topic 11: equilibrium II Flashcards
what is the dynamic equilibrium
when forward and backward reactions are occurring at equal rates. the concentrations of reactants and products stays constant and the reaction is continuous
what is the general formula of Kc (use the equation
mA + nB —> pC + qD
[ C]p [D]q
- m,n,p,q are the stoichiometric balancing numbers
- A,B,C,D stand for the chemical formula
what is the Kc
equilibrium constant
what is the unit used in Kc for the concentration
mol dm-3
what is the equation to work out the moles of reactants at equilibrium
moles of reactant at equilibrium = initial moles – moles reacted
what is the equation to work out the moles of products at equilibrium
moles of product at equilibrium = initial moles + moles formed
CH3CO2H +CH3CH2OH <—-> CH3CO2CH2CH3 + H2O
PRACTICAL: part 1 - preparing the equilibrium mixture
- use burettes to prepare a mixture in boiling tube of carboxylic acid, alcohol, and dilute sulfuric acid
- swirl and bung tube. leave the mixture to reach equilibrium for one week
CH3CO2H +CH3CH2OH <—-> CH3CO2CH2CH3 + H2O
PRACTICAL: part 2 - titrating the equilibrium mixture
- rinse a 250 cm3 volumetric flask with distilled water. use a funnel to transfer the contents of the boiling tube into the flask. rinse the boiling tube with water and add the washings to the volumetric flask
- use distilled water to make up the solution in the volumetric flask to exactly 250 cm3. stopper the flask, then invert and shake the contents thoroughly
- use the pipette to transfer 25.0 cm3 of the 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 sodium hydroxide solution
- add the sodium hydroxide solution from the burette until the mixture in the conical flask just turns pink. record this burette reading in your table
- repeat the titration until you obtain a minimum of two concordant titres
PRACTICAL: why does the pink colour of the phenolphthalein in the titration fade after the end-point of the titration has
been reached
because the addition of sodium hydroxide may make the equilibrium shift towards the
reactants
PRACTICAL: what does the sodium hydroxide react with
the sulfuric acid catalyst and any unreacted carboxylic acid in the equilibrium mixture
what is the partial pressure of a gas in a mixture
the pressure that the gas would have if it alone occupied the volume
occupied by the whole mixture
how do you work out the mole fraction of a gas
number of moles of a gas/ total number of moles of all gases
who do you work out the partial pressure of 1 gas
= mole fraction x total pressure
what is the unit of pressure
atm
work out the expression of Kp for the equation
N2(g) + 3H2(g) <—> 2 NH3(g)
p(N2) p(H2)^3
the substances in what state should be included in the Kp expression
only gases
what is the Kc value if the reaction doesn’t happen
Kc < 10^-10
what is the Kc value if reactants are predominate in an equilibrium
Kc is approximately 0.1
what is the Kc value when equal amounts of reactants and products
Kc = 1
what is the Kc value when products are predominate in equilibrium
Kc is approximately 10
what is the Kc value when the reaction goes too completion
Kc > 10^10
as ΔStotal increases,, what happens to the equilibrium constant
increases
what is the equation of ΔS
ΔS = R lnK
what is the effect of temperature on position of equilibrium and Kp
Both the position of equilibrium and the value of Kc or Kp will change it temperature is altered
- if temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction
- the position of equilibrium shifts left
- the value of Kc gets smaller as there are fewer products
what is the effect of concentration on position of equilibrium and Kc
changing concentration would shift the position of equilibrium but the value of Kc would not change
- increasing concentration of H2 would move equilibrium to the right lowering concentration of H2 and Cl2 and increasing concentration of HCl
- the new concentrations
would restore the equilibrium to the same value of Kc
what is the effect of concentration and pressure on rate
at higher concentrations and pressures, there are more particles per unit volume and so the particles collide with a greater frequency and there will be a higher frequency of effective collisions
why does increasing pressure not change Kp
- the increased pressure increases the pressure terms on bottom of Kp expression more than the top
- the system is now no longer in equilibrium so the equilibrium shifts to the right, increasing mole fractions of products and decreases the mole fractions of
reactants - the top of the Kp expression therefore
increases and the bottom decreases until the original value of Kp
is restored
what is the effect of pressure on the position of equilibrium
if pressure is increased the reaction will shift to oppose the change and move in the forward direction to the side with fewer moles of gas
- the position of equilibrium shifts right
- the value of Kp stays the same though as only temperature changes the value of Kp
what will happen to the Kp value if pressure is altered
Kp stays constant
what is the effect of catalyst on the equilibrium constant
a catalyst has no effect on the position of equilibrium or values of Kc and Kp but it will speed up the rate at which the equilibrium is achieved
it does not effect the position of equilibrium because it speeds up the rates of the forward and
backward reactions by the same amount
what is the effect of catalysts on the position of equilibrium
catalysts speeds up the rate allowing lower temperatures to be used (and hence lower energy costs) but have no effect on equilibrium
describe the haber process
N2 + 3H2 <—> 2NH3 exothermic
T= 450oC,P= 200 - 1000 atm, catalyst = iron
low temp gives good yield but slow rate:
compromise temp used
high pressure gives good yield and high rate:
too high a pressure would lead to too high
energy costs for pumps to produce the pressure
describe contact process
stage 1: S(s) + O2(g) <—> SO2(g)
stage 2: SO2(g) + ½O2 (g) <—> SO3 (g) H = -98 kJ mol-1
T= 450oC, 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 a pressure would lead to too high energy costs for pumps to produce the pressure
what does high pressure lead to
high energy costs for pumps to produce the pressure and too high equipment costs to have equipment that can withstand high pressures
what does recycling unreacted reactants do
recycling unreacted reactants back into the reactor can improve the overall yields of all these processes and
improve their atom economy
- industrial processes cannot be in equilibrium since the products are removed as they are formed to improve conversion of reactants
- they are not closed systems
