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Edexcel ALevel化学AS方程式总结

2022-01-05 | 编辑:y | 分享到
摘要:Edexcel ALevel化学AS方程式总结

  2022年1月 ALevel化学考试时间

  第一单元:1月12日

  第二单元:1月17日

  第三单元:1月24日

  第四单元:1月7日

  第五单元:1月14日

  第六单元:1月20日

  距离第一门化学考试还剩3天!

  今天带大家复习AS化学,第一单元、第二单元涉及到的化学反应方程式。

  第一单元

  ▄ Alkanes – with O2

  Complete combustion:CH4 + 2O2 → CO2 + 2H2O

  Incomplete combustion:CH4 + 1.5O2 →CO + 2H2O

  ▄ Alkanes – with Cl2, UV

  CH4 + Cl2 → CH3Cl + HCl

  Mechanism:Photochemical free radical substitution.

  ▄ Alkanes – improving the quality of fuels

  Catalytic Cracking:C10H22 → C5H12 + C5H10

  Conditions – Heat (600oC), Al2O3

  Isomerisation

  Reformation

Edexcel ALevel化学AS方程式总结

  ▄ Alkenes – with H2

  H2C=CH2 + H2 →CH3CH3

  Conditions - Heat, Ni

  ▄ Alkenes – with Br2

  CH3CH=CH2 + Br2→CH3CHBrCH2Br

  Colour change (orange to colourless)

  Mechanism:Electrophilic Addition

  ▄ Alkenes – with Bromine water

  CH3CH=CH2 + Br2 / H2O →CH3CHOHCH2Br

  Colour change (orange to colourless)

  Mechanism:Electrophilic Addition

  ▄ Alkenes – with HBr/dry/gas

  CH3CH=CH2 + HBr →CH3CH2BrCH3

  Major product Mechanism :Electrophilic Addition

  Explanation for major product:Secondary carbocationic intermediate is more stable than primary.

  ▄ Alkenes – with KMnO4 / H2SO4

  CH3CH=CH2 + [O] →CH3CHOHCH2OH

  Colour change (purple to colourless – H2SO4)

  Alkenes- Polymerisation

  Unit 2

  ▄ Group 2 metals – with H2O

  Mg:Mg + H2O → MgO + H2 (steam only)

  Ca,Sr,Ba:Ca + 2H2O→ Ca(OH)2 + H2

  ▄ Group 2 metals – with O2

  2Mg + O2 →2MgO

  ▄ Group 2 metals – with Cl2

  Mg + Cl2 →MgCl2

  ▄ Group 2 Oxides – with H2O

  MgO + H2O →Mg(OH)2

  ▄ Group 2 Oxides – with acids

  MgO + H2SO4→ MgSO4 + H2O

  ▄ Thermal stability

   Group 1 Carbonates:All stable to heat except for Li2CO3

  Group 2 Carbonates:MgCO3→ MgO + CO2

  Group 1 Nitrates

  Li:   2LiNO3 → Li2O + 2NO2 + ?O2

  Na, K, Rb, Cs :NaNO3→NaNO2 + ?O2

  ? Group 2 – Nitrates

  Mg(NO3)2→ MgO + 2NO2 + ?O2

  ▄ Solubility

  Group 1 and 2 Sulphates

  Decreases down the group - BaSO4 is insolubl

  Group 1 and 2 Hydroxides

  Increases down the group - Mg(OH)2 is insoluble

  ▄ Group 7 elements – with H2O

  Cl2 + H2O→HCl + HClO

  ▄ Group 7 elements – with NaOH

  In Cold Dilute Alkali: Cl2 + 2OH- →Cl- + ClO- + H2O  Reaction type - Disproportionation

  In Hot Conc. Alkali: 3Cl2 + 6OH- → ClO3- + 5Cl- + 3H2O Reaction type - Disproportionation

  ▄ Group 7 – Displacement reactions

  Cl2(g) + 2Br-(aq) → Br2(aq) + 2Cl-(aq)

  Observations:Green gas → orange solution

  ▄ Halide ions – with conc H2SO4:

  NaCl(s) + H2SO4 →NaHSO4(s) + HCl(g)

  Observations:Steamy white fumes

  NaBr(s) + H2SO4→NaHSO4(s) + HBr(g)

  2HBr(g) + H2SO4 → Br2(g) + SO2(g) + 2H2O(l)

  Observations Steamy white fumes and orange fumes

  NaI(s) + H2SO4→Products HI(g), I2(g) + H2S(g)

  NaI(s) + H2SO4→NaHSO4(s)+HI(g)

  Followed by oxidation of HI(g):

  2HI(g)+H2SO4→I2(g) + SO2(g)+2H2O(l)

  And:

  6HI(g)+ H2SO4→3I2(g)+S(s)+4H2O(l)

  And:

  8HI(g)+ H2SO4→4I2(g)+ H2S(g) +4H2O(l)

  Observations:Steamy white fumes and purple fumes

  ▄ Test for halide ions

  Ag+(aq) + Cl-(aq) →AgCl(s)

  Observations: White ppt – soluble in dilute ammonia

  ▄ Halogenoalkanes - with aqueous OH-

  CH3CH2Br + OH-→ CH3CH2OH + Br-

  Mechanism : Nucleophilic substitution (Sn1 or Sn2)

  ▄ Halogenoalkanes with ethanolic OH-

  CH3CH2Br + OH-→H2C=CH2 + Br- + H2O

  Mechanism: Elimination

  ▄ Halogenoalkanes - with CN-

  CH3CH2Br + CN-→CH3CH2CN + Br-

  Mechanism: Nucleophilic substitution

  ▄ Halogenoalkanes - with aqueous silver nitrate

  CH3CH2Br + H2O + Ag+→ CH3CH2OH + AgBr

  Fastest halogenoalkane :Iodo

  Explanation: C-I bond is weaker than C-Br and C-Cl

  ▄ Halogenoalkanes - with NH3

  CH3CH2Br + NH3→CH3CH2NH2 + HBr

  Conditions: Conc NH3 / heat / closed vessel

  ▄ Preparation of halogenoalkanes

  ①Chloroalkanes from alcohols :

  CH3CH2OH + Cl- → CH3CH2Cl + OH-

  Conditions: H2SO4 / NaCl / heat

  ②Bromoalkanes from alcohols:

  CH3CH2OH + Br-→CH3CH2Br + OH-

  Conditions: NaBr / H3PO4 / Heat

  Not H2SO4 / NaBr / heat as Br2 will form

  ③Iodoalkanes from alcohols:

  CH3CH2OH + I- → CH3CH2I + OH-

  Conditions: PI3 or P / I2

  Not H2SO4 / NaI / heat as I2 will form

  ▄ Alcohols – 1o Partial Oxidation:

  CH3CH2OH + [O] →CH3CHO + H2O

  Conditions: Distil product as it is formed

  ▄ Alcohols – 1o Complete Oxidation:

  CH3CH2OH + 2[O] →CH3CO2H + H2O

  Conditions: Heating under reflux

  ▄ Alcohols – 2o Oxidation:

  CH3CHOHCH3 + [O] →CH3COCH3 + H2O

  ▄ Alcohols – Dehydration:

  CH3CH2OH→ H2C=CH2 + H2O

  Conditions: Heating under reflux / NaOH – Aqueous

  ▄ Alcohols – Reaction with sodium:

  CH3CH2OH + Na →CH3CH2O-Na+ + H2

  Observation: Colourless effervescence 更多详情咨询老师即可~


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