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A-Level Chemistry [OCR B Salters] (sorry for any spelling mistakes - there are many) > [Year 2] O > Flashcards

[Year 2] O Flashcards

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1
Q

Describe the greenhouse effect.

A
  • Most UV radiation from the sun is absorbed by gasses in the earth’s atmosphere.
  • Some is reflected back into space.
  • Some reaches the surface of Earth which is absorbed.
  • This heats up the surface of the earth and radiates IR back towards space.
  • Some IR escapes through the ‘IR window’, as they are not absorbed by the gases in the atmosphere.
  • Some IR is absorbed by gases in the troposphere and reemitted in all directions included back towards earth.
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2
Q

Give examples of greenhouse gasses?

A

CO₂
H₂O
CH₄

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3
Q

What happens when greenhouse gasses absorb radiation?

A
  • Their bonds in the molecules vibrate more.
  • The vibrational energy is transferred to neighbouring particles during collisions.
  • These particles now have more kinetic energy and their temperature starts to rise.
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4
Q

How can polar molecules be dissolved?

A

Using polar solvents.

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5
Q

What is the difference between hydration and solvation?

A
  • Hydration is dissolving with water as a solvent.

- Whereas solvation is dissolving with a non-water solvent.

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6
Q

What happens when a salt dissolves in water?

A
  • Water is polar.
  • Its 𝛿+ Hs are attracted to the negative ions
  • Its 𝛿- O is attracted to the positive ions.
  • The water molecules surround the ions during hydration.
  • This ion-dipole interaction must be stronger/as strong as the bond broken for it to dissolve.
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7
Q

Why might molecules like Al₂O₃ not dissolve in water?

A
  • The bond between Al and O ions are stronger than the ion-dipole interaction that would form.
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8
Q

How do some non-polar molecules dissolve in water?

A

Through hydrogen bonds.

e.g. alcohols disolving in water

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9
Q

Why don’t polar molecules like haloalkanes dissolve in water?

A
  • they have weak dipoles.

- so water will form a stronger intermolecular bond than with the haloalkane.

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10
Q

How can you dissolve a haloalkane?

A

Use a solvent that can interact via pertinent dipole-dipole interactions.

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11
Q

How do non-polar substances dissolve in non-polar solvents?

A
  • by forming instantaneous dipole-induced dipole forces.
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12
Q

What is the standard lattice enthalpy?

A

The energy change when 𝟭 𝗺𝗼𝗹𝗲 of an ionic compound is formed from its 𝗴𝗮𝘀𝗲𝗼𝘂𝘀 ions under 𝘀𝘁𝗮𝗻𝗱𝗮𝗿𝗱 𝗰𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝘀.

(298K/100kPa)

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13
Q

What does the sign and magnitude of the standard lattice enthalpy tell you?

A
  • Always negative, as energy is released as bonds form.

- more negative = stronger bonding.

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14
Q

What is the enthalpy change of solution?

A

The energy change when 𝟭 𝗺𝗼𝗹𝗲 of an ionic compound is dissolved in the 𝗺𝗶𝗻𝗶𝗺𝘂𝗺 𝗮𝗺𝗼𝘂𝗻𝘁 𝗼𝗳 𝘀𝗼𝗹𝘃𝗲𝗻𝘁 to ensure 𝗻𝗼 𝗳𝘂𝗿𝘁𝗵𝗲𝗿 enthalpy change upon further dilution.

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15
Q

What is the enthalpy change of hydration?

A

The energy change when 𝟭 𝗺𝗼𝗹𝗲 of aqueous ions are formed from its 𝗴𝗮𝘀𝗲𝗼𝘂𝘀 ions under 𝘀𝘁𝗮𝗻𝗱𝗮𝗿𝗱 𝗰𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝘀.

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16
Q

How is the lattice enthalpy, enthalpy of solution and enthalpy of hydration linked?

A

enthalpy of solution = enthalpy of hydration - lattice enthalpy.

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17
Q

What happens if the enthalpy change of hydration is larger (in magnitude) than the lattice enthalpy?

A
  • The enthalpy change of solution will be negative.
  • So the enthalpy change of solution will release energy.
  • So the process is exothermic.
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18
Q

How can you find the enthalpy change of solution practically?

A

Calorimetry.

