Equilibria

Question 1

dynamic equilibrium

Answer 1

when the rates of the forward and back reactions are equal
- no net change in concentrations of reactants or products

Question 2

equilbrium constant

Answer 2

K - quantify the relative concentrations of reactants and products at equilibrium

aA +bB –> cC + dD
k = [C]c [D]d / [A]a [B]b

only effected by temperature

Question 3

if K is small…

Answer 3

then more reactants, equilibrium lies left

Question 4

if K is large …

Answer 4

then more prodcuts, reaction lies right

Question 5

if K is intermediate…

Answer 5

then similar magnitude

Question 6

reaction quotient

Answer 6

gives us an indication of the reaction progress at all points of chemical reaction, even before the equilibrium occurs

same equation as K

Question 7

if Q=K

Answer 7

then reaction is already at equilibrium

Question 8

if Q<K

Answer 8

need to increase Q, so increase product concentration, equilbrium shift to the right in order to get to equilibrium

Question 9

if Q>K

Answer 9

need to decrease Q so increase reactant concentration, equilibrium shift towards the reactants left

Question 10

changes in temperature and equilbrium

Answer 10

a change in temperature will cause the system to shift to counteract the change

increase in temp causes shift to endothermic reaction as increase heat will be absorbed

decrease in temp causes shift to exothermic reaction as heat will be released

Question 11

change in pressure for equilbrium/conc

Answer 11

a change in pressure will cause a shift to counteract change, only taking into occur the gaseous molecules though

increasing pressure shifts equilibrium to side with less moles gas

decreasing pressure shifts eq to side with more moles

this idea is same with concentration

Question 12

effective oxygen transport

Answer 12

ultises le chatiler principle
Hb(aq) + 4O2 (g) –> Hb(O2)4 (aq)

in the lungs, high pressure - system shifts to the least number of molecules so oxygen binds to haemoglobin

in body tissue, low pressure so system shifts to most moles so oxygen is released

Question 13

haemoglobin

Answer 13

a protein that transports oxygen from lungs to body tissue through the blood,

has 4 sub units 2 alpha and 2 beta chain

each sub unit has a heme group with iron ion that binds to one oxygen

Question 14

oxygen binding and shape of heme group

Answer 14

deoxygenated heme is domed/non planar

oxygenated heme is planar

binding of oxygen brings the Fe into the plane,called cooperative binding and distorts the shape to allow for oxygen to bind to the other 3 sites

Question 15

bohr effect

Answer 15

co2 and H+ promotes the release of oxygen as when a low ph
H+, the histidine residue is protonated which leads to the formation of salt bridges to distort the heme iron, favouring the domed conforrmation

co2 role - co2 binds to the amino group to produce a negatively charged group -NHCOO- to favour domed

Question 16

blood colour

Answer 16

the different conformations give the blood different colours

oxygenated haemoglobin absorbs blue/green light so appears red

deoxygenated haem absorbs red so give blue tinge

Question 17

porphyria

Answer 17

blood disorder which affects the ability of body to make haemologbin and ability to take up oxygen

leads to blue/purple blood and sensitivity to light

Question 18

gibbs free energy and equilbrium

Answer 18

for reactions at equilbrium
- no capacity to do work so gibbs free energy = 0

for forward
- Q is less than K so gibbs free energy G<0 negative so Q=K

for backwards
- Q is higher than K so gibbs is >0 positive

Question 19

vant hoff isotherm

Answer 19

delta rG = - RTlnK

used for determining if a reaction is sponanteous and in which direction

when Q<K, the forward reaction is favour, ln(Q/K) < 0 S, delta r G <0, so it is negative so forward reaction is favoured

when Q>K, backward reaction is favoured, ln(Q/K) >0 so delta r G >0, so positive, reverse reaction is favoured.

Question 20

example of van’t Hoff:
reaction between glucose and phosphate to form glucose 6 phosphate, concen at equilbrium 298k, [G] = 0.0080M, [P] = 0.0125M, [6GP] = 2.78X10-5M. Calculate K and delta r G, is it spontaneous ?

Answer 20

delta r G = -RTlnK

first K = 2.78x10-5 / 0.008, 0.0125 =0.278

delta r G = -8.314 x 298 x ln 0.278 = 3.17kjmol-1

forward reaction is not spontaneous but backward is as delta rG >0

Question 21

parallel coupling

Answer 21

reactions occur at same place and same time

Question 22

sequential coupling

Answer 22

one reaction occurs then using outcome of reaction to power the next

aims to exploit le chateliers principle regarding concentration change

Question 23

acid base equilbria

Answer 23

proton transfer is rapid so all acid base systems can be considered to be at equilbrium

• so dont need to worry about Q

Question 24

autoprotolysis of water

Answer 24

H2O + H2O –> H3O+ + OH-

ionic product of water at 25 degrees = kw = 1x10-14

Question 25

blood ph range

Answer 25

7.4 normal value
cannot deviate more than +- 0.4 units

Question 26

weak acids

Answer 26

HA + H2O –> H3O+ + A-
HA = weak acid
A- = conjugate base

only partially dissociates

Question 27

Ka

Answer 27

ka = [h3o+] x[a-] /[ha]

the larger ka is the more acidic
pka = -log10ka - the smaller this is the more acidic

Question 28

weak bases

Answer 28

B + H2O –> BH+ + OH-
BH = weak base

OH = conjugate acid

Question 29

conjugate acid base pairs in aqueous solution

Answer 29

when combining Ka and Kb, get [h3o+][oh-] which is equal to kw

therefore kw = ka x kb

andddddd, pkw = pka + pkb

this implies that a strong acid will have a weak conjugate base and vice versa

Question 30

temperature and ionisation of water

Answer 30

at a higher temperature, the ionisation of water increases because the reaction is endothermic so increase temp, equilibrium shifts to right to increase products

this increases h3o+ so at higher temps, the ph is also lower, more acidic as more acid

Question 31

buffers

Answer 31

to resist changes in ph
must have an acid component to react with OH

must have basic component to react with h3o+

uses weak acid and its salt to get conjugate base

Question 32

if acid is added to buffer solution

Answer 32

increase in h+ conc, cause a shift to the left to decrease
h+ will react with conjugate base to reform acid

equilibrium shifts to the left/backward

Question 33

if base is added to buffer

Answer 33

increase in oh- so equilbrium shifts to the right/forward to increase h+ conc

Question 34

henderson hasselbalch equation

Answer 34

ph = pka + log ([A-]/[HA])

rearranging Ka equation to make H+ the subject then taking logs of both sides

can use this to calculate the ph of buffers

Question 35

buffer capacity

Answer 35

the maximum amount of either strong acid or strong base that can be added before a significant change in pH will occur

Question 36

polprotic acids

Answer 36

can have multiple uses as has lots of hydrogens to donate

Question 37

titratation curve strong base and acid

Answer 37

starts ph low and ends high

100% ionised so products dominate

ph of 7 = equivialnce point

Question 38

molarity

Answer 38

to work out concentration
mol / volume in litres

Question 39

titration of weak acid and strong base

Answer 39

not so low ph starting,