Sunday 21 August 2016

Class XI: Thermodynamics (Chapter 6) notes

THERMODYNAMICS
It is the branch of chemistry which deals with energy changes in the chemical reactions and their feasibility.
System – Part of universe in which observations are made.
Surroundings -  Part of universe excluding system = Universe – System
Types of System –
a)         Open System -  There is exchange of energy and matter between system and surroundings. e.g. Open beaker.
b)         Closed system -  There is no exchange of matter but exchange of energy is possible between system and surroundings. e.g. presence of reactants in a closed vessel made up of conducting material.
c)         Isolated system -  There is no exchange of energy as well as matter between system and surroundings. e.g. Presence of reactants ina thermos flask or any insulated closed vessel.
State of the system -  The system must be described in order to make useful calculations by specifying quantitatively each of the properties such as pressure (P), volume (V) and temperature (T) as well as the composition of the system.
The state of the system is described by its measurable or macroscopic (bulk) properties.
State functions or state variables -  Those properties of a system which depend only on the state (initial and final state) of the system and do not depend upon the path of the system e.g. P.V.T.
Path functions -  Those properties of a system which do not depend upon the state (initial and final state) of the system but depend only upon the path of the system e.g. work done.
Types of processes :
a)      Isothermal process – that occurs at constant temperature.
b)      Isobaric process – that occurs at constant pressure
c)      Isochoric process – that occurs at constant volume
d)     Adiabatic  process – that occurs at constant heat i.e., there is no heat exchange between system and surroundings.
Internal energy (U or E) -  It is the energy possessed by the system. It depends upon –
a)      Heat passes into or out of the system
b)      Work is done on or by the system
c)      Matter enters or leaves the system
Internal energy is a state function as it depends upon initial and final state of the system
We cannot calculate the exact value of internal energy as it is the sum of many different kinds of energy such as mechanical energy, chemical energy, electrical energy, etc. whose values are difficult to calculate separately.
For adiabatic process, ΔU = U2 – U1 = Wad
FIRST LAW OF THERMODYNAMICS –
-           According to it, energy of an isolated system is constant
-           Energy can neither be created nor be destroyed
Mathematically, ΔU = q + w
Where U = change in internal energy, q = heat absorbed by the system, W = work done on the system.
Proof – Let initial internal energy = U1, final internal energy = U2. If ‘W’ work is done on the system and ‘q’ heat is absorbed. Then,
            U2 = U1 + q + W
            U2 – U1 = q + W
            ΔU = q + W
Sign Conventions –
Work done by the system = (.), Work done on the system = (+)
Heat absorbed  → q = (+) and U = (+), Heat evolved → q = (–) and U = (–)
During expansion W = (–). During compression W = (+)
For adiabatic process, ΔU = – Wad
For thermally conducting process, ΔU = – q
For closed system, ΔU = q + w
If the work is done by the system, ΔU = q – W  →  ΔU = q – W
APPLICATIONS :
1.         Work – (I = length, A = area) For 1 mole of a gas
            Change in volume  V = 1 × A
            As P = Force / area                             
            →        F on piston = Pext . A

            Work done = F × distance = P × A × I = P × (–ΔV).           Therefore, W = – P Δ V
2.         
W = – 2.303 nRT log V2 / V1
Also,  As Boyle’s law  :  P1/P2 = V2/V1
→   W = – 2.303 nRT log P1/P2

 Now  cases :
a)      For isothermal process (T = constant) i.e., for isothermal expansion of ideal gas into vacuum,
            W = 0 and q = 0  →  ΔU = 0
b)      For isothermal irreversible change, q = W = P (V2 – V1)
c)      For isothermal reversible change, q = – W = nRT InV2 / V1
d)     For adiabatic change, q = 0 → ΔU = Wad
Enthalpy (H) :
            It is the heat absorbed or heat evolved by the system.
            As U = qp – P Δ V (expansion)
            qp = ΔH = ΔU + P Δ V
Proof :  ΔU = q – PΔV
            U2 – U1 = qP – P (V2 – V1) at constant P
            qp = (U2 + PV2) – (U1 + PV1)
            Therefore, H = U + PV
            qp = H2 – H1 = ΔH ( it is also known as state function)
            Now, if H = U + PV
            ΔH = ΔU + ΔpV + pΔV
            As H = qp (at constant Pressure) i.e. ΔP = 0
            Therefore, ΔH = ΔU + PΔV
ΔH = (+) for endothermic reaction, ΔH = (.) for exothermic reaction.
Derivation ΔH = ΔU + ΔngRT
Proof  ΔH = ΔU + PΔV = ΔU + P (V2 – V1)
Now, if PV = nRT. So,  PV1 = n1RT  and PV2 = n2RT
ΔH = ΔU + n2RT – n1RT = ΔU + (n2 – n1) RT
ΔH = ΔU + ΔngRT
Where Δng = n2 – n1 for gaseous state = np – nr
–          Extensive Property – Property whose value depends upon the quantity of matter contained in the system. e.g. mass, volume, internal energy, enthalapy, heat capacity. etc.
          Intensive Property -  Property whose value does not depend on the quantity or size of matter present in the system. e.g. temperature, density, pressure
          Heat Capacity (C) – It is the amount of heat required to raise the temperature of a substance by IC.
            Specific Heat Capacity  (CS) –Amount of heat required to raise the temperature of 1gram of a substance by 1°C (1K).
            Molar Heat Capacity (Cm) –Amount of heat required to raise the temperature of 1 mole of a substance by 1°C.
                                                            Q = m C ΔT      where ΔT = T2 – T1

