ———————-(2)
from (1) & (2), ∆H = ∆U + ∆ng RT
The significance of this relation:
If ∆ng = 0
=>) ∆H = ∆U
e.g. C (s) + O2 (g) → CO2 (g) (do not consider solid)
=>) 1 -1 = 0
***∆ng → Solid & liquid not to be considered.
If ∆ng is +ve
=> ∆H > ∆U
e.g. C (s) + ½ O2 (g) → CO2 (g)
∆ng = ½
If ∆ng is -ve
=>) ∆H < ∆U
e.g. N2 (g) + 3 H2 (g) → 2NH3 (g)
∆ng = 2-4 = -2
For a gaseous reaction:
Question: 2A2 (g) + 5B2 (g)→ 2A2B5 (g)
At 270C the heat change at constant pressure is found to be -50160J. calculate the value of internal energy change.
Solution: ∆ng = 2-7 = -5
∆H = ∆U + ∆ng RT
∆U = ∆H – ∆ng RT
= 50160 – [-5×8.314×300]
= -50160 + 12471
= – 37689 J (ans)
Question: The enthalpy change of equation below is -93kg. Calculate ∆U for the reaction at 250C.
Solution: N2 (g) + 3H2 (g) →2NH3 (g) =>)∆ng = 2-4 = -2
∆H = ∆U + ∆ng RT
∆U = ∆H – ∆ng RT
= -93000 – [(-2) x 8.314 x 298] (Isothermal, ∆U=0; ∆H=0. adiabatic expansion, ∆H = 0, not ∆U)
= -93000 + 4955.144
= -88044 Joules
= -88.044 KJ (ans)
It is the amount of heat required to raise the temperature of a system through 1o C.
Heat capacity is represented by C.
Mathematically, C = q/(T2-T1) or q/∆T
Since heat capacity varies with temperature the temperature difference should be as small as possible. Thus, if dq=amount of heat absorbed in raising the temp by infinitesimally small dt.
Then heat capacity = dq/dt
The amount of heat required to raise the temp of 1gm of a substance to 10C is called specific heat.
The amount of heat required to raise the temp of 1mole of a substance through 10C is called moles heat capacity.
e. g. If 5.4 joule of heat is absorbed by 2g of Al and the temp rises from 298K to 300K, then heat capacity = 5.4/2=2.7 J/K
Sp. Heat = 2g – 5.4
1g = 5.4/2 = 2.7joule = 2.7/2 = 1.35 Jg-1K-1
Molar heat capacity=
1 mole of Al = 2kg x 5.4/2×2 x 27 = 36.45 Jmol-1k-1
∴The amount of heat required to raise the temperature of W gm of a substance to a temp difference ∆T is given as:
q = c x w x ∆t
C=sp. Heat as w is in gms or
q = c x n x ∆t
n = no. of moles
c = molar heat capacity
Types of heat capacity:
Heat capacity is of two types:
Heat capacity at constant volume i.e. Cv
Heat capacity at constant pressure i.e. Cp
The significance of Cv:
According to the 1st law of thermodynamics
Dq = du + Pdv
At constant volume dv = 0
∴dq = dv —————————(1)
But, c = dq/dt —————————(2) [by definition]
From (1) & (2), Cv = (dv/dt)v (constant volume)
Thus, Cv represent the internal energy change with temp.
Significance of Cp:
According to 1st law of thermodynamics,
dq = dU + Pdv ——————-(1)
H = U + PV (by definition)
dH = dU + Pdv ————(2)
from (1) & (2), dq = dH
by definition: C =
By definition: c = dq/dt
CP =( dH/dT)P
CP represents the enthalpy change with temp.
The relationship between CP & CV:
H = U + PV
But PV = RT (for ideal gas)
=> H = U + RT
Differentiate with terms of T,
CP = CV +R
[CP – CV = R] for 1mole of gas
[CP – CV = nR] for n mole of gas
Also, Cp/Cv= r
Value of r gives the atomicity of gas. It is 1.66 for monatomic gases. e.g. He, Ne
& temp 1.40 for diatomic gases e.g. O2, N2
& 1.33 for tri-atomic gases. e.g. O3
Question: Calculate the amount of energy required to raise the temperature of 60gm of H from 350 to 550C. Molar heat cThe molarty of H is : 24 jmol-1k-1
Solution: to find: n = 60/27
q = C x N x ∆T = 24 x 60/27 x 20 = 1066.67 J (ans)
Question: Define enthalpy of a reaction or heat of reaction. On what factors does it depend?
