topic 11 Flashcards
calculating moles of reactants at equilibrium
initial moles - moles reacted
calculating moles of products at equilibrium
initial moles + moles formed
working out equilibrium constant Kc - method part 1
- use burette to prepare a mixture in boiling tube of carboxylic acid, alcohol & dilute sulfuric acid
- swirl and bung tube. leave the mixture to reach equilibrium for one week
working out equilibrium constant Kc - method part 2
- rinse 250 cm³ volumetric flask with distilled water -
use a funnel to transfer contents of boiling tube into the flask. rinse boiling tube with water and add washing to volumetric flask - use distilled water to make up the solution to exactly 250 cm³. Stopper the flask and invert flask to shake contents thoroughly
- use pipette to transfer 25 cm³ of diluted equilibrium mixture to a 250 cm³ conical flask
- add 3 or 4 drops of phenolphthalein indicator to the conical flask
- set up the burette with NaOH solution
- add NaOH from burette until the mixture in conical flask just turns pink. Record the burette reading
- repeat the titration until you get 2 concordant results
working out initial moles of reactants
mass = density x vol
then
moles = mass / Mr
why is NaOH used
reacts with the sulfuric acid catalyst and any unreacted carboxylic acid in the equilibrium mixture
why does the pink colour fade after end-point has been reached
the addition of NaOH may make the equilibrium shift towards the reactants
partial pressure
the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture
partial pressure formula
mole fraction x total pressure
mole fraction formula
number of moles of a gas /
total no. of moles of all gases
Kp equation
partial pressure of product 1 ^ balanced number x partial pressure of product 2 ^ balanced number
DIVIDED
partial pressure of reactant 1 ^ balanced number x partial pressure of reactant 2 ^ balanced number
unit of Kp
atm^-2
what state used in Kp
only GASEOUS substances, all else must be ignored
effect of changing conditions on Kp or Kc
only TEMPERATURE changes KP or Kc values.
no change if pressure or concentration are altered or if catalyst is added
what does it mean if Kp is bigger
more products
what is said if Kc is small
equilibrium favours the reactants
effect of increased temperature on position of eqb and Kp value
- equilibrium shifts in the backward endothermic direction (left)
- Kp value gets smaller
effect of decreased temperature on position of eqb and Kp value
- equilibrium shifts in the forward exothermic direction
- Kp value increases as there are more products
effect of temperature on rate
as temp increases, more particles have energy greater than the activation energy so there are more successful collisions
effect of concentration on position of eqb and Kc value
equilibrium position would be shifted but Kc would NOT change
effect of pressure on position of eqb and Kp
equilibrium position would be shifted but Kp would NOT change - only varies with temp
effect of concentration and pressure on rate
at higher conc and pressure there are more particles per unit volume and so the particles collide with greater frequency, hence there will be a higher frequency of effective collisions
effect of catalysts on position of eqb and Kc
no effect on the position of eqb or the values of Kc and Kp
But it speeds up the rate at which eqb is achieved
why does a catalyst not effect position of eqb
because it speeds up the rates of the forward and backward reactions by the same amount
benefit of catalysts
speeds up the rate allowing lower temps to be used hence lower energy costs
importance of equilibrium to industrial processes - haber process
N2 + 3H2 ⇌ 2NH3
T = 450
P = 200 - 1000 atm
catalyst = iron
Low temp gives good yield but slow rate - compromise temp used
High pressures gives a good yield and high rate - too high a pressure would lead too high energy costs for pumps to produce the pressure
importance of equilibrium to industrial processes - contact process
stage 1: S + O2 —-> SO2
stage 2: SO2 + 0.5 O2 ⇌ SO3
T = 450
P = 1 to 2 atm
catalyst V2O5
low temp gives good yield but slow rate : compromise moderate temp used
High pressure gives slightly better yield and high rate : too high pressure would lead to too high energy costs for pumps to produce the pressure
using high pressures - disadv
too high energy costs for pumps to produce the pressure and too high equipment costsadvantage of recycling unreacted reactants
advantage of recycling unreacted reactants
improves overall yields of the processes
why can industrial processes not be in equilibrium
because the products are removed as they are formed to improve conversion of reactants - not closed systems
