TyroCity: Chemistry 12 Notes

Bond Energy Or Bond Enthalpy

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Bond energy of a bond is amount of energy required to break 1 mole of a bond in gaseous state or energy released when one mole of a bond is formed in gaseous state.

In terms of bond energy,
Hreaction = Heat absorbed – Heat released = bond energies of reactants – bond energies of products

Estimate the enthalpy change for the reaction

Sol:
The amount of energy released during formation of 2moles of H-Cl bond = 2 × 430 = 860 KJ

H reaction = Energy absorbed – Energy released
= 618 – 860
= -182 KJ

Calculate the enthalpy change for the hydrogenation of ethane gas.

Given, bond energy of H -H bond = 435 KJ mole-1

Bond energy of C – H bond = 413 KJ mole-1

Bond energy of C -C bond = 347 KJ mole-1

Bond energy of C=C bond = 615 KJ mole-1

The amount of energy absorbed during reaction
= 413 × 4 + 615 + 435
= 2702 KJ

Energy released during reaction
= 6 × 413 + 347
= 2825 KJ

H reaction = 2825 – 2702 KJ = -123 KJ

Half Life (t1/2)

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Half Life of a reaction is time required to reduce the initial concentration to half. The half-life of a reaction depends on the order of reaction. The variation of half-life with order is given as:

Integrated rate law expression

For Zeroth order reaction

For first order reaction

Integrating on both sides, we get,

So, half-life of first order reaction is independent of initial concentration.

Numerical
The half-life of a first order reaction is 50 mins. Calculate the time required to complete 75% of the reaction.
Given,

Again,
initial concentration (a) = 100 (let) then
at time t,

Concentration left (a-x) = 100-75 = 25
We have,

Calculate the half period of first order reaction when rate constant is 5 year-1.
We have,
For 1st order reaction
t1/2 = 0.693/5
= 0.1386 year

Laplace – Lavoiser Law Of Thermo Chemical Reaction

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

The amount of heat for the reverse of a reaction will be equal but will have opposite sign.

Thermodynamic Processes

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Isothermal process
Any physical or chemical process in which the temperature remains constant during the state change is called isothermal process.
Here, ∆T=0

Adbiatic process
Any physical or chemical process which takes place without flow of heat in or out of system during the state change is called adbiatic process.
Here, ∆Q=0

Isobaric process
Any physical or chemical process in which the pressure of the system remains constant during the state change is called isobaric process.
Here, ∆P=0

Isochoric process
Any physical or chemical process in which the volume remains constant during the state change is called isochoric process.
Here, ∆V=0

Enthalpy (H)

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

It is the sum of internal energy and product of pressure and volume.

i.e. H=E+PV…..(1)
Here, E, P & V all are state function so H is also the state function.
Being a state function it depends upon initial and final state and also its absolute value cannot be determined however change can be measured.
Now, ∆H=Hp-Hr…………(2)
Where, Hp= enthalpy of product
Hr= Enthalpy of reactant
∆H at constant pressure
We have, from first law of thermodynamics;
Q=∆E+W
Q=∆E+P∆V….(1)
Also, H=E+PV……..(2)
At constant pressure, Q=Qp
So, equation (1) becomes,

Qp=Ep-Er+P(Vp-Vr)
Qp=Ep+PVp-(Er+PVr)
Qp=Hp-Hr
Qp=∆H
It means, heat of reaction at constant pressure is equal to change in enthalpy of reaction.

Copper Sulphate(CuSO4)

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Copper Sulphate pentahydrate (CuSO4 5H2O) is commonly known as Blue Vitriol. It is obtained in lab by dissolving CuO, Cu(OH)2 or CuCO3 in dil H2SO4. The solution on crystallization gives blue triclinic crystals of CuSO4 5H2O.

CuO + H2SO4 → CuSO4 + H2O
Cu(OH)2 + H2SO4 → CuSO4 + H2O
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
CuSO4 crystallization → CuSO4 5H2O
Blue Vitriol
In large scale, CuSO4 is obtained by reacting scrap copper with hot and dilute H2SO4 in presence of wir.
2Cu + 2H2SO4 + O2 → 2CuSO4 + H2O
Hot & dil

Properties

Hydrated copper sulphate is blue crystalline solid while anhydrous CuSO4 is white amorphous solid.
It is soluble in water.
Action of heat:
Blue vitriol on heating losses 4 molecules of water at 1000c to give monohydrate form and at 2300c to give anhydrous CuSO4.
CuSO4 5H2O 100 → CuSO4H2O 230 → CuSO4
Blue blueish white white

