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CHEM > 2.1 Equlibrium, 2.2 Reaction Feasibility, 2.3 Kinetics > Flashcards

2.1 Equlibrium, 2.2 Reaction Feasibility, 2.3 Kinetics Flashcards

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

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2
Q

what happens to the pH of an acid when you dilute it

A

no of H+ ions decreases

pH increases

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3
Q

disorder in spontaneous processes

A

increase in disorder

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4
Q

conjugate base charge

A

-ve

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5
Q

conjugate acid charge

A

+ve

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6
Q

indicator if acid and base are weak

A

no indicator

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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

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8
Q

K if not stated

A

298 K

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9
Q

highest entropy state

A

gas

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10
Q

what is the best indicator to use determined by

A

the pH of the salt

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11
Q

reaction if delta G and delta entropy are negative

A

exothermic

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12
Q

1st order

A

change in reactant is proportional to change in rate

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13
Q

change in reactant is proportional to change in rate

A

1st order

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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

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15
Q

what can order of reaction be determined by

A

experiment only

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16
Q

rate determining step

A

slowest step

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17
Q

K > 1

A

more products than reactants

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18
Q

K < 1

A

more reactants than products

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19
Q

changing temperature in exothermic reactions

A

rise in temp: decrease in K
fall in temp: increase in K

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20
Q

changing temperature in endothermic reactions

A

rise in temp: increase in K
fall in temp: decrease in K

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21
Q

increase in temperature favours what reaction

A

endothermic

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22
Q

decrease in temperature favours what reaction

A

exothermic

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23
Q

order of reactants that do not take place in the slowest step

A

0 order

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24
Q

when do buffers form

A

when a salt of a weak alkali dissolves in a weak acid

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25
Q

overall order of reaction

A

sum of all orders

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26
Q

acid definition

A

Donates proton leaving a conjugate base

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27
Q

How does a buffer resist change in pH if alkali is added

A

OH- ions react with H+ ions and neutralise them

The H+ ions are replaced by the dissociation of the acid into H+ ions

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28
Q

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

A

0.36

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29
Q

equilibrium

A

when the rate of the forward reaction = the rate of the backwards reaction

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30
Q

what is K

A

equilibrium constant

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31
Q

CH3COOH ⇌ H+ + CH3COO-

How would the equilibrium constant be expressed?

A

K = [H+] [CH3COO-]​ / [​CH3COOH​]​

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32
Q

CH3COOH ⇌ H+ + CH3COO-

If sodium ethanoate solid was added, what would happen to

1) the position of equilibrium?
2) the pH

A

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)

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33
Q

base

A

a proton acceptor

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34
Q

What is formed when a base gains a proton?

A

conjugate acid

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35
Q

What is formed when an acid loses a proton?

A

conjugate base

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36
Q

difference between strong and weak acids/bases

A

Strong acids/bases are completely dissociated into ions in aqueous solutions

Weak acids/bases are partially dissociated into ions in aqueous solutions

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37
Q

examples of:

  • strong acids
  • weak acids
  • strong bases
  • weak bases
A
  • Hydrochloric acid, sulfuric acid or nitric acid
  • Ethanoic acid, carbonic acid and sulfurous acid
  • Solutions of metal hydroxides
  • Ammonia and amines
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38
Q

strong acid + strong base =

A

neutral solution

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39
Q

weak acid + strong base =

A

alkaline solution

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40
Q

strong acid + weak base =

A

acidic solution

41
Q

why is a solution of ammonium chloride is acidic

A
  • 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
Q

why is a solution of sodium carbonate is basic

A
  • 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
Q

how does an acid buffer work

A
  • 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
Q

what does a buffer do

A

pH of a buffer solution remains approximately constant when small amounts of acid, base or water are added

45
Q

how does a basic buffer work

A
  • The weak base removes excess hydrogen ions
  • The salt of this base provides the conjugate acid which combines with the hydroxide ions added
46
Q

what are indictors

A

weak acids where the weak acid has a different colour from its conjugate base

47
Q

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

(a) bromothymol blue
(b) methyl orange or bromothymol blue
(c) phenolphthalein

48
Q

why are no indiactors suitable for the titration of a weak acid and a weak base

A

The pH change around the equivalence point is fairly gradual

49
Q

relationshio between entropy and temperature

A

as temp increases, entropy increases

50
Q

What must ΔG° be if a reaction is to be feasible?

51
Q

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.

A

T = ΔH° /ΔS°

T = - 91800 /- 197.3

T = 465K or 192°C

52
Q

zero order reactant

A

changing reactant conc has no effect on the rate

53
Q

second order reactant

A

doubling concentration quadruples rate

54
Q
A

Rate = k [H2O2][I-][H+]0

55
Q

If the rate equation is = k [H2O2][I-], calculate the rate constant.

