2.1 Equlibrium, 2.2 Reaction Feasibility, 2.3 Kinetics Flashcards
(98 cards)
1
Q
what happens to equilibrium when alkali is added to a reaction with H+ ions on the product side
A
equilibrium shifts to the right
2
Q
what happens to the pH of an acid when you dilute it
A
no of H+ ions decreases
pH increases
3
Q
disorder in spontaneous processes
A
increase in disorder
4
Q
conjugate base charge
A
-ve
5
Q
conjugate acid charge
A
+ve
6
Q
indicator if acid and base are weak
A
no indicator
7
Q
standard enthalpy of formation
A
the enthalpy change that occurs when 1 mole of the substance is formed from its elements in their standard states
8
Q
K if not stated
A
298 K
9
Q
highest entropy state
A
gas
10
Q
what is the best indicator to use determined by
A
the pH of the salt
11
Q
reaction if delta G and delta entropy are negative
A
exothermic
12
Q
1st order
A
change in reactant is proportional to change in rate
13
Q
change in reactant is proportional to change in rate
A
1st order
14
Q
in equilibrium, if there are two solvents A and B, what happens when you add more B
A
the concentration of solute in B decreases
15
Q
what can order of reaction be determined by
A
experiment only
16
Q
rate determining step
A
slowest step
17
Q
K > 1
A
more products than reactants
18
Q
K < 1
A
more reactants than products
19
Q
changing temperature in exothermic reactions
A
rise in temp: decrease in K
fall in temp: increase in K
20
Q
changing temperature in endothermic reactions
A
rise in temp: increase in K
fall in temp: decrease in K
21
Q
increase in temperature favours what reaction
A
endothermic
22
Q
decrease in temperature favours what reaction
A
exothermic
23
Q
order of reactants that do not take place in the slowest step
A
0 order
24
Q
when do buffers form
A
when a salt of a weak alkali dissolves in a weak acid
25
overall order of reaction
sum of all orders
26
acid definition
Donates proton leaving a conjugate base
27
How does a buffer resist change in pH if alkali is added
OH- ions react with H+ ions and neutralise them
The H+ ions are replaced by the dissociation of the acid into H+ ions
28
Calculate equilibrium constant for..
N2O4 —(reversible)-> 2NO2
0.28 mol of N2O4 gas placed in an empty 1 litre flask and heated to 127 degrees. At equilibrium, 0.24 moles of NO2 present
0.36
29
equilibrium
when the rate of the forward reaction = the rate of the backwards reaction
30
what is K
equilibrium constant
31
CH3COOH ⇌ H+ + CH3COO-
How would the equilibrium constant be expressed?
K = [H+] [CH3COO-] / [CH3COOH]
32
CH3COOH ⇌ H+ + CH3COO-
If sodium ethanoate solid was added, what would happen to
1) the position of equilibrium?
2) the pH
1) Equilibrium would shift to the left
2) pH would increase
This is because of a reaction between H+(aq) and CH3COO-(aq) ions to make CH3COOH(aq)
33
base
a proton acceptor
34
What is formed when a base gains a proton?
conjugate acid
35
What is formed when an acid loses a proton?
conjugate base
36
difference between strong and weak acids/bases
Strong acids/bases are completely dissociated into ions in aqueous solutions
Weak acids/bases are partially dissociated into ions in aqueous solutions
37
examples of:
- strong acids
- weak acids
- strong bases
- weak bases
- Hydrochloric acid, sulfuric acid or nitric acid
- Ethanoic acid, carbonic acid and sulfurous acid
- Solutions of metal hydroxides
- Ammonia and amines
38
strong acid + strong base =
neutral solution
39
weak acid + strong base =
alkaline solution
40
strong acid + weak base =
acidic solution
41
why is a solution of ammonium chloride is acidic
- weak base and strong acid
- the solution will contain NH4+ and Cl- from the salt, H+ and OH- from the ions in the water
-some of the NH4+ and OH- will react to form an equilibrium of ammonium hydroxide
- this will leave an excess of H+ ions, therefore the solution will be acidic
42
why is a solution of sodium carbonate is basic
- strong base and weak acid
- the solutino will contain Na+ and CO2- from the salt, H+ and OH- ions from the water
- some of the H+ and CO2- will react to form an equilibrium mixture of carbonic acid
- this will leave an excess of OH- ions and therefore the solution will be basic
43
how does an acid buffer work
- The weak acid provides the hydrogen ions when these are removed by the addition of a small amount of base
- If an acid (H+) is added the conjugate base will form the weak acid.