Answer: Enthalpy of reaction i.e. ∆rH . It may be defined as the enthalpy change accompanying the conversion of definite no. of moles of reactants into products as indicated by the balanced equation.
e.g. CH4 (g) + 2O2 (g)→ CO2 (g) + 2H2O (l) ∆rH = -890.3 Kj
According to this reaction, when 1mole of CH4 is burnt with 2moles of O2 to produce 1mole of CO2 & 2moles of liquid water (H2O(l)). 890.3Kj of heat is evolved. Then the enthalpy of this reaction is -890.3Kj.
Factors affecting enthalpy of reaction:
Quantity of reactants: enthalpy of reaction depends upon the amount of reactants. Then the above equation is mul with 2:
2CH4 (g) + 4O2 (g)→ 2CO2 (g) + 4H2O (l)
∆rH = -1480.6 Kj
Allotropic modification: Enthalpy of reaction depends upon the allotropic form taken:
C (graphite) + O2 (g)→ CO2 (g) ∆H = -393.5 Kj
C (diamond) + O2 (g)→ CO2 (g) ∆H = -395.4 Kj
Physical states of reactants & products:
e.g. H2 (g) + ½ O2 (g) → H2O (l) ∆H = -286 Kj
H2 (g) + ½ O2 (g) → H2O (g) ∆H = -249 Kj
Temperature: enthalpy of reaction changes with temperature. The enthalpy of reaction at 1 bar pressure & at a specified temp is called standard enthalpy of reaction, represent as ∆H-
Whether the reaction is raised out at controlled temp or constant volume.
The concentration of solution: In case of reaction is solution enthalpy of reaction depends upon the concentration of the solution.
Let us discuss the various types of enthalpies of reaction:
Enthalpy of formation (∆fH): it may be defined as the enthalpy who’s accompanying the function of 1mole of a substance from its element.
e.g. formation of CO2.
C (s) + O2 (g) → CO2 (g) ∆fH = -393.1 KJ
e.g. formation of H2O.
H2 (g) + ½ O2 (g) → H2O (l) ∆fH = -286.1 KJ
formation of ethanol:
2C (s) + 3H2 (g) + ½ O2 (g) →C2H5OH (l) ∆H = -722 KJ
Standard enthalpy of formation:
It may be defined as the enthalpy change accompanying the formation of 1mole of a substance from its element. All the substances were in their standard states. i.e. 1 BAR pressure & a specified temp. it is represented by ∆fH(-)
The standard state corresponds to the pure stable state at 1 BAR pressure & at a specified temperature.
e.g. Standard state of ethanol at 298K is pure ethanol at 1BAR pressure & 298K.
similarly, the standard state of Fe at 500K is pure iron at 1bar pressure & 500K.
# According to the old convention, the standard temp = 298k.
The standard state of C is graphite and that of sulfur is rhombic sulfur.
Application of standard enthalpy of reaction: knowing ∆fH(-) of various substance, enthalpy of reaction can be calculated by using the relation:
∆fH(-) = ∑▒ (∆fH(-))product – (∆fH(-)) reactant
According to the convention, ∆fH(-) of elementary substance is taken as required.
Thus, for a reaction, aA + bB →cC + dD
∆fH(-) = [■(∆fH(-)of Cxc @+@∆fH(-)of Dxd )] – [■(∆fH(-)of Axa @+@∆fH(-)of Bxb )]
Let us calculate the enthalpy of formation of CH4 fro formula-
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) ∆rH(-) = 890Kj
Given, ∆fH(-) of CO2 = -393.5Kj
∆fH(-) of H2O = 286.0Kj
∆fH(-) = ∑▒ (∆fH(-))P – (∆fH(-)) R
-890.3 = [■(∆fH(-)of CO2 @+@∆fH(-)of H2O x2 )] – [■(∆fH(-)of CH4 @+@0 )]
∆fH(-) of O2 is 0 because it itself is an element.
-890.3 = -965.5 – x
x = -965.5 + 890.3 = -75.2 Kj (ans)
Enthalpy of combustion (∆CH):
It may be defined as the enthalpy change accompanying the combustion of 1mole of a substance in large excess of oxygen or air.
e.g. combustion of carbon.
C (s) + O2 (g) → CO2 (g) ∆H = -393.5Kj
e.g. combustion of methane.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = -890.3Kj
It should be noted that heat of combustion is always ‘-‘ve.