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19
Q

Describe calorimetry to find the enthalpy change of solution practically

A
  • In a polystyrene cup add the solvent and measure the initial temperature of the solvent,
  • measure out your solid solute in a weighing boat using scales.
  • add the solute to the solvent and stir. Put a lid over the polystyrene cup (leave a hole in the lid to keep the thermometer in.
  • Note down the maximum temperature reached after the solute was added.
  • use the equation E = mcΔT to work out the enthalpy.
    (mass in grams, T can be in any unit).
  • Work out the moles of solute added by doing moles = mass/Mr
  • Divide this by the energy to find the enthalpy change of solution.
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20
Q

Why might the enthalpy change of the solution from a practical be different than the actual value?

A
  • heat lost to surroundigns.
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21
Q

What affects the lattice enthalpy?

A
  • Size of ions (ionic radii)

- Size of change (bigger difference in charge = stronger bond).

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22
Q

How do the ionic radii affect the lattice enthalpy?

A
  • The smaller the ionic radii, the stronger the electrostatic attraction between ions.
  • as smaller ions can pack together more closely.
  • and so have a larger charge density.
  • Thus more energy is required to overcome the stronger forces.
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23
Q

How does the size of charge affect the lattice enthalpy?

A
  • Bigger the charge on an ion, the stronger the electrostatic attraction between ions.
  • Thus more energy is required to overcome these forces.
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24
Q

What is entropy?

A

The measure of disorder in a system.

  • It is the number of ways energy can be shared out in a system.
  • Given the letter S.
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25
Q

Why do gases have a larger entropy than solids?

A
  • Gases are more disordered.
  • So there are more ways to arrange the atoms/energy than in a solid.
  • And so it has a higher entropy.
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26
Q

How does the number of particles affect the entropy of a reaction?

A
  • If there are more moles of a particle then there are more ways energy can be distributed.
  • and so the entropy has increased.
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27
Q

How is the entropy change calculated?

A

ΔₛᵧₛS = S(products) - S(reactants)

Unit: JK⁻¹mol⁻¹

28
Q

How can the total entropy change be calculated?

A

ΔₜₒₜS = ΔₛᵧₛS + ΔₛᵤᵣᵣS

ΔₛᵤᵣᵣS = -ΔH/T (convert ΔH to Jmol⁻¹)

29
Q

What must be true for a reaction to be feasible?

A

The total entropy change must be positive or zero.

30
Q

What is solubility?

A

The maximum amount of solid that will dissolve in a solvent.

Units: gdm⁻³

31
Q

What is the ‘point of saturation’?

A

The point where water (or a solvent) can no longer hold any more solute.

32
Q

How do you get moldm⁻³ from gdm⁻³

A

divide by Mr of solid

as mole = mass/Mr

33
Q

What is the solubility product (Kₛₚ)?

A
  • The equilibrium constant for a sparingly soluble product in a saturated solution.
34
Q

What is the formula for Kₛₚ?

Use the general equation AₓBᵧ₍ₛ₎ ⇌ xAʸ⁺₍ₐᵩ₎ + yBˣ⁻₍ₐᵩ₎

A

Same as K𝒸 but only the aqueous products…

Kₛₚ = [Aʸ⁺₍ₐᵩ₎]ˣ [Bˣ⁻₍ₐᵩ₎]ʸ

(always include state symbols)

35
Q

How would you work through the following question…

Calculate the solubility product of Li₂CO₃ when the solubility of the is 13.3gdm⁻³ at 20ᵒC.

A
  1. write Kₛₚ expression using the reaction.
    Li₂CO₃₍ₛ₎ ⇌ 2Li⁺₍ₐᵩ₎ + CO₃²⁻₍ₐᵩ₎

Kₛₚ = [Li⁺₍ₐᵩ₎]² [CO₃²⁻₍ₐᵩ₎₎]

  1. work out the solubility of Li₂CO₃ in moldm⁻³
  2. 3/Mr(Li₂CO₃) = 13.3/74 = 0.18moldm⁻³
  3. We know 1 mol of Li₂CO₃ dissocaites to form 2 moles of Li⁺ and 1 mole of CO₃²⁻ therfore…

[Li⁺₍ₐᵩ₎] = 0.18 x 2 =0.36
[CO₃²⁻₍ₐᵩ₎] =0.18

  1. Put into equation

Kₛₚ = 0.36² x 0.18 = 0.023

  1. Work out units.

Kₛₚ = (moldm⁻³)² x (moldm⁻³) = (moldm⁻³)³ = (mol³dm⁻⁹)

Answer = 0.023mol³dm⁻⁹

36
Q

What is a feature of Brønsted-Lowry acid?

A
  • proton donors.
37
Q

How do H⁺ ions actually exist in water?

A

As hydroxonium ions (H₃O⁺)

38
Q

What is a feature of Brønsted-Lowry base?

