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Terms
Explanation
System
Part of the universe under investigation.
Open System
A system which can exchange both energy and matter with its surroundings.
Closed System
A system which permits passage of energy but not mass, across its boundary.
Isolated system
A system which can neither exchange energy nor matter with its surrounding.
Surroundings
Part of the universe other than system, which can interact with it.
Boundary
Anything which separates system from surrounding.
State variables
The variables which are required to be defined in order to define state of any system i.e. pressure, volume, mass, temperature, surface area, etc.
State Functions
Property of system which depend only on the state of the system and not on the path.
Example: Pressure, volume, temperature, internal energy, enthalpy, entropy etc.
Intensive properties
Properties of a system which do not depend on mass of the system i.e. temperature, pressure, density, concentration,
Extensive properties
Properties of a system which depend on mass of the system i.e. volume, energy, enthalpy, entropy etc.
Process
Path along which state of a system changes.
Isothermal process
Process which takes place at constant temperature
Isobaric process
Process which takes place at constant pressure
Isochoric process
Process which takes place at constant volume.
Adiabatic process
Process during which transfer of heat cannot take place between system and surrounding.
Cyclic process
Process in which system comes back to its initial state after undergoing series of changes.
Reversible process
Process during which the system always departs infinitesimally from the state of equilibrium i.e. its direction can be reversed at any moment.
Irriversible Process
This type of process is fast and gets completed in a single step. This process cannot be reversed. All the natural processes are of this type
Energy is exchanged between system and surround in the form of heat when they are at different temperatures.
Heat added to a system is given by a positive sign, whereas heat extracted from a system is given negative sign.
It is an extensive property.
It is not a state function.
It is the capacity for doing work.
Energy is an extensive property.
Unit : Joule.
Work (W):
Work = Force × Displacement i.e. dW = Fdx
Work done on the system is given by positive sigh while work done by the system is given negative sign.
Mechanical Work or Pressure-Volume Work: work associated with change in volume of a system against an external pressure.
Work done in reversible process: W=
W = – 2.303 nRT log v2/v1 = –2.303 nRT log p1/p2
Sum of all the possible types of energy present in the system.
ΔE = heat change for a reaction taking place at constant temperature and volume.
ΔE is a state function.
Value of ΔE is -ve for exothermic reactions while it is +ve for endothermic reactions.
Energy can neither be created nor destroyed although it can be converted from one form to another.
or
Energy of an isolated system is constant.
Mathematical Expression
Heat observed by the system = its internal energy + work done by the system. i.e. q = dE + w
For an infinitesimal process
dq = dE + dw
Where, q is the heat supplied to the system and w is the work done on the system.
For an ideal gas undergoing isothermal change ΔE =0.
so q= -w.
so, dE = - dw
ΔE = q - pdV
ΔE = qv
ΔE = 0
q = - pdV =-W
?q = 0
ΔE = W
H = E+PV
At constant pressure:
dH = dE + pdV
For system involving mechanical work only
dH = QP (At constant pressure)
For exothermic reactions:
dH = -ve
For endothermic reactions:
dH = +ve
dH = dE + dng RT
Where,
dng = (Number of moles of gaseous products - Number of moles of gaseous reactants)
C = q / dT
Cs = Heat capacity / Mass in grams
Cm = Heat capacity / Molar mass.
Cv = (dE/dT)v
Cp = (dE/dT)p
Cp – Cv = R
Variation Of Heat Of Reaction With Temperature:
dCP = (dH2 - dH1)/(T2-T1) & dCV = (dE2 - dE1)/(T2-T1
Bomb Calorimeter:
?
Heat exchange = Z × ΔT
Z–Heat capacity of calorimeter system
ΔT– Rise in temp.
Heat changes at constant volumes are expressed in ΔE and Heat changes at constant
pressure are expressed in dH.
Enthalpies
Definitions
Example
Enthalpy of Formation
Enthalpy change when one mole of a given compound is formed from its elements
H2(g) + 1/2O2(g) → 2H2O(l), ΔfH = –890.36 kJ / mol
Enthalpy of Combustion
Enthalpy change when one mole of a substance is burnt in oxygen.
CH4 + 2O2(g) →CO2 + 2H2O(l), ΔcombH = –890.36 kJ / mol
Enthalpy of Neutralization
Enthalpy change when one equivalent of an acid is neutralized by a base in dilute solution.
H+ (aq) + OH– (aq) → H2O(l) ΔneutH = –13.7 kcal
Enthalpy of Hydration
Enthalpy change when a salt combines with the required number of moles of water to form specific hydrate.
CuSO4(s) + 5H2O (l) → CuSO45H2O,
ΔhydH° = –18.69 kcal
Enthalpy of Transition
Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.
C (graphite) → C(diamond), ΔtransH° = 1.9 kJ/mol
Enthalpy of Sublimation
Enthalpy change when one mole of a solid substance sublime at constant temp. and 1 bar pressure
CO2(S) → CO2(g) ΔtfusH° = 6.00 kJ/mol
Enthalpy of fusion
Enthalpy change when one mole of a solid melts
H2O(S) → H2O (l) ΔtsubH° = 73.00 kJ/mol
?The total enthalpy change of a reaction is the same, regardless of whether the reaction is completed in one step or in several steps.
According to Hess’s law: ΔH = ΔH1 + ΔH2
Applying Hess’s law we get ΔH1 + 1/2 ΔH2 + ΔH3 + ΔH4 + ΔH5 = ΔHf (MX) (Lattice energy)
Lattice energy: The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions.
Statement:
It is impossible to take heat from a hot reservoir and convert it completely into work by a cyclic process without transferring a part of it to a cold reservoirs.
Mathematically:
ΔS = qrev/T
ΔS is entropy change.
Entropy is the degree of randomness thus it increases with increase in randomness of particles of the system i.e. ΔS is positive for melting of ice.
At equilibrium, ΔS = 0
For a spontaneous process, ΔS > 0
Entropy change in an isothermal reversible expansion of a gas
Spontaneous Processes: These type of physical and chemical changes occur of its own under specific circumstances or on proper initiations. For example: Flow of liquids from higher to lower level.
ΔG = ΔH - TΔS
ΔG = nRT ln Keq
ΔG = nFEcell
At equilibrium, ΔG = 0
For spontaneous process, ΔG < 0
Average amount of energy required to break one mole bonds of that type in gaseous molecules.
H–OH(g) → 2H(g) + ½O(g) ΔH = 498 kJ
O–H(g) → H2(g) + ½O2 (g) ΔH = 430 Kj
ΔHO–H = (498 + 430)/2 = 464 kJ mol–1
W = R (T2 – T1) ln v2/v1
q2 = RT2 ln v2/v1
W = q2
Efficiency (h). h =
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