A

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
Q

K equation

57
Q

K gives an indication of the position of equilibrium

A
  • K > 10^3 = lies to the right
  • 10^-3 < K < 10^3 = equilibrium
  • K < 10^-3 = lies to the left
58
Q

K value for pure liquids and solids

59
Q

what is K affected by

A

temperature

60
Q

strong acid strong base equation

A

HA + B —> BH+ + A-

  • HA: strong acid
  • B: strong base
  • BH: conjugate acid
  • A-: conjugate base
61
Q

when are conjugate acids and bases formed

A
  • conjugate base is formed by the acid donating a proton/H+
  • conjugate acid is formed by the base accepting a proton/H+
62
Q

relationship between strength of acid/base and strength of conjugate acid/base

A
  • the stronger the acid, the weaker the conjugate base
  • the stronger the base, the weaker the conjugate acid
63
Q

pH calculations for strong acids and bases

A

pH = -log10[H+]

pOH = -log10 [[OH-]

14 = pH + pOH

[H+] = 10^-ph

64
Q

calculate the pH of 0.2 mol nitric acid

A

= pH = -log10 [H+]

= -log10 [0.2]

= 0.7

65
Q

calculate the pH of 0.3 mol sodium hydroxide

A

= pOH = -log[OH]

= -log[0.3]

pOH = 0.5

14 = pH + POH

14 - 0.5 = 13.5

pH = 13.5

66
Q

calculate the pH of 0.5 mol hydrochloric acid

A

= pH = -log10 [H+]

= -log10 [0.5]

= 0.3

67
Q

calculate the H+ concentration when the pH of a solution is 9.6

A

= pH = -log10 [H+]

9.6 = -log10 [H+]

[H+] = 10^-9.6

= 2.5 x 10^-10 moles/l

68
Q

weak acid and weak base equation

A

HA + H20 —> H30+ + A-

69
Q

pH equations for weak acids and bases

A

pKa = -log10Ka

pH = 1/2pKa - 1/2log10c

Ka = [H+][A-] / [HA]

70
Q

Ex. calculate the pH of 0.2 mol/l ethanoic acid

A

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
Q

calculate the pH of 0.3 mol/l carbonic acid

A

pH = 1/2pKa - 1/2log10c

pH = 1/2 x 6.35 - 1/2 x 6.35 x 0.3

= 3.44

72
Q

calculate the pH of 0.47 mol/l benzoic acid

A

pH = 1/2pKa - 1/2log10c

pH = 1/2 x 4.2 - 1/2 x 4.2 x 0.47

= 2.27

73
Q

indicator K equations

A

HIn + H2O —> H3O+ + In-

KIn = [H+][In-] / [HIn] [H2O]

[In-] / [HIn] = [KIn] / [H+]

74
Q

endpoint of titration

A

when the pH changes rapidly. where you want the pH of the indicator to lie

75
Q

buffer equation

A

pH = pKa - log10 [acid] / [base]

76
Q

what can buffers be made from

A
  • weak acid and salt of the acid
  • weak base and salt of base
77
Q

what is pH of buffer determined by

A

the pKa of the weak acid/base (linked to its dissociation) and the ratio of acid to salt

78
Q

how to change pH of a buffer

A

change the composition of acid:salt

79
Q

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

A

= pH = pKa - log10 [acid] / [base]

= pH = 4.76 - log10 [0.04] / [0.06]

= 4.76 - (- 0.176)

= 4.94

80
Q

calculate the pH of a buffer solution consisting of 100cm of 0.1 mol/l propanoic acid and 0.2 moles of sodium propionate.

A

= pH = pKa - log10 [acid]/[base]

= 4.87 - log10 x 0.1/2

pH = 6.17

81
Q

rate equation

A

k[M]^m [N]^n

total order: m+n

83
Q

calculate rate equation:

A

k [A]^1 [B]^0 [C]^2

84
Q
A
  1. k [RX]
  2. k = 187.5
  3. units: s
85
Q

enthalpy change of a reaction

  • equation
  • what it shows
  • units
A

ΔH = ΔH (products) - ΔH (reactants)

→ tells us if reaction is exothermic or endothermic, (+ = endothermic, - = exothermic)

→ units are kJ/mol

86
Q

entropy change of a reaction

  • equation
  • what it shows
  • units
A

ΔS = ΔS (products) - ΔS (reactants)

→ tells us if disorder is increasing or decreasing

→ units are J K^-1 mol^-1

87
Q

free energy change of a reaction

  • equation
  • what it shows
  • units
A

ΔG = ΔG (products) - ΔG (reactants)

→ tells us if reaction is feasible (0 > G: feasible, G = 0: equilibrium, G > 0: not feasible)

→ kJ/mol

88
Q

relationship between enthalpy, entropy, and free energy

A

ΔG = ΔH - TΔS

89
Q

at what temperature does this reaction become feasible?

A

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
Q

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?
A
  1. ΔH = - 164 kJ/mol
  2. ΔS = - 162.5 J K mol
  3. ΔG = ΔH - TΔS
    ΔG = -164 - 773 (-0.1625)= -38.4 kJ/molreaction is feasible since G < 0
91
Q

how to change C into K

92
Q

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

A

3.20 x10-5

93
Q

order of reaction meaning

A

number of moles of reactant involved in rate determining step

94
Q

substance used to get a baseline with colorimeter

A

deionised water

95
Q

why would you dilute a sample solution for colorimeter readings

A

because the unknowns absorbance must not be out with the calibration range

96
Q

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.

A

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
Q

is H2O an acid or base

A

conjugate base

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