Calorific value: It may be defined as the amount of energy produced when 1gm of food or fuel is oxidized or undergo combustion.
Move the calorific value, better is the fuel.
Question: A cylinder of cooking gas is attempted to contain 11.2kg of butane. The thermo-chemical equation for the combustion of butane is:
C4H10 + 13/2 O2 → 4CO2 + 5H2O ∆H = -2658Kj
If a family needs 15000kj of energy per day. How long will the cylinder last?
Assuming that 30% of the gas is wasted due to incomplete combustion then how long will the cylinder last?
According to above equation:
58gm of C4H10 on combustion produces heat = 2658Kj.
11.2×103 gm of C4H10 on combustion produces heat = 2658/58x 11.2 x 103 =513.2 Kj x 103=513200
Daily consumption = 15000Kj
∴no. of days = 513200/15000 = 34 days
When 12g of C → CO heat produced = 111Kj
When, 3200g = 111/12×3200 = 29600 Kj
Total heat = 157600 + 29600 = 187200 Kj
Enthalpy of neutralization:
it may be defined as the enthalpy change accompanying the neutralization of 1gm of equivalent of acid by 1gm equivalent of base in a dilute aqueous solution:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) ∆H = -52.1Kj
CH3COOH (aq) + NaOH (aq) →CH3COONa (aq) + H2O (l) ∆H = -55.2Kj
NH4OH (aq) + HCl (aq) → NH4Cl (aq) + H2O (l) ∆H = -51.6Kj
It should be noted that enthalpy of neutralization of strong acid & strong base is always constant i.e. 57.1Kj.
Explanation: According to Arrhenius theory, strong acid, strong bases & salts completely longer in aqueous solution. Therefore the no. of H ion produced by the 1gm equivalent of an acid is the same as no of all ions produced by the ionization of 1gm equivalent of a base. Thus the overall reaction is the formation of H2O from its ions.
e.g. let us consider the neutralization of NaOH & HCl.
K+(aq) + OH- (aq) + H+ (aq) + Cl- (aq) → K+ (aq) + Cl- (aq) + H2O (l)
OH- (aq) + H+ (aq) → H2O (l) ∆H = -57.6Kj
However, if one of the acid or base is weak then enthalpy of neutralization is less than 57.1 Kj.
Reason: According to Arrhenius concept, weak acids & weak bases are not completely ionized in aqueous solution. Some energy is utilized to dissociate a weak acid or weak base.
Thus enthalpy of neutralization of weak acid & strong base or weak base & strong acid = the difference between the enthalpy of formation of H2O from its ion and the enthalpy of dissociation of a weak acid or weak base.
H2 + ½ O2 → H2O (elements) ∆H = -286Kj
Neutralization, H+ + OH- → H2O (lons) ∆H = -57.1Kj
e.g. let us consider the neutralization of CH3COOH & NaOH
CH3COOH ↔ CH3COO- (aq) + H+ (aq) ∆H = +1.9Kj
NaOH → Na+ (aq) + OH- (q)
H+ (aq) + OH- (aq) → H2O (l) ∆H = -57.1Kj
CH3COOH + Na+ (aq) + OH- (aq) →CH3COO- (aq) + Na+ (aq) + H2O (l) ∆H = -55.2Kj
Let us consider the neutralization of NH4OH & HCl:
NH4OH□(↔┴water ) NH4+ (aq) + OH- (aq) ∆H = +5.5Kj
HCl → H+ (aq) + Cl- (aq) ∆H = -57.1Kj
NH4OH + H+ (aq) + Cl- (aq) →NH4+ (q) + Cl- (aq) + H2O (l) ∆H = -51.6Kj
Question: what the heat would be released when-
a. 0.5 mole of HCl is neutralized by 0.5 moles of NaOH.
b. an aqueous solution containing 0.5 moles of HNO3 acid is mixed with 0.3mole of NaOH.
c. 200cc of 0.1M H2SO4 is mixed with 150cc of 0.2 M KOH.