Relationship between CP and CV for ideal gases –
For constant volume, qV = CVΔT = ΔU  and  at constant P, qP = CPΔT = ΔH
For 1 mole of a gas, ΔH = ΔU + R ΔT
On putting the values of ΔH and ΔU, we get       CPΔT = CVΔT + R ΔT
VP = VC + R     Therefore,  CP – CV = R
We can measure energy changes (ΔH and ΔU) associated with chemical or physical processes by an experimental technique called calorimetry. The process is carried out in calorimeter immersed in a known volume of a liquid.
Knowing that heat capacity of liquid in which calorimeter is immersed and heat capacity of calorimeter, it is possible to determine the heat evolved in the process of measuring temperature changes under two different values,  a) at constant volume,  b) at constant pressure.
                                                            ΔH = HP – HR
Enthalpy changes -  Reactants  →  Products
The enthalpy changes accompanying the reaction is known as reaction enthalpy (ΔH)
Standard enthalpy of reaction -  The reaction enthalpy changes for a reaction when all the participating substances are in their standard states [standard temperature and pressure (1 bar) [Δ,H0]
DEFINITIONS
1.         Standard enthalpy of fusion  (ΔfusH0) -  It is the heat evolved or absorbed by the system when one mole of a solid substance melts in standard state.
2.         Standard enthalpy of vapourisation (vapH0) -  Amount of heat require to vapourize mole of a liquid at constant temperature and under standard pressure (1 bar)
3.         Standard enthalpy of sublimation (ΔsubH0) -  Amount of heat absorbed or evolved when sublimes at constant temperature and standard pressure (1 bar)
            ΔH is directly proportionsal to the intermolecular interactions in substance
4.         Standard enthalpy of formation (ΔtH0) -  Amount of heat absorbed or evolved when 1 mole of compound is formed from its elements in their most stable states (reference state) at 25°C and 1 bar.
Example - H2 (g) + 1/2O2 (g) → H2O (1) ΔtH0 = 285.8KJ/mol; C (graphite) + 2H2 (g) → CH4 (g)
5.         Standard enthalpy of combustion – (ΔCH0) Enthalpy change when 1 mole of a substance is burnt in presence of air completely.
6.         Enthalpy of atomization (ΔaH0) – It is enthalpy change on breaking 1 mole of bonds completely to obtain atoms in the gas phase.
Example -    CH4 (g)   →   C (g)  + 4H (g)                     Na (s)  →  Na (g)
7.         Bond enthalpy (ΔbondH0) -  Energy is required break a bond and released to form a bond.
            The amount of heat absorbed or released due to break or form one mole of bonds of reactants is known as bond enthalpy.               ΔrH0 = BER – BEP
8.         Enthalpy of solution (ΔsolH0) – It is the enthalpy change when 1 mole of a substance is dissolved in specified amount of solvent.

 9.         Lattice enthalpy – It is the enthalpy change which occurs when 1 mole of an ionic compound dissociates into its ios in gaseous state (ΔLattice H or U)
10.       Heat of hydration -  Amount of enthalpy change when 1 mole of the anhydrous salt combines with required number of moles of water so as to change into the hydrated salt.                                         CuSO4 + aq → CuSO4.5H2O
11.       Heat of neutralization of an acid by a base -  It is the heat change when 1 gram equivalent of the acid is neutralized by a base, the reaction is carried out in dilute aqueous solution.
            When 1 gram equivalent of HCI is neutralized by NaOH or vice-versa, 57.1 kJ of heat is produced
–          Heat of neutralization is taken for 1 gram equivalent of acid and base because neutralization involves combination of 1 mole of H+ ions with 1 mole of OH ions. 1 gram of any acid on complete dissociation gives 1 mole of H+ ions but 1 mole of an acid may not give 1 mole of H+ ions.
Example -  1 mole H2SO4 → 2 moles of H+ ions on complete dissociation,
            1 gram equivalent H2SO4  → 1 mole of H+ ion

Hess’s law of constant heat summiation –
According to it, if a reaction takes place in several steps then its standard enthalpy is the sum of the standard enthalpies of the intermediate reaction into which the overall reaction may be divided at same temperature.                                                         OR
According to it, if a reaction takes place in one step or in many number of steps, the amount of energy released or absorbed (enthalpy change) always remain constant at constant temperature.
Example -        C (graphite, s) + O2 (g) → CO2
            Step 1 : C (s) + ½ O2 → CO (g);   Step 2 :  CO (g) + ½ O2 →  CO2 (g)  i.e. ΔH = H1 + H2


No comments:

Post a Comment