Anhydrous CuSO4 gives back blue vitrial in contact with water so anhydrous CuSO4 is used to detect water.
Anhydrous CuSO4 on further heating decompose forming CuO.
CuSO4 above 230 → CuO + SO3
Black

Action of water:

CuSO4 gives acidiec salt solution due to hydrolysis of Cu++ ion as it. Is salt of weak base and strong acid.
CuSO4 (aq) → Cu++ + SO4
Cu++ + 2H2O → Cu (OH)2 + 2H+

Reaction with KI:

KI reduces CuSO4 to cuprous iodide which appears as white ppt while KI is oxidized to I2

Reaction with NH3:

when NH3 is added to CuSO4 solution, a bluish white ppt of Cu(OH)2 is formed which dissolves in excess ammonia forming tetra amine copper (II) sulphate which is commonly known as Schweizer’s reagent.
CuSO4 + NH4OH → Cu(OH3)4SO4
Bluish white
CuSO4 + NH4OH + (NH4)2SO4 → [Cu(NH3)4]SO4 + 4H2O
Tetra amine copper (ii) sulphate (deep blue)

Uses

It is used as electrolyte in purification of copper & electroplating of Cu.
It is used in making schweizer’s reagent used for mfg of paper.
It is used as pesticide in controlling aphids and fungal growth.

Mohr’s Salt Or Ferrous Ammonium Sulphate FeSO4(NH4)2 SO4.6H2O)

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Hydrated ferrous ammonium sulphate is commonly known as mohr’s salt .It is prepared by crystallization of equimolar solution of FeSO4 and (NH4)2SO4.

FeSO4(aq) + (NH4)2SO4 crystallization → FeSO4 (NH4)2SO4.6H2O.
Mohr’s salt
This double salt is more stable than FeSO4 alone and is used as reducing agent in redox fitration. During redox reaction Fe+2 ions mohr’s salt is oxidised to Fe+3 ions.

(Properties same as FeSO4. 7H2O)

Determination Of Order Of Reaction By Initial Concern Method

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

For the reaction,

2NO + Cl2 → 2NOCl

Following datas were obtained:

Determine the order of the reaction with respect to NO, Cl2 and overall order. Write the rate law expression and determine the value of rate constant.

Soln:

Let the order w.r.t. NO and Cl2 be in and n. The rate low expression will be,

Dividing eqn (ii) by (i) we get,

Dividing eqn (iii) by (ii), we get,

order w.r.t. NO = 2

order w.r.t. Cl2 = 1

Overall order of r × n = 2 + 1 = 3

Then, the rate law expression,

Rate = k [NO]2 [Cl2]

Putting the value of m and n in equation (i), we get,

The experimental data for the reaction

Determine the order of the r × n wrt A and B and overall r × n. Write the rate law expression and determine the value of rate of constant.

Solution:

Let the order w.r.t. A and B be m and n. The rate law expression will be,

Dividing eqn (ii) by (i) we get,

Ferric Chloride (FeCl2 or Fe2Cl6)

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Preparation:

Anhydrous FeCl3 is obtained by passing dry Cl2 gas over red hot iron.
2Fe + 3Cl2 → 2FeCl3

Hydrated ferric chloride is obtained by dissolving iron in aquaregia or by dissolving Fe2O3 or Fe (OH)3 in HCl. On crystallization FeCl3 solution gives yellow crystals of FeCl3 6H2O.
2Fe + 9HCl + 3HNO3 → 2FeCl3 + 3NOCl + 6H2O
Fe2O3 + 6HCl → 2FeCl3 + 3H2O
Fe (OH)2 + 6HCl → FeCl3 + 3 H2O

FeCl3 (ag) crystallization → FeCl3.6H2O
Yellow crytals

Properties:

Anhydrous FeCl3 is black amorphous solid while hydrated FeCl3 is yellow crysaline solid.
At lower temperature (4000c) the vapor density of ferric chloxde corresponds to the molecular formula Fe2Cl6. Fe2Cl6 is chlorine bridged dimeric structure.