Solution: No. of moles of Na+ ion = M x volume = 0.2
No. of moles of OH- ion = 0.03
H+ + OH- → H2O (l)
0.03moles are neutralized
When 0.03moles of H+ ion combines with 0.03 moles of OH- ion, heat produced = 57.1 x 0.03 = 1.713 Kj
Enthalpy of the solution: (∆ solutionH)
It may be defined as the enthalpy change accompanying the dissolution of 1mole of a substance in a large excess of solvent so that further dilution doesn’t produce any heat change.
e.g. diss of KCl
KCl (s) + aq → KCl (aq) ∆H = +18.6Kj
e.g. diss of CuSO4
CUSO4 (s) +aq → CuSO4 (aq) ∆H = -66.5Kj
Enthalpy of hydration (∆hydration H):
It may be defined as the enthalpy change that takes place when 1mole of a substance combines with the required no. of molecules of H2O to form its specific hydrate.
Enthalpy of fusion: (∆fusion H):
It may be defined as the enthalpy change accompanying the conversion of 1mole of solid into liquid at its melting point.
e.g H2O (s) → H2O (l) ∆H = +6.0Kj at 273K
Enthalpy of vaporization (∆vaporizationH):
It may be defined as the enthalpy change accompanying the conversion of 1mole of liquid into vaporous at its boiling point.
H2O (l) → H2O (g) ∆H = +40.6Kj at 393K
Enthalpy of sublimation (∆ sublimationH):
It may be defined as the enthalpy change accompanying the conversion of 1mole of solid directly into vapor at a temp below its melting point.
I2 (s) → I2 (g) ∆H = +62.4Kj
It should be noted that-
∆ subH = ∆fusH + ∆vapH
Hess’s law of constant heat summation:
According to this law, the heat evolved or absorbed during a process is the same whether the process is carried out in 1 step or I several steps. In another word, enthalpy change depends upon the initial & the final state & is independent of the path followed.
Illustration: let us consider a process A→D in 1 process & suppose Q1 joule of heat is produced.
Now suppose the above process is carried out in three steps i.e.
A→B→C→D & let q1, q2 & q3 be the heat evolved in each step so that q1+q2+q3 = Q2.
According to Hess’s law Q1=Q2.
It means by converting A→ D in 1 step & reconverting D→A in 3 steps, (Q1-Q2) joule of heat is produced. Thus, by repeating this process a number of time a large amount of energy can be created which is against the 1st law of thermodynamics. Law of conservation of energy. so Q1=Q2
Experimental rectification: carbon can be converted to CO2 in 2 different ways:
C (s) + O2 (g) → CO2 (g) ∆H = -393.5Kj
C (s) + ½ O2 (g) → CO (g) ∆H = -110Kj
CO (g) + ½ O2 → CO2 (g) ∆H = -283.5Kj
C (s) + O2 (g) → CO2 (g) ∆H = -393.5Kj
Heat produced in both the case is same. This process Hess’s law.
Important Implication of Hess’s law: is that thermo-chemical equilibrium can be added, sub, mul with any integer & divided with any integer.
Applications of Hess’s law:
To calculate the enthalpy of formation there are many substances whose enthalpy of formation cannot be determined exp. However, it can be calculated with the help of Hess’s law.
Let us calculate the enthalpy of formation of CO.
The required equation is:
C (s) + ½ O2 (g) → CO ∆H = ?
Experimentally, enthalpies of combustion of C & CO are determined i.e.
C (s) + O2 (g) → CO2 (g) ∆H = -393.5Kj ————–(1)
CO (g) + ½ O2 (g) → CO2 (g) ∆H = -283.5Kj ————–(2)
Subtract (2) from (1) to get the required equation:
i.e. C(s) + O2 (g) – CO (g) – ½ O2 (g) → CO2 (g) – CO2 (g) ∆H = -393.5 – (-283.5) = – 110Kj
C (s) + ½ O2 (g) → CO
# To calculate the enthalpy of transition: It may be defined as the enthalpy change accompanying the conversion of 1mole of 1allortropic form into the other.
The transition of one allotropic for into other form is a very slow process & enthalpy change can’t be determined exp. However, this can be calculated with the help of Hess’s law.
Let us calculate the enthalpy of transition of graphite to diamond. The required equation is:
C (graphite) → C (diamond) ∆H = ?
Example- enthalpies of combustion of graphite & diamond are determined:
C (graphite) + O2 (g) → CO2 (g) ∆H =-393.5Kj ———-(1)
C (graphite) + O2 (g) → CO2 (g) ∆H =-395.4Kj ———-(2)
H2O (l) → H2O (g)
∆ng = 1-0=1
∆H0 = ∆U0 + ∆ng RT
∆U0 = ∆H0 – ∆ng RT
= 40.6 – 1×8.314×373/1000
= 40.6 – 3.1
= 37.5 Kj
Note- the value for evaporation is taken at 373K (i.e. at BP) vap -298K (room temp)
The concept of bond enthalpy:
It may be defined as the amount of energy required to break 1mole of bonds of a particular bond is gaseous molecule into gaseous atoms.
e.g→ H2 + 435.0 Kjmol-1 → H (s) + H (g)
It is also called the enthalpy of atomization.
