Equilibrium Flashcards
1st year equilibrium: mention _____in answer
Effect on yield
1st year equilibrium: 3 important points
- dynamic equilibrium
- homogeneous equilibrium
- equilibrium law
dynamic equilibrium?
( equilibrium bit) Rate of forwards and backwards reactions are equal (dynamic bit) and reactions continue at a molecular level
homogeneous equilibrium?
all particles in the equilibrium system are in the same state
equilibrium law?
at eq., the [products]/ [reactants] = constant
• amounts of r and p don’t change
Kc general?
- The eq. constant
- Kc tells us the relative proportion of reactants and products present at eq.
- [r] and [p] DO NOT CHANGE AT EQ.
- the units for Kc must be worked out for each question
if there are no units?
must write no units,, can’t just leave it blank
for the equilibrium constant?
need the conc at EQUILIBRIUM
Equilibrium constant is constant unless
the temperature is changed.
Kp vc Kc?
Kp for gases
Kc for liquids but Kc can be used for gases too
Mole fraction, x, equation
Mole fraction = moles of the gas/ total n of all gases
all x values add to?
1
partial pressure is?
the contribution a gas makes to the total pressure
partial pressure symbol
p
total pressure symbol
P
Partial pressure equation?
p = xP pp = mole fraction x total pressure
the sum of the partial pressure =?
total pressure
Kp =
pproducts/ preactants
when working out, K ignore?
solids, leave them out of the equation
Effect of changing temp on K?
- ony thing that changes K
- an ⬆ in temp shifts the eq. in the endothermic direction
- a ⬇ in temp shifts the equilibrium in the exothermic direction
- value of K has to change w temp
effect of increasing temp on K?
- of the forwards reaction is endothermic, and the temp is ⬆
- the equilibrium yield of the products would ⬆
- eq. yield of the reactants would decrease
- so K would decrease
the opposite is true for exothermic reactions when temp increases
when explaining the impact of changing conc/ pressure on K include these in the answer
- Kc/Kp expression
- theoretically what would happen
- what actually happens - equilibrium shift
- impact on Kc - none
impact of adding a catalyst on the value of K?
- does not change
- bc does not impact position of eq.
- the forwards and backwards reactions are sped up equally so eq is established faster
Kp and Kc are?
temp dependent constants
what happens when the pressure is increased but there are the same amount of moles on either side of the equation?
both sides impacted the same so no effect on K
does the mole fraction change when the pressure is increased at a constant temp?
Yes. Increasing the P would cause the equilibrium to shift to the side with the fewest moles.
so the partial pressure would change and so the mole fraction must also change in the same way.
Mpa to pa ?
x 10-6
dm3 to m3?
dm3 / 1000
(long q) the steps?
- initial n
* n at eq.
(long q) the steps?
- initial n
- n at eq.
- ( total n at eq.) then x
- p = xP
- plug p into Kp equation
why does the K value not change w changing [] or P?
- equilibrium shift (le Chatelier’s principle)
- the changes brought about by this balance
- so K doesn’t change
theoretically, if P was increased which side increases tge most
the side with the most moles.
- if this side is the products, then K would theoretically increase
- if this side is the reactants, you are / by a bigger number so K would theoretically decrease
Really, when P is changed what happens?
- if P is ⬆, then the side w the fewest n favoured and more of the thing on that side is made, less of the thing on the other side (smaller p)
- so K doesn’t change (essentially the changes balance)
what happens when the conc of a substance is increased?
- eq. shifts to use up the extra substance
- so the conc of this substance ⬇
- and the conc of substances on the other side of the arrow ⬆
- so K does not change