A
  • proton acceptors
39
Q

Give a general equation for a Brønsted-Lowry acid in water?

A

HA₍ₐᵩ₎ + H₂O₍ₗ₎ ⇌ H₃O⁺₍ₐᵩ₎ + A⁻₍ₐᵩ₎

40
Q

Give a general equation for a Brønsted-Lowry base in water?

A

B₍ₐᵩ₎ + H₂O₍ₗ₎ ⇌ BH⁺₍ₐᵩ₎ + OH⁻₍ₐᵩ₎

41
Q

Where is the position of equilibrium in weak acids, and what does this mean?

A
  • More to the left.
  • So backwards reaction is favoured.
  • So dissociate poorly
  • Therefore not too many H⁺ produced.
42
Q

Where is the position of equilibrium in strong acids, and what does this mean?

A
  • More to the right.
  • So forward reaction is favoured strongly.
  • So dissociate almost completely.
  • Therefore many H⁺ produced.
43
Q

Where is the position of equilibrium in strong bases, and what does this mean?

A
  • More to the right.
  • So forward reaction is favoured strongly.
  • So dissociate almost completely.
  • Therefore many OH⁻ produced.
44
Q

Where is the position of equilibrium in weak bases, and what does this mean?

A
  • More to the left.
  • So backwards reaction is favoured.
  • So dissociate poorly.
  • Therefore not too many OH⁻ produced.
45
Q

What happens to protons in acid-base reactions?

A

They are exchanged.

46
Q

What is the role of water in the following reaction…

HA₍ₐᵩ₎ + H₂O₍ₗ₎ ⇌ H₃O⁺₍ₐᵩ₎ + A⁻₍ₐᵩ₎

A
  • acts as a base.

- as it accepts protons.

47
Q

What is a conjugate acid?

A

A species that has gained a proton.

48
Q

What is a conjugate base?

A

A species that has lost a proton.

49
Q

What are the conjugate pairs in the following reaction…

HA + B ⇌ BH⁺ A⁻

A
  • HA and A⁻ are pairs with HA as the acid and A⁻ as the base.
  • B and BH⁺ are pairs with B as the base and BH⁺ as the acid.
50
Q

What assumptions can you make about the concentrations in…

  • A neutral solution?
  • A strong acid?
A
  • Neutral solution: [H⁺] = [OH⁻]

- Strong Acids: [H⁺] = [Acid]

51
Q

What is the pH equation?

A

pH = -log₁₀[H⁺]

[H⁺] = 10⁻ᵖᴴ

52
Q

What must you factor in when finding the pH of acids like H₂SO₄

A
  • Two protons are produced for every 1 acid.

- Therfore [H⁺] = 2[Acid]

53
Q

How does pure water exist in reality?

A

In equilibrium with its ions…

2H₂O ⇌ H₃O⁺ + OH⁻

which simplifies to…

H₂O ⇌ H⁺ + OH⁻

54
Q

Use the equilibrium of pure water to derive Kw.

A

H₂O ⇌ H⁺ + OH⁻

Kc = [H⁺] [OH⁻] / [H₂O]

Water dissociated very weakly, therefore, we can assume the concentration of water is a constant.

So we can multiply out [H₂O] on both sides to get…

Kw = [H⁺] [OH⁻]

55
Q

What assumptions can we make when using Kw for pure water and Kw at standard conditions?

A

Kw = 1x10⁻¹⁴ (this is at standard conditions and will change if the temperature changes.)

In pure water… [H⁺] = [OH⁻]
therefore…
Kw = [H⁺]²

56
Q

How do you work out the pH of strong bases?

A
  • strong bases dissociate to produce OH⁻.
  • Kw = [H⁺] [OH⁻]
  • therefore…
    [H⁺] = Kw / [OH⁻]
  • then put into pH equation
57
Q

What constant is used to help work out the pH of weak acids?

A

Ka

58
Q

Use the equilibrium of weak acids to derive Ka.

A
  • HA ⇌ H⁺ + A⁻
  • Kc = [H⁺] [A⁻]/[Ha]
  • Only a small amount of weak acids dissociates so… [HA₍ₐᵩ₎]equilibrium ≈ [HA₍ₐᵩ₎]start
  • Ka = [H⁺] [A⁻]/[Ha]ₛₜₐᵣₜ
  • The dissociation of acid is much greater than of water so we can assume all the H⁺ comes from the acid so…
    [H⁺₍ₐᵩ₎] ≈ [A⁻₍ₐᵩ₎]
  • Ka = [H⁺]²/[Ha]ₛₜₐᵣₜ
59
Q

How do you work out the pH of a weak acid?