In the case of a diatomic molecule, bond enthalpy and bond dissociation enthalpy have the same meaning.
In case of the polyatomic molecule, the average of the bond dissociation enthalpies is called the bond enthalpy.
e.g→ H – O – H (g) → H – O (g) + H (g) ∆H =498Kj
H – O (g) → H (g) + O (g) ∆H =430Kj
B.E O-H = (498+430)/2 = 464 KJmol-1
Application of bond enthalpy:
Knowing the bond enthalpies of various bonds, enthalpy of reaction (∆rH) can be calculated by using the reaction:
∆rH = ∑▒〖(B.E)〗reac – ∑▒〖(B.E)〗product
2. Bond enthalpies can be calculated with the help of Hess’s law.
e.g. let us calculate the B.E of C-H bond in CH4.
The required equation is:
CH4 (g) → C (g) + 4H (g) ∆H=?
∆H = (∆fH(-))p – (∆fH(-)) R
∆H = [■(∆fH(-)of C @+@∆fH(-)of H x4 )] – [∆fH(-)of CH4 ]
B.E C-H = ∆H/4
Let us calculate the B.E of C-C bond in ethane (C2H6).
C2H6 →2C (g) + 6H (g)
∆H = [■(∆fH(-)of Cx2 @+@∆fH(-)of H x6 )] – [∆fH(-)of C2H6 ]
∆H = B.E C-C + B.E C-H x 6
B.E C-C = ∆H – 6 x B.E C-H
Calculate enthalpy change for the reaction:
2C2H2 (g) + 5O2 (g) → 4CO2 + 2H2O (g)
Given average bond enthalpies of various bonds as:
C-H = 424 Kjmol-1
C≡ C = 814 Kj
O = O 499Kj
2 H-C ≡ C-H + 5 O = 0 → 4O=C=O +2H-O-H
∆rH = (B.E)R – (B.E)P
= [■(B.E of C≡C ×2@B.E of C-H ×4@B.E of O=O ×5)] – [■(B.E of C=O ×8@+@B.E of H-O×4)] = -2573 KJ
Spontaneous and non-spontaneous processes:
Spontaneous process: A process which takes place either by itself or after proper initiation under a given set of conditions is called a spontaneous process.
The term spontaneity doesn’t mean instantaneity. It simply means that the process has an urge to take place. I.e. it is a flexible process or probable process.
[A] Where no initiation is required:
Evaporation of H2O
H2O (l) → H2O (g)
Dissolution of NaCl in H2O
NaCl + aq → Na+ (aq) + Cl- (aq)
Combination of NO & O2 to form NO2
2NO + O2 → 2NO2
[B]Where initiation is required:
Combustion of C once ignited.
C (s) + O2 (g)→ CO2 (g)
Combustion of CH4, once ignited
CH4 + 2O2→CO2 + 2H2O
Decomposition of CaCO3 once heated. CaCO3 (s) → CaO (s) + CO2 (g)
Non-spontaneous process: A process which is forbidden but is made to take place by supplying energy from the external source. As soon as the supply of energy is cut off, the process stops. Such a process is called a non-spontaneous process.
e.g. 1. The flow of water uphill.
2. the flow of heat from cool to the hot body as in refrigeration.
3. The process of electrolysis.
4. Mixing of sand in water.
The criterion for the spontaneity of the process:
As some processes are spontaneous while others are non-spontaneous, it means there must be criteria responsible for the spontaneity of the processes.
The overall tendency of a process to occur in a particular deterioration is called a driving force. It is a resultant of the two tendencies.
The tendency to acquire mind. energy. It is a well-known fact that every system in this universe tends to acquire minm. energy. It is because of lesser energy, more stability.
e.g. 1. H2O flows from upper level to lower level.
2. Heat flows from hot body to cold body.
It is suggested that lowering of energy may be the criterion responsible for the spontaneity of process. i.e. more a process to be spontaneous ∆ H = ‘-‘ve. It is justified by the occurrence of the full exothermic reaction.
