topic 11

Question 1

calculating moles of reactants at equilibrium

Answer 1

initial moles - moles reacted

Question 2

calculating moles of products at equilibrium

Answer 2

initial moles + moles formed

Question 3

working out equilibrium constant Kc - method part 1

Answer 3

1. use burette to prepare a mixture in boiling tube of carboxylic acid, alcohol & dilute sulfuric acid
2. swirl and bung tube. leave the mixture to reach equilibrium for one week

Question 4

working out equilibrium constant Kc - method part 2

Answer 4

1. 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
2. use distilled water to make up the solution to exactly 250 cm³. Stopper the flask and invert flask to shake contents thoroughly
3. use pipette to transfer 25 cm³ of diluted equilibrium mixture to a 250 cm³ conical flask
4. add 3 or 4 drops of phenolphthalein indicator to the conical flask
5. set up the burette with NaOH solution
6. add NaOH from burette until the mixture in conical flask just turns pink. Record the burette reading
7. repeat the titration until you get 2 concordant results

Question 5

working out initial moles of reactants

Answer 5

mass = density x vol
then
moles = mass / Mr

Question 6

why is NaOH used

Answer 6

reacts with the sulfuric acid catalyst and any unreacted carboxylic acid in the equilibrium mixture

Question 7

why does the pink colour fade after end-point has been reached

Answer 7

the addition of NaOH may make the equilibrium shift towards the reactants

Question 8

partial pressure

Answer 8

the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture

Question 9

partial pressure formula

Answer 9

mole fraction x total pressure

Question 10

mole fraction formula

Answer 10

number of moles of a gas /
total no. of moles of all gases

Question 11

Kp equation

Answer 11

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

Question 12

unit of Kp

Answer 12

atm^-2

Question 13

what state used in Kp

Answer 13

only GASEOUS substances, all else must be ignored

Question 14

effect of changing conditions on Kp or Kc

Answer 14

only TEMPERATURE changes KP or Kc values.

no change if pressure or concentration are altered or if catalyst is added

Question 15

what does it mean if Kp is bigger

Answer 15

more products

Question 16

what is said if Kc is small

Answer 16

equilibrium favours the reactants

Question 17

effect of increased temperature on position of eqb and Kp value

Answer 17

• equilibrium shifts in the backward endothermic direction (left)
• Kp value gets smaller

Question 18

effect of decreased temperature on position of eqb and Kp value

Answer 18

• equilibrium shifts in the forward exothermic direction
• Kp value increases as there are more products

Question 19

effect of temperature on rate

Answer 19

as temp increases, more particles have energy greater than the activation energy so there are more successful collisions

Question 20

effect of concentration on position of eqb and Kc value

Answer 20

equilibrium position would be shifted but Kc would NOT change

Question 21

effect of pressure on position of eqb and Kp

Answer 21

equilibrium position would be shifted but Kp would NOT change - only varies with temp

Question 22

effect of concentration and pressure on rate

Answer 22

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

Question 23

effect of catalysts on position of eqb and Kc

Answer 23

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

Question 24

why does a catalyst not effect position of eqb

Answer 24

because it speeds up the rates of the forward and backward reactions by the same amount

Question 25

benefit of catalysts

Answer 25

speeds up the rate allowing lower temps to be used hence lower energy costs

Question 26

importance of equilibrium to industrial processes - haber process

Answer 26

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

Question 27

importance of equilibrium to industrial processes - contact process

Answer 27

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

Question 28

using high pressures - disadv

Answer 28

too high energy costs for pumps to produce the pressure and too high equipment costsadvantage of recycling unreacted reactants

Question 29

advantage of recycling unreacted reactants

Answer 29

improves overall yields of the processes

Question 30

why can industrial processes not be in equilibrium

Answer 30

because the products are removed as they are formed to improve conversion of reactants - not closed systems