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 and forward rxns occur simultaneously
- rate of both reactions are equal
physical equilibrium
equilibrium set up in physical processes
e.g. vaporization ↔ condensation
chemical equilibrium
equilibrium set up in chemical processes
e.g. decomposition of CaCO3
closed system
- system in which neither matter nor energy is gained or lost from the system
- this allows an equilibrium to be reached
what happens when bromine is placed in a sealed container at room temp?
- most of its particles will have enough energy to evaporate
- but this causes conc of bromine vapour to increase in the closed system
- vapour can’t escape so many of its particles will condense back
- this is possible because bromine is a volatile liquid with a b.pt close to room temp
equilibrium constant
symbol: Kc
- tells us the equilibrium position
- i.e. the proportion of reactants and products in the equilibrium mixture
- at a given temp, ratio of conc of products (raised to the power of molar coefficients) : conc of reactants (raised to the power of molar coefficients) is constant
NOTE: in aqueous rxns, the conc of the solvent won’t appear in the equilibrium constant expression as its conc doesn’t change
homogeneous reaction
reaction in which all constituents (reactants and products) are of same state
Meaning of Kc’s value
- if Kc > 0, products are favoured over reactions
- vice versa
- if Kc > 1, the reaction almost goes to completion
- if Kc < 1, the reaction is barely proceeding
e. g. if Kc = 4, conc. of products is 2x that of reactants - DOES NOT tell us anything about how quickly equilibrium will be reached
situation where Kc won’t apply
non-reversible rxns
Reaction quotient
symbol: Q
- measures relative amount of reactants and products during a rxn at a particular point in time
- helps figure out which direction a rxn is likely to proceed based on the pressure and 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: forward rxn favored
- vice versa
effect of inverting the rxn on Kc
inverts Kc value
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
- temp
- catalyst
Factors affecting equilibrium: concentration
- if reactant conc. > product conc., forward rxn favored
- vice versa
Factors affecting equilibrium: pressure
- ↑ pressure = rxn resulting in ↓ no of moles favored
- vice versa
factors affecting equilibrium: temp
- ↑ temp = endothermic side favored
- vice versa
- change in temp = change in Kc
why does only temp change Kc?
- pressure and conc affect Q (NOT Kc)
- the equilibrium shifts to oppose the change
- this results in equal rates on both sides again, leaving Kc unchanged
- however, activation energies differ between forward and backward rxn
- and the energy change is not the same on both sides
- altering temp allows one rxn to proceed faster than the other
- this alters Kc
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:
450°C at 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
conditions required to achieve equilibrium
closed system
concentrations of reactants and products at equilibrium
- constant but not necessarily equal
- depends on the reaction equation and the conditions
why do different reactions have different equilibrium positions?
- depends on the amount of gibbs free energy (∆G*) – i.e. work – available in the system
- the farther the reaction is from equilibrium, the more negative (i.e. viable) the forward reaction is
- ∆G* = 0 at equilibrium
*generally comes with a theta to denote standard conditions