C (s) + O2 (g) → CO2 (g) ∆H =-393.5Kj
H2 (g) + ½ O2 (g) → H2O (l) ∆H =-286Kj
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H =-890.3Kj
Thus ∆H –ve may be the criterion for the spontaneity.
Limitations of this concept:
[A] Many endothermic reactions are also spontaneous.
e.g. 1. Evaporation of H2O.
H2O (l) → H2O (g) ∆H =+40.6Kj
2.dissolution of NH4Cl in water.
NH4Cl + aq (s) → NH4 (aq) + Cl- (aq) ∆H =+15.1Kj
3.Decomposition of HgO.
HgO (s) → Hg (l) + ½ O2 (g) ∆H =+90.3Kj
[B] Reaction with ∆H=0 is also occurring.
e.g. 1. Mixing of gases
2. Spreading a drop of ink in H2O.
3. Expansion of an ideal gas in a vacuum.
[C] Reversible reactions are also occurring. In such reaction, if the forward reaction is exothermic backward reaction will be endothermic or vice versa.
[D] Exothermic reaction carried out in the closed vessel does not proceed to completion the process until an equation is established.
Thus ∆H alone is not the soul & sufficient criterion responsible for the spontaneity of the process. Some other forces are also required to explain the spontaneity & that factors are:
B) The tendency to acquire max Randomness:
Examinations of all the above endothermic reactions reveal that they precede with the inc I randomness.
Evaporation of H2O.
H2O (l) → H2O (g) ∆H =+40.1Kj
Less random more random
H2O molecular in the gaseous state is more random than H2O molecules in a liquid state. Three evaporation of H2O proceeds with the increase in randomness.
e.g.- Diss of NH4Cl in H2O.
NH4Cl (s) + aq → NH4+ (aq) + Cl-1 (aq) ∆H =+15.1Kj
(less random) (more random)
In the solid state NH4+ & Cl-1 ions have fixed arrangements. While in liquid solution these ions move freely. Hence diss of NH4Cl in H2O proceeds with increase in randomness.
Decomposition of HgO:
HgO (s) → Hg (l) + ½ O2 (g) ∆H =+90.3Kj
(less random) (more random)
Liquid Hg & gaseous O2 are in the more random state than solid HgO. Their decomposition of HgO proceeds with the increase in randomness.
Thus, the tendency to acquire max randomness also contributes towards spontaneity of process. But like ∆H tendency to acquire max randomness is not the only factor responsible for spontaneity. Both these factors taken together explain the spontaneity of process.
The concept of entropy & entropy change:
Entropy is a thermodynamic factor which is a measure of the degree of randomness it is represented by ‘S’.
Gaseous state is a most random state while solid state is the least random state.
i.e. S (g) > S (l) > S (s)
Important characteristics of entropy:
Important characteristics are:
It is a state function.
It is an extensive property.
Entropy change during a process is given as:
∆S = ∑▒SP – ∑▒SR
Sp > Sr, then ∆S is +ve . i.e. Entropy increase or randomness increase.
e.g. H2O (l) → H2O (g)
H2O (s) → H2O (l)
2NH3 (g) → 2N2 (g) + 3H2 (g) more moles, more entropy.
NaCl (s) + aq → Na+ (aq) + Cl- (aq)
if Sp < Sr, then ∆S is -ve . i.e. Entropy decrease.
e.g. H2O (g) → H2O (l)
H2O (l) → H2O (s)
N2 (g) + 2H2 (g) → 2NH3
Ag+ (aq) + Cl- (aq) → AgCl (s)
Entropy change is defined as the amount of heat absorbed reversibly & isothermally divided by the temp at which heat is absorbed.
∆S = (q rev,isothermally)/T
5. The units of entropy are JouleK-1mol-1 or JK-1mol-1
6. The total entropy change of universe-
∆Stotal = ∆Ssys + ∆Ssurroundings
7.For an irreversible process
∆Stotal = ∆Ssys + ∆Ssurroundings > 0
8.For reversible process
∆Stotal = 0
The entropy of fusion (∆fusS):
It may be defined as the entropy change accompanying the conversion of 1mole of solid into liquid at its melting point.
∆fusS = SL – Ss = ∆fusH/T
Where, SL – the molar entropy of liquid
SS – the molar entropy of solid
∆fusH – the entropy of fusion
Tf – fusion or melting temp.
e.g. Let us calculate the entropy of fusion of ice (∆fusH=6.025 KJmol-1)
∆fusS = 6.025/273 x 1000J = 6025/273 = 22.1
Since ∆fusH is always +ve,
∆fusS will also be +ve.
