[Year 2] O Flashcards

Question

Why do gases have a larger entropy than solids?

Answer 1

- 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.

Answer 2

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

Answer 3

ΔₛᵧₛS = S(products) - S(reactants) Unit: JK⁻¹mol⁻¹

Answer 4

ΔₜₒₜS = ΔₛᵧₛS + ΔₛᵤᵣᵣS ΔₛᵤᵣᵣS = -ΔH/T (convert ΔH to Jmol⁻¹)

Answer 5

The total entropy change must be positive or zero.

Answer 6

The maximum amount of solid that will dissolve in a solvent. Units: gdm⁻³

Answer 7

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

Answer 8

divide by Mr of solid as mole = mass/Mr

Answer 9

- The equilibrium constant for a sparingly soluble product in a saturated solution.

Answer 10

Same as K𝒸 but only the aqueous products... Kₛₚ = [Aʸ⁺₍ₐᵩ₎]ˣ [Bˣ⁻₍ₐᵩ₎]ʸ (always include state symbols)

Answer 11

1. write Kₛₚ expression using the reaction. Li₂CO₃₍ₛ₎ ⇌ 2Li⁺₍ₐᵩ₎ + CO₃²⁻₍ₐᵩ₎ Kₛₚ = [Li⁺₍ₐᵩ₎]² [CO₃²⁻₍ₐᵩ₎₎] 2. work out the solubility of Li₂CO₃ in moldm⁻³ 13. 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 4. Put into equation Kₛₚ = 0.36² x 0.18 = 0.023 5. Work out units. Kₛₚ = (moldm⁻³)² x (moldm⁻³) = (moldm⁻³)³ = (mol³dm⁻⁹) Answer = 0.023mol³dm⁻⁹

Answer 12

- proton donors.

Answer 13

As hydroxonium ions (H₃O⁺)

Answer 14

- proton acceptors

Answer 15

HA₍ₐᵩ₎ + H₂O₍ₗ₎ ⇌ H₃O⁺₍ₐᵩ₎ + A⁻₍ₐᵩ₎

Answer 16

B₍ₐᵩ₎ + H₂O₍ₗ₎ ⇌ BH⁺₍ₐᵩ₎ + OH⁻₍ₐᵩ₎

Answer 17

- More to the left. - So backwards reaction is favoured. - So dissociate poorly - Therefore not too many H⁺ produced.

Answer 18

- More to the right. - So forward reaction is favoured strongly. - So dissociate almost completely. - Therefore many H⁺ produced.

Answer 19

- More to the right. - So forward reaction is favoured strongly. - So dissociate almost completely. - Therefore many OH⁻ produced.

Answer 20

- More to the left. - So backwards reaction is favoured. - So dissociate poorly. - Therefore not too many OH⁻ produced.

Answer 21

They are exchanged.

Answer 22

- acts as a base. | - as it accepts protons.

Answer 23

A species that has gained a proton.

Answer 24

A species that has lost a proton.

Answer 25

- 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.

Answer 26

- Neutral solution: [H⁺] = [OH⁻] | - Strong Acids: [H⁺] = [Acid]

Answer 27

pH = -log₁₀[H⁺] [H⁺] = 10⁻ᵖᴴ

Answer 28

- Two protons are produced for every 1 acid. | - Therfore [H⁺] = 2[Acid]

Answer 29

In equilibrium with its ions... 2H₂O ⇌ H₃O⁺ + OH⁻ which simplifies to... H₂O ⇌ H⁺ + OH⁻

Answer 30

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⁻]

Answer 31

Kw = 1x10⁻¹⁴ (this is at standard conditions and will change if the temperature changes.) In pure water... [H⁺] = [OH⁻] therefore... Kw = [H⁺]²

Answer 32

- strong bases dissociate to produce OH⁻. - Kw = [H⁺] [OH⁻] - therefore... [H⁺] = Kw / [OH⁻] - then put into pH equation

Answer 33

- 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

Answer 34

- Ka = [H⁺]²/[Ha]ₛₜₐᵣₜ - [H⁺]² = Ka x [Ha]ₛₜₐᵣₜ - [H⁺] = √(Ka x [Ha]ₛₜₐᵣₜ) - then put into pH equation

Answer 35

Another way of measuring the strength of an acid equations: pKa = -log₁₀Ka Ka = 10⁻ᵖᴷᵃ

Answer 36

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

Answer 37

- 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.

Answer 38

- 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.

Answer 39

- 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

Answer 40

- 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