At higher temperature (7000c), Fecl3 decompose forming ferrous chloride.
FeCl3 above 700 → 2FeCl2 + Cl2

Hydrolysis
The aqueous solution of FeCl3 is acidic due to hydrolysis to form weak base and strong acid.
FeCl3 + 3H2O → Fe(OH)3 + 3HCl
Weak base strong acid
Reaction with potassium Ferro cyanide:
Ferric chloride reacts with potassium ferro cyanide forming a prussiara blue colouration of ferric ferro cyanide.
4FeCl3 + 3 K4 [Fe (CN)6]0 → Fe4 [Fe (CN)6]3 + 12KCl
Prussian blue ferroc ferro cyanide
Reaction with ammonium thiocyanate:
Ferric chloride gives a blood red coloration of Ferric thiocyanate with ammonium thiocyanate.
FeCl3 + 3 NH4CNS → [Fe (CNS)3] + 3NH4Cl
Ferric thiocyanate
Blood red
Reaction with ammonium hydroxide:
Ferric chloride gives a reddish brown ppt of ferric hydroxide with Ammonium Hydroxide solution.
FeCl3 + 3NH4OH → Fe (OH)3 + 3NH4Cl
Ferric hydroxide
Reddish brown

Uses:

It is used in medicine as astrigents and antiseptic.
It is used as mordant in dying.
It is used as a catalyst in friedal craft’s reaction.
It is used to etch metals like Cu, Ag, during making block.

Free Energy Change And Net Useful Work

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Work other than pressure volume work (PΔV) is net useful work. So, total work of a process is,

W = Wnet + PΔV

Wnet = W – PΔV

From Gibb’s helmuntz equation,

ΔG = H – TΔS ……………. (i)

We have,

H = E + PΔV
S = Q /T

TΔS = Q

Putting the values of H and TS in eqn (i), we get
ΔG = ΔE + PΔV – Q

From first law of thermodynamics
Q = ΔE + W

Putting the value of Q in above enq we get,
G = Δ E + PΔV – ΔE – W
G = PΔV – W
G = – (W – PΔV)
G = – Wnet
Wnet is – ΔG
Thus, the net useful work is the amount of decrease in free energy of a system under constant temperature and pressure.

Relation between free energy change and cell potential

The net work done by galvanic cell in carrying charge is

W = charge × potential difference = nf Ecell

We have,
ΔG = -Wnet
ΔG = -nfEcell

Relation between G and equilibrium constant

The variation of G is given by Nernst equation as,
G = ΔG + RT in Q

Where, Q = reaction quotient

Where, the reaction is at equilibrium
g = k (equilibrium constant)
ΔG = O
O = G° + RT in k
ΔG° = -RT InK
ΔG = -2.303 RT logK

The overall reaction for corrosion of iron by oxygen is,

4Fe (s) + 3O2 (g) → 2Fe2O3 (s) Rust

Calculate the free energy change and equilibrium constant for this reaction at 250°c.

Given,

ΔS° and ΔH° = -543 JK-1 and -1652 KJ mol-1
= -1652000 Jmol-1

Δ G° = Δ H° – TΔS°
= 1652000 – 298 × (-543)
= -1490186 J

Again,

ΔG° = -RTInK
= -2.303 RTlogK
1490186 = – 2.303 × 8.314 × 298 × log K
logK = 261.17
K = 10261.17
K = 1.48 × 10261.17

Calculate the free energy at standard condition for galvanic cell having following cell reaction.

2Al (s) + 3 Cu ++ (aq) → 2Al+++ (aq) + 3Cu (s)

E° Cu++/Cu = 0.34 V → E° Al+++/A1 = -1.66V

So, the standard cell potential,

E°cell = E°cathode – E°anode

= E°cu++/cu – E°Al+++/A1
= 0.4 – (-1.66)
= 2V

Again, Δ G = -nfEcell
= – 6 × 96500 × 2
= 1.158 × 106J

Heat Treatment Of Steel

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

Steel can be given desire property by heat treatment there are three different heat treatment.

1. Annealing:
This method is done to obtain soft steel. When steel is heated to red hot (about 11000c) and then cooled & lowly, steel becomes soft and the process is called annealing. In annealed steel, carbon is present in free state.

2. Quenching or Hardening:
This process is done to obtain hard steel and brittle steel. When steel is heated to red not and then cooled rapidly by plunging into water or oil, gives hard and brittle steel and the process is called Quenching. In quenched steel, carbon is present in combined form as iron carbide (Fe3C) or Cementite.

3. Tempering:
This process is done to obtain hard and malleable steel Quenched steel is heated below red hot (about 7000c ) and then cooled slowly to obtain mild steel and the process is called tempering. In tempered steel, carbon is present in both combined and free state.

Sign Convention Of Heat And Work

Chemistry 12 Notes — Sun, 08 Apr 2012 05:41:42 +0000

1. Heat (q)
Case I:
When heat is absorbed by the system, q = +ve.

Case II:
When heat is released by the system, q = -ve.

2. Work (w)
Case I: (Work of Expansion)
When work is done by the system, w = +ve.

Case II: (Work of Compression)
When work is done on the system, w = -ve.