Hence, SL > Ss
Entropy of vaporization (∆vapS):
It may be defined as the entropy change accompanying the conversion of 1mole of liquid into vapors at its boiling point. It is represented as ∆vap & is given by reaction:
∆vapS = Sg – SL = ∆vapH/Tb
Sg – the molar entropy of gas.
SL – the molar entropy of liquid.
∆vapH – the entropy of vapor
Tb – boiling temp.
Since ∆vapH is always +ve so ∆vapS will be +ve.
Hence, Sg > SL
Question: Let us calculate the entropy of the vapor of H2O.
Solution: ∆vapH = +40.6 KJ
∆vapS = ∆vapH/T = 40600/273 J = 108.8 JK-1mol-1
Quesiton: A heated Cu block at 130oC losses 340Joule of heat to surrounding which are at room temp of 520C. Calculate-
a) Entropy changes of system
b) entropy changes of surroundings
c) total entropy change
∆Ssys = qsys/T = (-340)/403 = -0.843 JK-1
∆Ssurr = qsurr/T = 340/305 = 1.11 JK-1
∆Stotal = ∆Ssys + ∆Ssurr = -0.84 + 1.11 = +0.27 JK-1
limitations of 1st law of thermodynamics:
This law fails to predict the feasibility of the process. i.e. whether the given reaction is spontaneous or not.
Definitions of the 2nd law of thermodynamics:
This law can be stated in a number of ways-
All naturally occurring processes are thermodynamically irreversible.
2. Without the help of an external force, a spontaneous process can’t be reversed.
e.g. > heat by itself can’t flow from cold to a hot body.
3. Complete conversion of heat into work is not possible without leaving off somewhere else.
4. all naturally occurring process proceeds with an increase in entropy or randomness.
5. combining 1st & 2nd law of thermodynamics we can say that “energy of the universe is conserved, while entropy constantly increases & tends towards max.
The concept of free energy & force energy change:
It may be defined as the max amount of energy available to the system for conversion into useful work or non-expansion work. It is represented by ‘G’ and is given by the reaction:
G = H – TS
‘G’ is also called Gibb’s force energy.
Now , H = U + PV
G = U + PV – TS
Change in free energy at constant temp and pressure is given by reaction:
∆G= ∆U + P∆V – T∆S
(∆G)TP = ∆H – T∆S [∆H= ∆U + P∆V]
→It is the measure of sponthe taneity of process.
It should be noted that G is a state form & is an extensive property.
The physical significance of ∆H:
According to the 1st law of thermodynamics,
∆U = q + w (w → w.d on system)
This work includes both expansion & non-expansion work.
i.e ∆U = q + Wexp + Wnon-exp
∆U = q -P∆V + Wnon-exp [Wexp = – P∆V]
q = ∆U+-P∆V – Wnon-exp
q = ∆H – Wnon-exp —————-(1)
By definition → ∆S = q/T
=> q = T∆S ——————–(2)
From (1) & (2),
T∆S = ∆H – Wnon-exp
Wnon-exp = ∆H – T∆S
Wnon-exp = ∆G [∆G=∆H – T∆S ]
Or, – ∆G = – Wnon-exp
Thus, decrease in free energy (-∆G) is a measure of useful work (non-expansion) done by the system.
Driving spontaneity criterion for entropy
∆Stotal = ∆Sssystem + ∆Sssurroundings ———–(1)
Let us consider a process at constant temp & const pressure.
Suppose ‘q’ amount of heat is loosed by the surroundings and gained by the system, then by definition:
∆Sssurroundings = (-q)/T = (-∆Hsys)/T ————-(2) [qp = ∆H]
From (1) & (2), ∆Stotal = ∆Sssystem – ∆Hsys/T
T∆Stotal = T∆Sssystem – ∆Hsystem
-T∆Stotal = ∆Hsystem – T∆Sssystem
-T∆Stotal = (∆G)T,P [∆G= ∆H – T∆S]
Or, (∆G)T,P = -T∆Stotal
We know that :
If ∆Stotal is +ve, pthe rocess is spontaneous. Thus ∆G becomes ‘-‘ve.
If ∆Stotal is ‘-‘ve , pthe rocess is non-spontaneous. Thus ∆G becomes ‘+‘ve.
