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Chemistry > Rates, Equilibrium & pH > Flashcards

Rates, Equilibrium & pH Flashcards

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1
Q

When does a dynamic equilibrium exist?

A
  • rate of forward = rate of backward
  • concentrations of reactants and products doesn’t change
  • system is closed (temp, pressure and conc. unaffected by outside influences)
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2
Q

What does le Chatelier’s principle say?

A

When a system in dynamic equilibrium is subjected to an external change, the equilibrium will shift to counteract the change.

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3
Q

What way does a catalyst shift the equilibirum?

A

Catalysts increase the rates of the forward and backward reaction to the same extent, so they have no effect on the position of the equilibrium.

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4
Q

What does the magnitude of K tell us?

A
  • K = 1 means equilibrium is halfway between product and reactants.
  • K < 1 means equilibrium lies towards reactants (1x10-2 indicates an equilibrium well in favour of the reactants)
  • K > 1 means equilibrium lies toards products (100 indicates an equilibrium well in favour of the products)
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5
Q

What does mole fraction mean?

A

The mole fraction is the ratio between the moles of a gas and the total moles of gas in the closed system. It is calculated by dividing the moles of a gasous substance present at equilibrium by the total moles of gas present at equilibrium.

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6
Q

What does partial pressure mean?

A

Partial pressure is the fraction of the total pressure that a particular gaseous substance contributes (or the pressure the gas would exert if it were alone in the volume occupied). It is calculated by multiplying the mole fraction by the total pressure of the system.

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7
Q

What is the table used for Kc?

A

I

C

E

[E] = E/V

([E] only needed when no. of terms on top and bottom isn’t the same).

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8
Q

What is the table used for Kp?

A

I

C

E

x (E/total moles)

p (x * total pressure)

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9
Q

Explain the effect on K (Kp or Kc) of changing the temperature of a system if the forward reaction is exothermic.

A
  1. Increasing temp:
  • K decreases
  • [reactants] increases and [products] decreases (or partial pressures for Kp) to reflect the lower value of K
  • Equilibrium shifts to the left
  1. Decreasing temp:
  • K increases
  • [reactants] decreases and [products] increases to reflect the higher value of K
  • Equilibrium shifts to the right
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10
Q

Explain the effect on K (Kp or Kc) of changing the temperature of a system if the forward reaction is endothermic.

A
  1. Increasing temp:
  • K increases
  • [reactants] decreases and [products] increases (or partial pressures for Kp) to reflect the higher value of K
  • Equilibrium shifts to the right
  1. Decreasing temp:
  • K decreases
  • [reactants] increases and [products] decreases to reflect the lower value of K
  • Equilibrium shifts to the left
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11
Q

Explain the effect on Kp and the position of the equilibrium when:

  1. Increasing pressure
  2. Decreasing pressure

for a reaction with more gaseous moles on the right.

A

1.

  • No change to Kp
  • when increasing pressure, the ratio of p(products)/p(reactants) is greater than Kp AND Kp hasn’t changed.
  • To restore ratio, p(products) decreases and p(reactants) increases
  • Equilibrium shifts to the left

2.

  • No change to Kp
  • When decreasing pressure, the ratio of p(products)/p(reactants) is less than Kp AND Kp hasn’t changed.
  • To restore ratio, p(products) increases and p(reactants) decreases
  • Equilibrium shifts to the right
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12
Q

Explain the effect on Kc and the position of the equilibrium when:

  1. Increasing concentration of reactants
  2. Increasing concentration of products
A

1.

  • No change to Kc
  • when increasing [reactants], the ratio of [reactants]/[products] is less than Kc AND Kc hasn’t changed.
  • To restore ratio, [products] increases and [reactants] decreases
  • Equilibrium shifts to the right

2.

  • No change to Kc
  • when increasing [products], the ratio of [reactants]/[products] is greater than Kc AND Kc hasn’t changed.
  • To restore ratio, [products] decreases and [reactants] increases
  • Equilibrium shifts to the left
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13
Q

What are Bronsted-Lowry acids and bases?

A

Acids: Species that donate a proton

Bases: Species that accept a proton

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14
Q

What are conjugate pairs?

A
  • Species linked by the transfer of a proton (always on opposite sides of a reaction). The species that is formed when a substance looses a proton is called the conjugate base, and the species that is formed when a substance gains a proton is called the conjugate acid.
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15
Q

Write an equation for the equilibrium formed when Bronsted-Lowry acids and bases react, and explain how you would label the conjugate pairs.