A
  • Ka = [H⁺]²/[Ha]ₛₜₐᵣₜ
  • [H⁺]² = Ka x [Ha]ₛₜₐᵣₜ
  • [H⁺] = √(Ka x [Ha]ₛₜₐᵣₜ)
  • then put into pH equation
60
Q

What is pKa?

A

Another way of measuring the strength of an acid

equations:
pKa = -log₁₀Ka
Ka = 10⁻ᵖᴷᵃ

61
Q

What is a buffer?

A

A chemical that resists a change in pH when small amounts of acid or base are added.

62
Q

What are acidic buffers?

A
  • something that resist the change of pH in order to keep a solution below pH7
  • it is made from a weak acid and it’s salt.
63
Q

Ethanoic acid (CH₃COOH) can be used as a buffer with sodium ethanoate (CH₃COO⁻Na⁺).

Describe what happened when an acid is added?

A
  • Ethanoic acid is a weak acid in equilibrium…
    CH₃COOH₍ₐᵩ₎ ⇌ CH₃COOH₍ₐᵩ₎ + H⁺
  • Because it is a weak acid its equilibrium lies far to the left so the concentrations are…
    CH₃COOH₍ₐᵩ₎ ⇌ CH₃COOH₍ₐᵩ₎ + H⁺
    [High] [Low] [Low]
  • Sodium ethanoate dissociates fully in equilibrium…
    CH₃COO⁻Na⁺ ⇌ CH₃COO⁻ + Na⁺
  • So its equilibrium lies far to the right and ist concentration are…
    CH₃COO⁻Na⁺ ⇌ CH₃COO⁻ + Na⁺
    [Low] [High] [High]
  • When an acid is added its H⁺ react with a negative ion.
  • This is CH₃COO⁻ from the sodium ethanoate equation, as there is a high concentration of these.
  • This produces more CH₃COOH using the additional H⁺ returning the pH to normal and shifting the equilibrium to the left.
64
Q

Ethanoic acid (CH₃COOH) can be used as a buffer with sodium ethanoate (CH₃COO⁻Na⁺).

Describe what happened when a base is added?

A
  • Ethanoic acid is a weak acid in equilibrium…
    CH₃COOH₍ₐᵩ₎ ⇌ CH₃COOH₍ₐᵩ₎ + H⁺
  • Because it is a weak acid its equilibrium lies far to the left so the concentrations are…
    CH₃COOH₍ₐᵩ₎ ⇌ CH₃COOH₍ₐᵩ₎ + H⁺
    [High] [Low] [Low]
  • Sodium ethanoate dissociates fully in equilibrium…
    CH₃COO⁻Na⁺ ⇌ CH₃COO⁻ + Na⁺
  • So its equilibrium lies far to the right and its concentration are…
    CH₃COO⁻Na⁺ ⇌ CH₃COO⁻ + Na⁺
    [Low] [High] [High]
  • When a base is added its OH⁻ ions react with a positive charge.
  • This is not the Na⁺ however as this will create a base and will dissociate back to OH⁻.
  • So this reacts with the H⁺.
  • Because there is a low contraction of H⁺ as this is used up, it shifts the equilibrium of the ethanoic acid equation to the right.
  • This drives the forward reaction and so produces more H⁺ to return the pH to normal
65
Q

How would you work through the following question…

Calculate the pH of a buffer that contains 2.35x10⁻² mol of methanoic acid and 1.84x10⁻² mol of sodium methanoate in a 1dm³ solutions. The value of Ka at 25°C for methanoic acid is 1.78x10⁻⁴

A
  • Write Ka expression from the equilibrium equation.
    HCOOH₍ₐᵩ₎ ⇌ H⁺₍ₐᵩ₎ + HCOO⁻₍ₐᵩ₎
    Ka = [H⁺][HCOO⁻] / [HCOOH]
  • we assume that salts dissociate fully and weak acids dissociate poorly so…
    [salt] = [A⁻] AND [HA₍ₐᵩ₎]equilibrium ≈ [HA₍ₐᵩ₎]start
  • [H⁺] = Ka x [HCOOH] / [HCOO⁻]
  • [H⁺] = 1.78x10⁻⁴ x 2.35x10⁻²) / 1.84x10⁻²
  • [H⁺] = 2.27x10⁻⁴ moldm⁻³
  • plug into pH equitation
  • pH = 3.64

A-Level Chemistry [OCR B Salters] (sorry for any spelling mistakes - there are many) (14 decks)

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