If ∆Stotal =0 , process is equilibrium & ∆G=0.
b. Spontaneity from Gibb’s Helmholtz equation:
∆G= ∆H – T∆S
if both energy & entropy factor favfavorse process i.e. ∆H is –ve & ∆S is +ve, the process is spontaneous. Hence, ∆G= -ve
if energy factor favor & entropy factor opposes the process i.e. ∆H is –ve & ∆S is –ve, then if ∆H > T∆S → process spontaneous. In this condition , ∆G →-ve
∆H < T∆S → process non-spontaneous. In this condition ∆G →+ve.
∆H = T∆S → process is in equilibrium. In this condition ∆G = 0.
If entropy factors favor and energy factor opposes the process. i.e both ∆H & ∆S are +ve. Then if:
∆H> T∆S → non-spontaneous. In this condition ∆G →+ve.
∆H< T∆S → spontaneous. In this condition ∆G →-ve.
∆H = T∆S → process is in equilibrium. In this condition ∆G = 0.
When both energy & entropy factor oppose the process.
i.e. ∆H = +ve, T∆S = -ve
Process always non-spontaneous & ∆G = +ve.
To sum up:
∆G = -ve → process spontaneous
∆G = +ve → process non-spontaneous
∆G = 0 → process in equilibrium
Effect of temp on the spontaneity of the process:
∆H does not change much with temp but ∆S changes with temp and thus the spontaneity also changes with temp.
[A] Exothermic reactions: for these reactions, ∆H is ‘-‘ve. Thus energy factor favors the process. ∆S may be ‘+’ve or ‘-‘ve .
1. if ∆S is the ‘+’ve – i.e. entropy factor also favors process.
Such processes are spontaneous under all condition (no role of T)
i.e. ∆S ‘+’ve & ∆H ‘-‘ve.
2. if ∆S is ‘-‘ve i.e. (entropy factors opposes process)
∆G may be ‘+’ve or ‘-‘ve depended upon the change of T∆S.
For such a process to be spontaneous the nag of T∆S should be less than ∆H, thus temp should be low.
Hence exothermic reactions are forward at low temperature.
[B] Endothermic reactions: For these reactions, ∆H is ‘+’ve. Thus energy factor opposes the process. ∆S may be ‘+’ve or ‘-‘ve.
1. if ∆S is ‘-‘ve. Entropy factor opposes the process. Such processes are non-spontaneous under all conditions.
2. if ∆S is ‘+’ve. Entropy factor favor process.
∆G may be ‘+’ve or ‘-‘ve depending upon the mag of T∆S.
For such a process to be spontaneous the mag of T∆S> ∆H, thus temp should be high.
Hence endothermic reactions are favors at high temperature.
Question: Define standard free energy of formation and standard free energy change ∆G(-). How is ∆G(-) calculate?
Answer: Standard free energy of formation (∆fG(-)).
It may be defined as the free energy change accompanying the formation of 1mole of a substance from its elements, all the substances being in their standard state i.e. at 1bar pressure & specified temp. it is represented by ‘∆fG(-)’.
Standard free energy change (∆G(-)) : It may be defined as the free energy change accompanying the formation of definite no. of moles of reactants →products. It is represented ∆G(-).
Calculation of ∆G(-) or ∆rG(-):
When ∆fG(-) of reactants & products are given then-
∆rG(-) = ∑▒〖(∆fG〗-)products – ∑▒〖(∆fG〗-)reactants
when ∆H(-) & ∆S(-) are given then-
∆G(-) = ∆H(-) – T∆S(-)
[∆rG(-) = -2.303 RT log K, where K = equilibrium constant=1.33] —–Important
∆G = -2.303 RT log 1.33
= -2.303 x8.314 x 333 x log1.33
= -19.14 x 333 x log1.33
=-19.14 x 333 x 0.1198
= – 763.8 KJmol-1 (ans)
Question: how is ∆U calculated experimentally.
Answer: ∆U is calculated experimentally by Bomb calorimeter. It consists of a constant volume vessel. It is provided with a cup or which a known weight of a substance is taken. It is heated (burned) electrically and oxygen as of high pressure is made to pass through it.
This vessel is placed in another insulated vessel containing water. Since heat is produced during combustion, temp of water rises which is noted.
Then ∆U is calculated by using the relation:
∆U = (Q × ∆T ×M)/W
Where Q = heat capacity of the calorimeter
∆T = change in temp.
W = weight of substance taken
M = MM of substance.