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Chemistry Summary  CH(1) Gases Factors affects on gas behaviour: 1)Temperature(T) (k)=(c+273) 2)Pressure(P)(atm)=(760torr(mmHg)) or (1.01325x105pa(n/m2)) 3)Volume(V)(liter) 4)Number of moles(n)(moles)=(mass/molar mass) Gas laws: 1)Boyle’s law:...

Chemistry Summary  CH(1) Gases Factors affects on gas behaviour: 1)Temperature(T) (k)=(c+273) 2)Pressure(P)(atm)=(760torr(mmHg)) or (1.01325x105pa(n/m2)) 3)Volume(V)(liter) 4)Number of moles(n)(moles)=(mass/molar mass) Gas laws: 1)Boyle’s law: studies the relation between V and P When temp(T) and no.moles(n) are constant The volume(V) and pressure(P) are inversely 𝟏 V 𝑽∝ 𝑷 𝑷𝟏 𝑽𝟐 = 𝑷𝟐 𝑽𝟏 P 2)Charle’s law: studies the relation between V and T When pressure(P) and no.moles(n) are constant The volume(V) and temp(T) are directly 𝑽∝𝑻 V 𝑻𝟏 𝑽𝟏 = 𝑻𝟐 𝑽𝟐 t(c)+273=T(k) G.R: why T, V are directly proportional? Because when T inc the kinetic energy of molecules inc So the spaces between them inc so V increase. 3)Avogadro’s law: studies the relation between V and n When pressure(P) and temp(T) are constant The volume(V) and no.moles(n) are directly 𝑽∝𝒏 V 𝒏𝟏 𝑽𝟏 = 𝒏𝟐 𝑽𝟐 n G.R: why when n inc V must inc? Because when n inc at const V The space between molecules dec so attraction force between molecules inc so P inc 4) ideal(general)gas law: PV=RnT (R is contant = 0.08206 litre.atm/nk) 5) combined gas law: 𝑷𝟏 𝑽𝟏 𝑷𝟐 𝑽𝟐 = 𝑻𝟏 𝑻𝟐 6) Daltion’s law for partial pressure: (𝒏𝟏 +𝒏𝟐 +⋯ )𝑹𝑻 Pt=P1+P2+…= 𝑽 𝑷𝟏 𝒏𝟏 = = 𝑿𝟏 (𝒎𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒈𝒂𝒔) 𝑷𝒕 (𝒏𝟏 + 𝒏𝟐 + ⋯ ) Ideal gas Van der waal (real)Non-ideal gas 𝑻 𝒉𝒊𝒈𝒉, 𝑽 𝒉𝒊𝒈𝒉, 𝑷 ≤ 𝟏𝟎 𝒂𝒕𝒎 𝑻 𝒍𝒐𝒘, 𝑽 𝒍𝒐𝒘, 𝑷 > 𝟏𝟎 𝒂𝒕𝒎 𝑷𝑽 = 𝒏𝑹𝑻 𝒂(𝒏𝟐 ) (𝑷 + ( 𝟐 )). (𝑽 − (𝒏𝒃)) = 𝒏𝑹𝑻 𝑽 Volume of molecules Attraction force Between molecules Can neglect volume of molecules Must consider volume of molecules Weak attraction force Strong attraction force Causes of deviation from ideal As we neglect volume of molecules with respect to P,V in ideal gas 7) gas diffusion: is gradual mixing between 2 or more different gases 8) gas effusion(u): Escape of gas through hole into the vaccum 𝟑𝑹𝑻 𝒖=√ 𝑴 T(temp k) R(const=8.314 j/mol.k) M(molar mass kg)  Ch(2) Corrosion Corrosion : is destruction of material properties due to reaction with environment Types of corrosion: 1) chemical(dry-direct-gas) 2) electro-chemical(indirect-wet): Zn Zn++ H+ Cl- HCl Anode: lose electrons Zn=> Zn+++2e Cathode: gain electrons 2H++2e=>H2  Electrochemical cell 1)Electrolytic: change electric energy to chemical energy (non-spontaneous) 2) Galvanic: change from chemical to electric (spontaneous) like: Daniel cell: e- Zn Cu ZnSo4 CuSo4 anode cathode Salt bridge: to balance the charge Types of cathodic environment Poor o2 rich o2 + O2+4H+ +4e=> acidic 2H +2e=> H2 2H2O Non-acidic 2H20+2e=> O2+2H2O +4e 2oH-+H2 => 4OH- Factors affect on corrosion: 1) Air: moist and humid-temperature-pollution 2) water: temperature-pollution- salts 3)soil: pollution- salts-water-gases 4) Gas: Acidity(SOx-NOx-H2S) Activity series  CH(3) Forms of corrosion Uniform Non-uniform(localized) 1)atmospheric: 1)erosion-corrosion: Corrosion due to atmosphere Due to mechanical wear Factors affect on it: and chemical reaction 1) type on environment Factors:1)turbulence 2)temperature 2)presence of suspended solids 3)time 3)hardness 4) type of material 2) stress- corrosion cracking 5) surface condition: Due to mechanical wear only Roughness-presence of Factors:1) time foreign materials-formation 2)stress /3)type surround of oxide layer 4)type of metal 