44
what does a buffer do
pH of a buffer solution remains approximately constant when small amounts of acid, base or water are added
45
how does a basic buffer work
- The weak base removes excess hydrogen ions
- The salt of this base provides the conjugate acid which combines with the hydroxide ions added
46
what are indictors
weak acids where the weak acid has a different colour from its conjugate base
47
Which of these indicators
phenolphthalein-pH range 8.2-10.0
bromothymol blue-pH range 6.0-7.6
methyl orange-pH range 3.0-4.4
would be suitable for
(a) strong acid and strong base titration
(b) strong acid and weak base titration
(c) weak acid and strong base titration
(a) bromothymol blue
(b) methyl orange or bromothymol blue
(c) phenolphthalein
48
why are no indiactors suitable for the titration of a weak acid and a weak base
The pH change around the equivalence point is fairly gradual
49
relationshio between entropy and temperature
as temp increases, entropy increases
50
What must ΔG° be if a reaction is to be feasible?
negative
51
If ΔH°(reaction) = - 91.8 kJ mol-1 and ΔS°(reaction) = -197.3 J K-1mol-1, calculate the temperature at which the reaction just becomes feasible.
T = ΔH° /ΔS°
T = - 91800 /- 197.3
T = 465K or 192°C
52
zero order reactant
changing reactant conc has no effect on the rate
53
second order reactant
doubling concentration quadruples rate
54
Rate = k [H2O2][I-][H+]0
55
If the rate equation is = k [H2O2][I-], calculate the rate constant.
k = rate/[H2O2][I-]
Substitute any of the experimental data
k = 9.2x10-6 / (0.02x0.02)
0.023 mol-1 l s-1
56
K equation
57
K gives an indication of the position of equilibrium
- K > 10^3 = lies to the right
- 10^-3 < K < 10^3 = equilibrium
- K < 10^-3 = lies to the left
58
K value for pure liquids and solids
1
59
what is K affected by
temperature
60
strong acid strong base equation
HA + B —> BH+ + A-
- HA: strong acid
- B: strong base
- BH: conjugate acid
- A-: conjugate base
61
when are conjugate acids and bases formed
- conjugate base is formed by the acid donating a proton/H+
- conjugate acid is formed by the base accepting a proton/H+
62
relationship between strength of acid/base and strength of conjugate acid/base
- the stronger the acid, the weaker the conjugate base
- the stronger the base, the weaker the conjugate acid
63
pH calculations for strong acids and bases
**pH = -log10[H+]**
**pOH = -log10 [[OH-]**
**14 = pH + pOH**
**[H+] = 10^-ph**
64
calculate the pH of 0.2 mol nitric acid
= pH = -log10 [H+]
= -log10 [0.2]
= 0.7
65
calculate the pH of 0.3 mol sodium hydroxide
= pOH = -log[OH]
= -log[0.3]
pOH = 0.5
14 = pH + POH
14 - 0.5 = 13.5
pH = 13.5
66
calculate the pH of 0.5 mol hydrochloric acid
= pH = -log10 [H+]
= -log10 [0.5]
= 0.3
67
calculate the H+ concentration when the pH of a solution is 9.6
= pH = -log10 [H+]
9.6 = -log10 [H+]
[H+] = 10^-9.6
= 2.5 x 10^-10 moles/l
68
weak acid and weak base equation
HA + H20 —> H30+ + A-
69
pH equations for weak acids and bases
**pKa = -log10Ka**
**pH = 1/2pKa - 1/2log10c**
**Ka = [H+][A-] / [HA]**
70
Ex. calculate the pH of 0.2 mol/l ethanoic acid
pH = 1/2pKa - 1/2log10c
pH = 1/2 x 4.7 - 1/2 x 4.7 x 0.2
- 4.7 is ppKa value for ethanoic acid in data booklet
- c is concentration
= 2.73
71
calculate the pH of 0.3 mol/l carbonic acid
pH = 1/2pKa - 1/2log10c
pH = 1/2 x 6.35 - 1/2 x 6.35 x 0.3
= 3.44
72
calculate the pH of 0.47 mol/l benzoic acid
pH = 1/2pKa - 1/2log10c
pH = 1/2 x 4.2 - 1/2 x 4.2 x 0.47
= 2.27
73
indicator K equations
**HIn + H2O —> H3O+ + In-**
**KIn = [H+][In-] / [HIn] [H2O]**
**[In-] / [HIn] = [KIn] / [H+]**
74
endpoint of titration
when the pH changes rapidly. where you want the pH of the indicator to lie
75
buffer equation
pH = pKa - log10 [acid] / [base]
76
what can buffers be made from
- weak acid and salt of the acid
- weak base and salt of base
77
what is pH of buffer determined by
the pKa of the weak acid/base (linked to its dissociation) and the ratio of acid to salt
78
how to change pH of a buffer
change the composition of acid:salt
79
calculate the pH of buffer solution formed by mixing 40cm of 0.1 mol/l ethanoic acid and 60cm of 0.1 mol/l sodium ethanoate solution
= pH = pKa - log10 [acid] / [base]
= pH = 4.76 - log10 [0.04] / [0.06]
= 4.76 - (- 0.176)
= 4.94
80
calculate the pH of a buffer solution consisting of 100cm of 0.1 mol/l propanoic acid and 0.2 moles of sodium propionate.
