Reactions and Conditions Flashcards
Half-Life Equation
t1/2=Ln2/k
Iodine Peroxodisulfate reaction and how the catalyst works
Overall
S2O82- (aq) + 2I- (aq) = 2SO42- (aq) + I2(aq)
2Fe3+(aq) + 2I-(aq) = I2(aq) + 2Fe2+(aq)
2Fe2+(aq) + S2O82-(aq) = 2Fe3+(aq) + 2SO42-(aq)
The Fe3+ can react with either I- or S2O82- first
Fe3+ ion catalyst
Overall Order
Sum of individual orders
Processes in Haber Process
Diffusion to surface Adsorption to surface and weakens covalent bonds and enables the reaction to take place Reaction Desorption from surface Diffusion away
Nitration of Benzene
Reflux cHNO3 cH2SO4
Alkylation and acylation Benzene + chloroalkane/ acyl chloride
AlCl3 anhydrous
Oxidation of side chain in benzene
Reflux Alkaline KMnO4 then H2SO4
Electrophilic substitution with Phenol
Phenol + Br2 = Tribromophenol + 3HBr. White ppt
Occurs with phenylamine
Nitration of Phenol
dilute HNO3
concentrated HNO3 produced trinitrohenol
Formation of Amines with NH3
Haloalkane + Hot ethanolic NH3 reflux
Hot excess ammonia to avoid secondary and tertiary amines
Phenylamine
Nitrobenzene reduced to phenylamine cHCl reflux Sn catalyst
Diazotization
Phenylamine + Nitrous acid + HCl = diazonium chloride + 2H2O
HNO2 Nitrous acid
Less than 10 degrees C
Coupling diazonium ion with phenol
An alkaline solution of Phenol
Alkaline can react with OH on phenol to form O- Na+ section
Formation of Amides
Acyl chlorides +NH3/amines = Amide + HCl
Concentrated NH3 with acyl chlorides for primary amides
Amines can be used to produce a secondary amide
YOU CANNOT SIMPLY USE A CARBOXYLIC ACID AND AN AMINE. ONLY POLYMERS AND PROTEINS CAN DO THAT. CARBOXYLIC ACIDS ARE NOT REACTIVE ENOUGH
Hydrolysis of Amides
Heat under reflux HCl/NaOH
Hydrolysis of peptide bonds
With acid 6 moldm-3 HCl, 24 hours, 110 celsius
With alkaline NaOH and reflux
SimplifiedNernst equation
E= Eo + 0.059/z log [oxd]/[red]
Full Nernst Equation
E = Eo + RT/zF log Ox^n/Red^n
Standard Electrode Potential
The voltage of a half cell measured under standard conditions with a standard hydrogen electrode as the other half-cell
Enthalpy change of hydration
The enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution. STP
Enthalpy change of solution
The energy absorbed or released when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution. STP
Lattice Energy
Lattice energy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions.
Differences between the predicted structure of benzene
Does not undergo addition reactions like ethene
All the bond lengths are the same
The enthalpy of hydrogenation is lower than the expected
Reactivity of Phenol in regard to dilute vs conc HNO3 and Br2
Reacts readily with bromine water. Reacts with dilute nitric acid to form 2-nitrophenol or 4-nitrophenol. With concentrated HNO3 trinitrophenol is formed.
Why is phenol more reactive than benzene
The lone pairs on the oxygen atom overlap with the pi system and delocalise in the benzene ring. Electron density increases and thus more susceptible to attack from electrophiles.
Species that can be reduced multiple times
Iron, Vanadium
Volume and moles
n=V/24
Purpose of Salt Bridge
To complete the electrical circuit and so that ionic balance is maintained
Full Nernst Equation
E=Eθ + RT/zF ln[oxidised]/[reduced]
For when the temperature is not 25 degrees
Simplified Nernst Equation
E=Eθ + 0.059/z log10[oxidised]
For metal/metal ion electrodes
Copper Ions with hydroxide ions colour and what type of reaction is it?
What can the hydroxide be replaced for?
[Cu(H20)6]2+ + 2OH- = Cu(OH2)2(H2O)4 (s) + 2H2O
Precipitation reaction
No ligand is formed
Pale blue precipitate
The exact same reaction can occur by using ammonia. When excess ammonia has been added it forms the deep blue solution.
Cu(OH2)2(H2O)4 precipitate with concentrated ammonia solution
Cu(OH2)2(H2O)4 + 4NH3 = [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-
Pale blue precipitate disappears and a deep blue solution is formed
Copper ions with concentrated hydrochloric acid
[Cu(H20)6]2+ + 4Cl- = [CuCl4]2- + 6H2O
Yellow solution