Chemistry ch5 protons
If a reaction is endothermic, an increase in temp will shift equilibrium to RHS to decrease temp
Position of equilibrium: how far a reaction has gone & the proportion of prod to react in mixture
Kc=equilibrium constant for an equilibrium with concs in moldm-3 at a given temp
Kc=[PROD]/[REACT] Haber: N2+2H2>2NH3> mol-2dm+6
For an exothermic reaction, an increase in temp decreases Kc/Kp Endothermic, increase in temp increases Kc/Kp 1/2O2>(0.1)pwr1/2
Enthalpy comb: energy req to completely combust 1mol of atoms/molecules
Enthalpy of form: energy change when 1mol of a compounds is formed from its elements under standard conditions(25c & 1atm)
Enthalpy change=-mct/n
Enthalpy change of lattice formation: enthalpy change when 1mol of an ionic solid is formed from its gaseous ions under s.c (Na+ + Cl-)
Lattice breaking: changing 1mol of an ionic compound into its gaseous ions under s.c (Na+ +Cl->NaCl)/ has the same value of L.F but is always +ve
Standard 1st I.E: enthalpy change when 1mol of e- are removed from 1mol of gaseous atoms to form 1mol of gaseous ions
Standard E.A: enthalpy change when 1mol of gaseous atoms gain 1mol of e- to form 1mol of -ve ions Na+1/2Cl2>NaCl
L.E= keep HF same change sign of everything else
atmNa>1st I.E Na>atmCl>1stE.A Cl>L.E
More energy is needed to remove a 2nd e- due to increase in nuclear charge
The more -ve the enthalpy of formation, the more stable the compound
Solubility of ionic compounds: a substance will dissolve if enough energy is released when new bonds are made between it & water to make up for that used in breaking original attractions
Enthalpy change of hyd: enthalpy change when 1mol of gaseous ions is hydrated under s.c forming an infinitely dilute solution always -ve/exo as bonds form with water
Enthalpy change of sol: enthalpy change when 1mol of a substance dissolves in excess water under s.c to form an infinitely dilute solution +ve/endo, energy to break the lattice comes from electrostatic attraction between polar molecules
Hsol=Hlatt-Hhyd / Hhyd=Hhydcation+Hhydanion
Substances with a large +ve Hsol are likely to be insoluble ….
More charge means more attraction between ions> greater lattice breaking enthalpy (Mg2+O2->Na+F-)
The smaller the distance between ions the stronger the attraction between them giving a higher lattice enthalpy
For a compound to be soluble Hhyd must be greater than lattice breaking
D-block element: those in which the last e- are added to d-orbitals of the outer d-sub shell
Transition element: d-block elements that can form at least one stable ion with a partially filled d-orbital
Properties: metals,high melting point,from complex ions, act as catalysts, variable oxidation states
Ligands: (NH3,H2O)species with a pair of e- which can dative bond to a central ion/e- pair donor to form complex ions
Transition elements readily form complex ions as they can accept a pair of e- due to partially filled d-orbital
aq Cu (CuSO4) + aq ammonia (NH4OH) dropwise> Cu(OH)2 pale blue ppt
Excess ammonia> ammonia molecules displace water molecules> [Cu(NH3)4(H2O)2]2+ pale blue> deep blue
aq Cu> [Cu(H2O)6]2+(blue) +4Cl- (HCl)> [CuCl4]2- (yellow)+6H2O solution appears green
Colour in transition elements: e- absorb energy & jump from a low to a higher energy level, ligands split d-orbitals 2 to a higher 3 lower energy levels, colour is due to non-absorbed frequencies of light, transition metal ions are coloured as transition between split d-orbitals due to absorbed frequencies
Cu is white due to no d-d transition as it’s full all light reflects , if all absorbed appears black
different ligands surround a central ion result in diff energy split d-orbitals
| [Cu(H2O6]2+ | pale blue |
| [Cu(NH3)4(H2O)2]2+ | deep blue |
| [CuCl4]2- | yellow/green |
Reactions with OH-:
[Cr(H2O)6]3+ + 3OH-> Cr(H2O)3(OH)3 excess> +OH-> [Cr(OH)6]3- grey/green>bottle green
[Fe(H2O)6]2+ +2OH-> Fe(H2O)4(OH)2 & oxidises to Fe(H2O)3(OH)3 dark green>brown
[Cu(H2O)6]2+ +2OH->[Cu(OH)2(H2O)4]
[Cu(H2O)6]2+ +4NH3>[Cu(NH3)4(H2O)4] pale>deep blue
As catalysts: provide a surface for reactant molecules to come together with correct orientation & weakening their bonds by drawing e- density towards them
due to their partially filled d-orbitals which can accept ligands & variable oxidation states
They are important in industry as they make process more efficient, products are obtained quicker with less energy needed to produce e.g Fe catalyst 500c 250atm in Haber process/Fe in Hb
Cu [Ar] 4s1 3d10 / Cr [Ar] 4s1 3d5
Rate equations: can only be determined by experiment & can’t be deduced from the stoichiometric equation/
Order of reaction: sum of the powers to which the concs are raised in the rate equation
Change iny/change in x
Rate constant ‘k’: constant of proportionality in rate equation linking rate to conc of reactants
k value will increase if temp is increased reaction speeds up
Zero order: Rate=k in moldm-3s-1 1st order rate=k[A] n s-1 2nd order rate=k[A]2 in mol-1dm+3s+1 3rd order=mol-2dm+6s-1 k=rate/[A][B]
RDS is the limiting/slowest step & controls the overall rate of reaction
Reaction mechanism:describes the 1 or more steps involved in a reaction showing how various bonds are broken & made
The ‘OH-‘ion doesn’t appear in rate equation as it’s only involved in fast step & doesn’t affect overall rate
If the OH- we’re taking part in the slow step, increasing their conc would increase the rate / zero order reactants enter after RDS/ order of reaction= no. Of particles participating in RDS
Principle: measure how reactant or prod conc changes with time,keep all factors const e.g temp/pressure, draw graph of conc against time & measure gradient to find rate (t=0)
Entropy: freedom of movement possessed by molecules or atoms within a system
DeltaG=deltaH-(TdeltaS) / DeltaS=P-R /deltaG must be -ve for a reaction to be spontaneous
If deltaG is -ve then products predominate & KcKp is large/higher than 1
Always divide delta s by 1000/ T must be in kelvin> from degrees>kelvin +273
0=H(TS)>T=G/S