= pH = pKa - log10 [acid]/[base]
= 4.87 - log10 x 0.1/2
pH = 6.17
81
rate equation
k[M]^m [N]^n
total order: m+n
82
83
calculate rate equation:
k [A]^1 [B]^0 [C]^2
84
1. k [RX]
2. k = 187.5
3. units: s
85
enthalpy change of a reaction
- equation
- what it shows
- units
**ΔH = ΔH (products) - ΔH (reactants)**
→ tells us if reaction is exothermic or endothermic, (+ = endothermic, - = exothermic)
→ units are kJ/mol
86
entropy change of a reaction
- equation
- what it shows
- units
**ΔS = ΔS (products) - ΔS (reactants)**
→ tells us if disorder is increasing or decreasing
→ units are J K^-1 mol^-1
87
free energy change of a reaction
- equation
- what it shows
- units
**ΔG = ΔG (products) - ΔG (reactants)**
→ tells us if reaction is feasible (0 > G: feasible, G = 0: equilibrium, G > 0: not feasible)
→ kJ/mol
88
relationship between enthalpy, entropy, and free energy
ΔG = ΔH - TΔS
89
at what temperature does this reaction become feasible?
**ΔH = ΔH (products) - ΔH (reactants)**
= (130 + (-394)) - (-348 + (-110))
= +194 kJ/mol
**ΔS = ΔS (products) - ΔS (reactants)**
= (161 + 214) - (44 + 198)
= +133 J K^-1 mol^-1
**ΔG = ΔH - TΔS**
0 = ΔH - TΔS
TΔS = ΔH
T = ΔH / ΔS
T = 194 / 0.133
→ divide entropy by 1000 to get it in kJ
T = 1458 K
90
calculate:
1. the standard enthalpy change
2. the standard entropy change
3. the free energy change at 500 C. is the reaction at this temperature feasible?
1. ΔH = - 164 kJ/mol
2. ΔS = - 162.5 J K mol
3. ΔG = ΔH - TΔS
ΔG = -164 - 773 (-0.1625)
= -38.4 kJ/mol
reaction is feasible since G < 0
91
how to change C into K
add 273
92
ΔG = 25.6 kJ mol at 298K
ΔG = 2.3 x RT x logK
R = 8.31 x 10^-3 kJ mol^-1
calculate equilbrium constant
3.20 x10-5
93
order of reaction meaning
number of moles of reactant involved in rate determining step
94
substance used to get a baseline with colorimeter
deionised water
95
why would you dilute a sample solution for colorimeter readings
because the unknowns absorbance must not be out with the calibration range
96
The mass of the screw was 1·43 g.
concentration of Cu2+ in the 0.5 solution = 0.032 mol l-1
Calculate the percentage by mass of copper in the screw.
Absorbance (diluted sample) = 0.34
concentration of Cu2+ in original sample = 0.064 mol l-1
no. of mol in 250cm3 = volume x concentration = 0.25litres x 0.064 mol l-1 = 0.016mol
mass = no. of mol x gfm = 0.016 x 63.5 = 1.016g
=71%
97
is H2O an acid or base
conjugate base
98
