Chapter 19 - Equilibrium Flashcards
what is a generalised equation for Kc
aA + bB –(REVERSIBLE)–> cC + dD
Kc = [C]^c [D]^d / [A]^a [B]^b
when (in terms of Kc) is equilibrium to the left/right/centre
Kc < 1
equilibrium to the left
Kc > 1
equilibrium to the right
Kc = 1
equilibrium central
how can we work out the units of Kc
write out the expression for Kc including only the units with powers
cancel accordingly
what is a homogeneous equilibrium
“A homogeneous equilibrium contains equilibrium species that are all in the same state or phase”
what is a heterogeneous equilibrium
“A heterogeneous equilibrium contains equilibrium species in different states or phases”
what do we need to change about Kc if it’s a heterogeneous equilibrium
exclude any species in liquid or solid phase because their concentrations essentially remain constant
what is the generalised method for calculating Kc where the system is not already at equilibrium
1) write out full equation and check for any solids/liquids to exclude from Kc
2) write out all known initial concentrations
3) use molar ratios and any information given about equilibrium moles to predict moles remaining/formed at equilibrium for all species
4) Find equilibrium concentrations using volume
5) calculate Kc
NOTE: be careful when doing moles formed, the molar ratios need to be in relation to the CHANGE in moles not just the moles
how can we measure equilibria where all the species are in their gaseous phase
We can use Kp, this uses partial pressures instead of concentrations
as pressures and concentrations are very similar, Kc and Kp are directly linked
what can we say about amounts of gases if they are under the same conditions
the same number of moles will take up the same volume
how can we calculate the mole fraction of a gas
mole fraction(A) = moles of A / total moles
you can replace moles with volume because its a fraction and moles/volume are interchangeable
what should all mol fractions sum to
1
define partial pressure
“In a gas mixture the partial pressure, p, of a gas is the contribution it makes towards the total pressure, P”
how can we calculate partial pressure
partial pressure = mole fraction of that gas x total pressure
p(A) = x(A) * P
what should all partial pressures sum to
the total pressure
how can we calculate Kp
for a reaction
aA(g) + bB(g) –(REVERSIBLE)–> cC(g) + dD(g)
Kp = (p(D))^d * (p(C))^c / (p(A))^a * (p(B))^b
calculate units as normal
what are the 3 main points to remember from Le Chatalier’s principle
1) if the concentration of a species increases, the equilibrium position shifts to decrease that concentration
2) if pressure increases, equilibrium shifts to the side with fewer moles of gas
3) if temperature increases, equilibrium shifts to the endothermic side
what is the only thing that changes an equilibrium constant
temperature
NOT pressure, concentration or catalyst
what is the effect of temperature (in terms of Kp and Kc) on an equilibrium if the forwards reaction is exothermic
- equil constant decrease as temperature increases
- therefore yield of products decreases as temp increases
if Kc decreases then [C] and [D] MUST decrease and [A] and [B] MUST increase
- this makes equil shift left
what is the effect of temperature (in terms of Kp and Kc) on an equilibrium if the forwards reaction is endothermic
- equil constants increase with increasing temperature
- therefore yield of products increase as temperature increases
as temp increases and Kc increases [C] and [D] MUST increase and [A] and [B] MUST decrease
what is the effect of pressure and concentration of the equilibrium constants and why does the position of equilibrium move
- Kp and Kc is unaffected by changes in conc. and pressure
- position of equilibrium changes due to Le Chatalier’s BECAUSE Kc and Kp MUST stay the same
for an equilibrium where Kc (tries to) decrease due to a concentration change, what happens to the concentrations and position of equilibrium
- if Kc would decrease when the concentration of a species is changed then to bring Kc back to its initial value, more products MUST be made and more reactants MUST be used up
- so equilibrium moves right, there are the same overall ratios, just more stuff
the same works vice versa
how does an increase in pressure (in terms of Kp) affect a dynamic equilibrium where there are equal numbers of moles of gas on each side
all partial pressures increase by the same amount
but there are the same orders on the top and bottom of the equation so there is no overall change in equilibrium
how does an increase in pressure (in terms of Kp) affect an equilibrium where there are different numbers of moles of gas on each side of the equation
- when total pressure increases, all partial pressures increase by the same amount
- but the orders on the top/bottom of the equation are different
- this would force Kp to change so the position of equilibrium moves to counter this and keep Kp constant
what is the effect of a catalyst on Kc/Kp
- it has no effect
- a catalyst increases the rate of the forwards and backwards reactions by the same amount
How do the rates of the forwards and backwards reactions change in an equilibrium where pressure increases (and uneven gas distribution)
- the rates of both reactions increase because concentration of reactant particles increase as pressure increases (in gases)
- so frequency of successful collisions increases
- however the reaction with more moles of gaseous reactants increases more because there are more reactant terms
what to ALWAYS check and recheck when doing equilibrium calcs
that you don’t have a liquid or solid in the Kc expression
what tend to be the units for kp
some power of (Pa) e.g. Pa^-1
or (atm)^n