A
  • HA + B ⇌ BH+ + A-
  • HA and A- are a conjugate pair: HA is the conjugate acid of A- and A- is the conjugate base of the acid HA
  • B and BH+ are a conjugate pair: B is the conjugate base of BH+ and BH+ is the conjugate acid of the base B
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16
Q

What are alkalis?

A

Bases that release hydroxide ions in water.

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17
Q

Write ionic equations showing the role of H+ in reactions between acids and:

  1. Metals (e.g. Ca)
  2. Carbonates
  3. Metal oxides
  4. Alkalis
A
  1. Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)
  2. CO32-(aq) + 2H+(aq) → H2O(l) + CO2(g)
  3. 2H+(aq) + O2-(s) → H2O(l)
  4. H+(aq) + OH-(aq) → H2O(l)
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18
Q

What is the expression for Ka?

A

Ka = [H+][A-]/[HA]0

[HA]0 can be used because only a tiny amount of HA dissociates (i.e. [HA] >> [H+] or equilibrium sits very far to the left), so the initial concentration is close enough to the equilibirum concentration that to 3 s.f. they can be used interchangeably.

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19
Q

What is the simplified expression for Ka and why can we use it?

A

Ka = [H+]2/[HA]0

Assume that the dissociation of the acid is much greater than the dissociation of the water, so all the H+ ions in solution come from the acid, therefore: [H]+ = [A-]

20
Q

How do you find the pH of a weak acid?

A

[H]+ = sqrt(Ka * [HA]0)

*remember to use Ka not pKa*

21
Q

How do you find the concentration of a weak acid?

A

[HA]0 = [H+]2/Ka

22
Q

What are the limitations of using the Ka approximations?

A

For stronger ‘weak’ acids, there is more dissociation, so the assumption that:

[HA]0 = [HA]equilibrium is no longer valid.

23
Q

How are Ka and pKa related?

A

pKa = -log10(Ka)

24
Q

How are pH and [H+] related?

A

pH = -log10([H+])

25
Q

What is the ionic product of water, and how does it relate to pH?

A

The equilibirum constant for the dissociation of water:

Kw = [OH-][H+]

For pure water, [OH-] = [H+], therefore Kw = [H+]2 , so if you know the Kw of pure water at a certain temperature, you can find pH.

26
Q

How do you calculate the pH of strong n-protic acids?

A

Strong acids fully dissociate, donating all their protons. Therefore:

[H]+ = n*[Acid]0

27
Q

How do you calculate the pH of strong n-acidic bases?

A

Strong bases fully dissociate, donating n moles of hydroxide ions per mole of base:

[OH]- = n * [Base]

Plug this into expression for Kw and rearrange for [H+] to calculate pH.

28
Q

What is a buffer solution?

A

A system that minimises pH changes when small amounts of acid or base are added.

29
Q

What components exist in an acidic buffer solution?

A

Equilibrium between a weak acid and its conjugate base:

HA ⇌ H+ + A-

30
Q

Explain the relative quantities of the components present in a buffer.

A

High concentration of HA and A- relative to H+:

  • When a small amount of acid is added, [H+] increases, so equilibrium shifts to left, with most of the added H+ reacting with A- to form HA, reducing [H+] to close to the original value.
  • When a small amount of base is added, the hydroxide ions react with H+ to form water, decreasing [H+]. Equilibrium shifts to the right, with some HA dissociating into H+ and A-, raising [H+] to close to the original value.

As the amount of HA and A- is so much bigger than H+ in each case, the change in their concentrations is negligible.

31
Q

What are the two methods of making an (acidic) buffer?

A
  1. Mix a weak acid with its conjugate salt (salt of its conjugate base):
    • The salt fully dissociates: ANa → A- + Na+
    • The weak acid only partially dissociates, so the concentration of HA and A- is much higher than H+
  2. Mix an excess of weak acid with a strong alkali:
    • All the base gets used up when it reacts with the weak acid:
      • HA + OH- → A- + H2O
    • The excess weak acid partially dissociates, forming a solution with a high conc. of HA and A- and a relatively low conc. of H+
32
Q

How do you calculate the pH of a buffer solution (include assumptions)?

A

Assume that:

  • The salt of the conjugate base is fully dissociated i.e. [A-]equilibrium = [Salt]0
  • HA is only slightly dissociated i.e. [HA]equilibrium = [HA]0

Then calculate the pH as you would for any weak acid, using the original expression for Ka, not the simplified one.