2) Galvanic: contact between materials Factors:1) potential difference 2)distance from contact point 3)area (cathode/anode) Rusting of Iron: Anode: Fe=> Fe+++2e Cathode: O2 rich non-acidic O2 +2H2O+4e=>4OH+ Equation: O2 2Fe+O2+2H2O=>2Fe(OH)2==>2Fe(OH)3 H2 o Oxide layers Protective Non-protective Non-active active Adherent non-Adherent Non-porous porous Example:Al2O3 Example :Fe(OH)3  CH(4) Corrosion protection Methods of protection: 1) material selection 2) design: Avoid galvanic corrosion Avoid stress corrosion cracking Avoid atmospheric corrosion 3)coating: Anodic coating: more active cover less active Cathodic coating : less active cover more active  CH(5) Cement industry Concrete: Sand, rock, water, cement. Cement: Clay:Al2O3.2SiO.2H2O Lime stone: CaCo3 Scheme of cement manufacturing: 2) 3) storing and mixing limestone Grinding lime stone in ball and the clay in the silo mill 1) crushing lime stone in jaw crusher Grinding clay in different ball mill because they are different in hardness 5)clinker grey balls(1-10mm) 4) 6)cooling Refractory-bricks Reaction in the rotary kiln 1400c 7) grinding at ball mill with adding gypsum steel 8) cement Rotary kiln: 1 2 3 4 3to6 Rotary kiln properties: 1)It’s inclined from 3to6 degrees to slide the raw material slowly 2)it rotates to prevent coagulation of materials 3) it has a fan to get rid from unwanted gases Zones of the rotary kiln 1)drying zone(100-400’c): Evaporation of physical water 2)Pre-heating zone(400-700): (400-550):evaporation of chemical water (550-700):decomposition of clay:Al2O3/2SiO2 3) Calcination zone(700-1000): Decomposition of lime stone CaCO3CaO+CO2 4)burning zone(1000-1500): 2CaO+SiO2=> 2CaO.SiO2 3CaO+SiO2=>3CaO.SiO2 3CaO+Al2O3=>3CaO.Al2O3 4CaO+Al2O3+Fe2O3=>4CaO.Al2O3.Fe2O3 Impurities in soil Importance of Fe2o3: 1) clincker 2) fluxing material  CH(6) Fuel and Combustion Fuel: are materials combust(react with O2 +heat) To produce enough amount of energy Like: hydrocarbons (with Sulphur impurities) Main combustion reactions: 1)(complete) C+O2=> CO2+33.7 MJ/kg 2)(incomplete)C+(1/2)O2=>CO+10.5 MJ/kg G.R: why we prefer complete combustion than incomplete combustion? Because the complete produce CO2 less toxic than CO from incomplete , and complete produce more energy. 3)H2+(1/2)O2=>H2O+65.4mJ/kg G.R why H2 fuel is most important fuel? Because it produce H2O non-toxic and very high amount of energy. 4)S+O2=>SO2+9.11 MJ/kg G.R why it’s important to decrease Sulphur in fuel? Because it produce little amount of energy and produce SO2 toxic gas in which SO2=o2>SO3=rain H2O>H2SO4 If we got a molecule and we want to get needed O2 For example: Complete combustion of ( C) Balance equation C + O2 => Co2 No.moles |1 1 1 Molar mass |12 32 44 mass |12 32 44 divide on smaller mass |1 8/3 11/3 Which mean every 1kg C we need 8/3 kg O2 to produce 11/3 kg CO2 Balance the equation C2H6+(7/2)O2=>2CO2+3H2O No.moles | 1 7/2 2 3 Molar mass | 30 32 44 18 Mass | 30 112 88 54 Divide on smaller mass | 1 56/15 44/15 9/5 Balance equation |C3H7OH+(9/2)O2 => 3CO2+4H2O No.moles | 1 9/2 3 4 Molar mass | 60 32 44 18 Mass | 60 144 132 72 Divide on smaller mass | 1 2.4 2.2 1.2 Some examples: Air supply: Theoretical air: calculated from equation: [(%c*%complete*8/3)+(%C*%incomplete*4/3) +(%H*8)+(%S*1)-(%O2)]*(100/23)*fuel mass Excess(X): Used for calculating air only not to calculate O2 Extra air added to theoretical Actual air: (1+X)*theoretical For example: to get energy from fuel we use that equation: [(%c*%complete*33.7)+(%c*%incomplete*10.5) +(%H*65.4)+(%S*9.11)]*fuel mass