TOPIC 10 - EQUILIBRIUM I + II Flashcards
Closed system
Is one where no substances are added to or lost from the system.
Dynamic Equilibrium
When the rate of the forward reaction is equal to the rate of the reverse reaction and the concentration of the reactants and products remain in constant only reached in a closed system.
- system will always go ahead to establish equilibrium. Reaction still continues even when quantity remains constant.
Le Chatelier’s Principle
If dynamic eq. disturbed by a change in conditions the position of the equilibrium moves to counteract to oppose the change in the system.
2 things must the reaction be for equilibrium to be reached?
- The reaction must be closed (i.e. nothing being added or removed)
- The reaction must be reversible
3 changes that affect position of equilibrium?
- Pressure
- Temperature
- Concentration of reactants/products (material either added or removed)
ALWAYS TALK ABT RATE OF REACTION
and obtaining greater yield
What is the effect on equilibrium position when reactants are added?
Shifts to right to get rid of added reactants. So increase in yield (but not greatly if by-product formed).
Opposite is reactants are removed, shifts to right to produce more reactants.
When do changes in pressure affect the position of the equilibrium? What is the effect on equilibrium position when the pressure is increased?
For reactions that involve gases
Increase in press. shift equil to side with fewer molecules - moles to reduce the press in the system so it occupies a smaller volume.
If same amount on either side, rate will be same and no equil. shift.
For some reactions increase in pressure= darkening due to compressing molecules in a smaller volume
What is the effect on equilibrium position when the temperature is increased on a forward exothermic reaction?
The equilibrium will move to the left (in the endothermic backwards direction) to lower the temperature. so syst. counteract by absorbing heat and problem is slow ROR. And decrease in yield.
Catalysts
Have NO EFFECT on equil. Catalyst will reduce time required to estabilish equil by lowering the activation energy. only increase ROR and also can lower energy costs
catalysts increase the ROR without being chemically altered or used up in a reaction by lowering the activation energy of the reaction.
Kc/ Equil Law
the concentration of products/concentration of reactants.
- small letters = moles
= [C]^c [D]^d / [A]^a [B]^b all [] is mol/dm3 when attempting it can make 1/mol/dm3 = dm3/mol
show how far to the left or right the equilibrium is to the left/right
Kc > 1 = forwards react dominates. productsproducts
heterogenous systems
heeterogenous catalyst
A system in which there are at least two different states present.
Catalyst diff state to the reactants.
homogeneous system
A system in which everything is in the same physical state
Why are solids or liquids not included in the Kc expression?
Because their concentration is constant. Only works for homogeneous equilibria
Why aren’t catalysts included in the Kc expressions?
Because they don’t affect the equilibrium concentrations of the products or reactants
Industrial process - Harbour process:
3H2 + N2 —> 2NH3
- low temp = 450, high press = 200-100atm and iron catalyst. Need these conditions in order to maintain a greater yield, especially compromising temp maximises the yeild whilst ROR fast.
Industrial process - contact process - producing sulfur trioxide
Stage 1: S (s) + O2(g) –>SO2 (g) not reversible
Stage 2: SO2 (g) + ½ O2 (g) —> SO3 (g) ^H = -98 kJ mol-1 reversible - high press favour right side
catalyst = V2O5 but slow rate and compromise temp used - give slightly better yield at a higher rate but high pressure - too high energy costs - good yield at high rates
Industrial process - hydration of ethene to produce ethanol
CH2=CH2 (g) + H2O (g) —> CH3CH2OH(l) catalyst = concen H3PO4
p = 70 atm temp= 300
high pressure also leads to unwanted polymeriastion of ethene = polyethene
rate of reaction over time graph
there are 2 rates of reaction lines A + B (reversible react)C + D - A+B reaction falls with time - C+D reaction increases (non-zero rate) = the 2 rates of react become equal
calc Kc
moles of reactant at equilib = initial moles - moles reacted
moles of product at equilib = initial moles + moles formed
CH3COOC2H5 + H20 = CH3COOH + C2H5OH n: 1 1 0 0 n at equilib: 0.7 0.7 0.3 0.3 c: 0.7/v 0.7/v 0.3/v 0.3/v v - total volume always then use Kc eq
calc Kc
N2 + 3H2 = 2NH3 n: 1.5 4.0 0 30% N reacted 0.3*1.5 = 0.45 moles formed n at equil: (1.5-0.45) (4-0.45*3) (0+0.45*2) times by n in eq c = n/v(1.5dm3) in Kc eq
Partial pressure
of a gas in a mixture of gases is the pressure that the gas would exert if it alone occupied the volume of the mixture
Kp - used for gas syst
calc Partial pressure
mole fraction*total pressure
Tp = p1 +p2 (no. of diff gases X mole fract = n of a gas/ total n of ALL gases
1 mole of PCL5 vapour heated to 500K in sealed vessel. Press at 6 atm contains 0.6 mol of Cl2. calc Kp
PCL5 = PCL3 + Cl2
i. 1 0 0
c/f. 0.4 0.6 0.6
total n = 1.6
e. f/1.6 each gas individually
Kp = Partial press reactant/ partial pressure of prod
= f/1.6 * 6atm) each one and units in atm
1 mole of PCL5 vapour heated to 500K in sealed vessel. Press at 6 atm contains 0.6 mol of Cl2. calc Kp
PCL5 = PCL3 + Cl2
i. 1 0 0
equil mole c/f. 0.4 0.6 0.6
total n = 1.6
e. f/1.6 each gas individually
Kp = Partial press reactant/ partial pressure of prod
= f/1.6 * 6atm) each one and units in atm
- use eq to check mole ratio to find the equilibrium mole
Kp eq
= partial pressure of each individual gas of products ^n / pp of reactants of each gas ^n
if a pp left in the denominator and pp incr of products and the pp of reactants would have to compensate so Kp is constant just like Le Chateliers principle
so depsite press changes the Kp remains constant
Effect of temp on Kc and Kp
Catalyst, concen and press has no effect on Kc and Kp so stays the same - restores it the same
- Kp and Kc ONLY changed by TEMP
- altering temp changes rate for both forward and backward reaction
- the equilib produces a new equilibrium constant
- direction of movement depends on sign of the enthalpy
In EXO FORWARD REACT:
increase in temp - K decrease
decrease temp = K increase
Kc/Kp value gets smaller due to fewer products - favours products
larger the Kp/Kc = greater amount of products
all above opposite for ENDO
increasing pressure effects on Kp
increase pressure of the terms on the bottom Kp expression and is now no longer in equilibrium and shifts to the right so therefore the top of the Kp expression increases and bottom decreases until original Kp value is RESTORED
Industrial processes cannot be in equilibrium why?
the products are removed as they are formed to improve conversion of reactants and they are not closed systems.
could potentially improve overall yeild and atom economy when recycling unreacted reactants
CO2 poisoning
-Hb has a higher affinity to CO - blocking the attachment to O2
treated by breathing pure O2 to displace CO
Hb(CO)4 + 4O2 = Hb(O2)4 + 4CO
Qc (reaction quotient)
the direction can be predicted by calc Qc
- same expression as Kc
Qc < Kc - the reactants if reaction disturbed the equil move to right and if reaction increases there is an increase of products re-establishes Kc value