33
Q

Explain how the body controls pH in the blood (including equations).

A
  • pH needs to be kept between 7.35 and 7.45
  • pH controlled using a carbonic acid-hydrogen carbonate buffer system:
    • H2CO3 ⇌ H+ + HCO3-
    • H2CO3 ⇌ H2O + CO2
  • The levels of H2CO3 are controlled via respiration: breating out CO2 shifts equilibrium to the right.
  • The levels of HCO3- are controlled by the kidneys, with excess being extreted in the urine.
34
Q

Sketch and explain the shape of the titration curves for :

  • a strong acid/strong base titration
  • a strong acid/weak base titration
  • a weak acid/strong base titration
  • a weak acid/strong weak titration
A
  • The initial pH depends on the strength of the acid
  • At the start, small amounts of base have little impact on the pH of the solution
  • All the graphs then have a vertical section (expect weak/weak), which is the equivalence (end) point, where the moles of base added is equal to the original moles of acid, and all the acid has been neutralised. At this point, small additions of base cause sudden, large increases in pH.
  • The final pH depends on the strength of the base: the stronger the base, the higher the pH.
35
Q

How do you calulate Ka from the equivalence point?

A
  • If you add half the volume of base needed to reach equivalence point, the moles of base is equal to half the original moles of acid. By this point, half of the moles of acid will have been neutralised, so the concentrations of acid and base will be equal.
  • As we are assuming [HA] = [A-] (ignoring dissociation from weak acid, only counting A- formed in reaction with base), this means [A-] = [H+], so pKa = pH.
36
Q

How do you choose a suitable indicator for a titration?

A

The entire pH range needs to lie within the vertical part of the pH curve.

37
Q

What types of titrations is methyl orange suitable for?

A
  • pH range: 3.1 - 4.4
  • strong acid/strong base
  • strong acid/weak base
38
Q

What types of titration is phenolphthalein suitable for?

A
  • pH range: 8.3 - 10
  • strong acid/strong base
  • strong base/weak acid
39
Q

How do indicators work?

A
  • Differently coloured conjugate pairs
  • As pH changes during a titration, the equilibrium concentrations of the conjugate pairs will also change.
  • Colour changes depending on whether majority of indicator is protonated or deprotonated.
40
Q

How do you calibrate a pH meter?

A
  • Place bulb into distilled water and allow reading to settle. Adjust reading to 7.
  • Repeat with standard solutions of pH 4 and pH 10, rinsing probe between each reading.
41
Q

Outline how a colorimeter can be used to continuously monitor the progress of a reaction.

A
  1. Prepare standard solutions of the substance of which you are measuring the concentration.
  2. Plot calibration curve.
  3. Measure aliquots of reaction mixture taken at regular time intervals to determine concentration at these times.
42
Q

How does an iodine clock reaction work?

A
  1. A known amount of delaying agent and a few drops of indicator are added to an excess of reactants and a stopclock is started.
  2. As the reactants react to form a product, the delaying agent instantly reacts with the product, meaning there is no product in the mixture until the delaying agent is used up.
  3. Once all the delaying agent has been used up (a set amount of product has been formed), the indicator (starch) turns blue-black and the stopclock is stopped.
43
Q

What shapes are the concentration-time and rate-concentration graphs for:

  • Zero order reactions?
  • First order reactions?
  • Second order reactions?
A
  • Straight line with -ve gradient & y-intercept at []0, horizontal line w/ y-intercept at k
  • Exponential decay curve w/ regular half-lives, straight line through origin w/ gradient equal to k
  • More rectangular decay curve w/ increasing half-lives, parabola w/ y-intercept at 0
44
Q

What is the rate determining step and how is it related to the rate equation?

A

The step with the slowest rate in a multi-step reaction. The order of a reaction wrt a reactant shows the number of molecules of that reactant that are involved in the rds.

45
Q

What do the variables in the Arrhenius equation represent?

A

k = Ae-Ea/RT

Where:

k = rate constant

A = pre-exponential factor

Ea = activation energy in kJmol-1

R = gas contant (8.31 JK-1mol-1)

T = temperature in Kelvin

46
Q

How can you calculate the activation energy of a reaction using the Arrhenius equation?

A
  1. Write Arrhenius equation in logarithmic form: ln k = -Ea/RT + ln A
  2. Plot ln k against 1/T, to get a graph with gradient -Ea/R and y-intercept at ln A
  3. Multiply gradient by -R to get activation energy.

Chemistry (11 decks)

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