Topic 7: Equilibrium Flashcards
reversible rxn
rxn in which products can react with one another under suitable conditions to produce rxns
dynamic equilibrium
- conc. of reactants and products don’t change over time
- backward & forward rxns are occurring simultaneously
- with the rate of both reactions equal
physical equilibrium
equilibrium set up in physical processes
e.g. melting of solids, evaporation of bromine
chemical equilibrium
equilibrium set up in chemical processes
e.g. decomposition of CaCO3
what happens when bromine is placed in a sealed container at room temp?
- bromine is a volatile liquid with a boiling point close to room temp
- significant number of particles will have enough energy to evaporate
- concentration of bromine vapour increases in the closed system
- as the vapour can’t escape, many of its particles will condense back
equilibrium law
at a given temp, the ratio of conc of products (raised to the power of molar coefficients) to conc of reactants (raised to the power of molar coefficients) is a constant
this constant is called Kc
NOTE: in aqueous rxns, the conc of the solvent won’t appear in the equilibrium constant expression as its conc won’t change
Meaning of Kc’s value
- if Kc > 0 at a given temp, products are favoured over reactions
- if Kc
situation where Kc won’t apply
non-reversible rxns
Reaction quotient
- denoted by Q
- measures relative amount of reactants & products during a rxn at a particular point in time
- helps in figuring out which direction a rxn is likely to proceed using pressure/conc of reactants
difference between Kc and Q
Kc: describes rxn at equilibrium
Q: describes rxn not at equilibrium
Meaning of Q’s value
- if Kc > Q: rxn is in favour of products (forward)
- if Kc = Q: equilibrium
- if Kc
effect of inverting the rxn on Kc
inverts the value of Kc
1/Kc
effect of doubling rxn coefficients on Kc
square the expression Kc
Kc^2
effect of halving reaction coefficients
square root of Kc
Kc^(1/2)
effect of adding together 2 rxns on Kc
multiply the 2 Kc values
Kc1 x Kc2
Le Chatelier’s principle
a system of equilibrium, when subjected to change, will respond to minimise the effect of the change
Factors affecting equilibrium
- concentration
- pressure
- change in temp
- catalyst
Factors affecting equilibrium: concentration
- increase in product conc: backwards reaction favoured
- decrease in product conc: forwards reaction favoured
- increase in reactant conc: forwards reaction favoured
- decrease in reactant conc: backwards reaction favoured
factors affecting equilibrium: pressure
- increase in pressure = shift towards side with less no of moles
- decrease in pressure = shift towards side with higher no of moles
factors affecting equilibrium: change in temp
- increase in temp = shift to endothermic side (ΔH = +tive)
- decrease in temp = shift to exothermic side (ΔH = -tive)
- due to temp having diff effects on forward/backward rxns due to diff activation energies
factors affecting equilibrium: catalyst
- catalysts have no overall effect on equilibrium position
- it only speeds up the attainment of equilibrium
Haber’s Process
N2 (g) + 3H2 (g) 2NH3 (g) [ΔH = -93 kJ/mol]
conditions for Haber’s process, Contact process, methanol production
Favourable:
- low temp
- high pressure
Actual:
- moderate conditions
450 degrees and 200 atm
Why are moderate conditions taken instead of favourable conditions?
low temp = slow rate of rxn = more time taken to produce yield = inefficient
high pressure = risk of explosion = costly to rebuild infrastructure
Contact process
- Combustion of sulphur-containing compounds
S(s) + O2 (g) -> SO2 (g) - Oxidation of SO2 to SO3
2SO2 (g) + O2 2SO3 (g) [ΔH = -196 kJ/mol]
catalyst: vanadium (V) oxide - SO3 + H2SO3 (concentrated) -> H2S2O7 (oleum)
H2S2O7 + H2O -> H2SO4
Alternatively,
SO3 + H2O -> H2SO4
But it’s highly exothermic and produces corrosive mist so it’s generally not used.
Production of Methanol
CO (g) + 2H2 (g) CH3OH (g) [ΔH = -90 kJ/mol]
- catalyst: Al2O3 OR CuO3 OR ZnO3
