Unit 1.7 Equilibria reactions

Question 1

What are reversible reactions?

Answer 1

Reactions which can be readily reversed, allowing the products to be easily turned back into reactants

Examples:
1. blue copper sulphate + heat -> white copper sulphate + steam
CuSO₄.5H₂O(s) -> CuSO₄(s) + H₂O(g)
To reverse, add a few drops of water to the anhydrous copper sulfate:
2. white copper sulphate + water -> blue copper sulphate
CuSO₄(s) + H₂O(l) -> CuSO₄.5H₂O(s)

First reaction needed heat, ∴ it’s endothermic
Second reaction released heat, ∴ it’s exothermic

Question 2

How to write a reversible reaction?

Answer 2

How to write reversible reaction arrow? = Half arrow forward, half arrow backwards, both bottom & above each other (if u know u know)
In addition, u can show which is favoured more based on the length (,’:))

Writing a reversible reaction:
Sealed tube of ammonium chloride heated, will decompose into ammonia and hydrogen chloride gas
1. NH₄Cl(s) -> NH₃(g) + HCl<g)
If tube remained sealed, once enough products formed, they’ll react and recombine to form the ammonium chloride. Behold, the reverse reaction
2. NH₃(g) + HCl(g) -> NH₄Cl(s)
Now you can combine these to create a reversible reaction equation:
NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

Question 3

What is dynamic equilibrium?

Answer 3

When the forward and reverse reactions occur at the same rate
- When a reversible reactions ‘appears’ to have stopped
- It has reached dynamic equilibrium
- It didn’t actually stop but the forward reaction is happening at the same rate as the reverse reaction
Forward reaction = Reactants
Reverse reaction = Products

Further detailed explanation:
- When one molecule of ammonium chloride splits
- One molecule of ammonia joins with one molecule of hydrogen chloride
- The amount/moles of the reactant and product will therefore remain the same
- ‘Appears’ that the reaction has stopped

Question 4

What are the 2 key features highlighted of dynamic equilibrium?

Answer 4

1. The rate of the forward reaction is equal to the rate of the reverse reaction, the system is in constant motion at the molecular level
2. The concentrations of the reactants and products remain constant, the macroscopic properties remain constant
   For the features of dynamic equilibrium to exist, the system must be closed, so that nothin can escape or enter

Question 5

How to tell if equilibrium has been reached with[/out] graph?

Answer 5

• Concentration of reactants (or products) remain the same
• In a graph, practically both forward & reverse reaction are like “in-phase”, damn physics style

In addition to all this, u can state the position of equilibrium.
Basically stating which side favours the equilibrium.
So u can either get more reactants or more products based on which one has more concentration during the dynamic equilibrium.

Question 6

In industry, which type dynamic equilibrium reaction would be favoured the most? and why?

Answer 6

• One where it favours the products (the right hand side)
• Why? So that there’s more product
• More product = more profit

Question 7

What is the Le Chatelier’s principle
(pretty big, not for long)

Answer 7

Le Chatelier’s principle - If a system at equilibrium is subjected to change, the equilibrium tends to shift so as to minimise the effect of that change.

Changes he’s referring to:
- Pressure
- Temperature (endothermic & exothermic)
- Concentration of reactants/products

Guess u just needa know how to describe what happens to a system if the equilibrium is disturbed by a change in external conditions (not my words)

Question 8

Le Chatelier’s principle in terms of temperature

Answer 8

• To find out the effect of changing temp., find out whether the reaction is exothermic (△H = -Ve) or endothermic (△H = +Ve)
  Example:
  Exothermic reaction = 2NO₂(g) ⇌ N₂O₄(g) △H = -24kJ mol^-1
  If heat increases:
• System will shift to minimise the change by tryna reduce the temperature
• Heat will be taken in
• Endothermic, reverse reaction will be favoured
• Equilibrium will move towards the left hand side (the reactants)
  If heat decreases:
• System will shift to minimise the change by producing heat
• Heat will be taken out
• Exothermic, forward reaction will be favoured
• Equilibrium will move towards the right hand side (the products)

Example 2:
Endothermic reaction = N₂(g) + O₂(g) ⇌ 2NO(g) △H = +180kJ mol^-1
If heat increases:
- System will shift to minimise the change by tryna reduce the temperature
- Heat will be taken in
- Endothermic, forward reaction will be favoured
- Equilibrium will move towards the right hand side (the products)
If heat decreases:
- System will shift to minimise the change by producing heat
- Heat will be taken out
- Exothermic, reverse reaction will be favoured
- Equilibrium will move towards the left hand side (the reactants)

In general:
Endothermic + heat = more of that side
Exothermic - heat = more of that side
Use ur 100% brain power O_o

Question 9

Le Chatelier’s principle in terms of pressure

Answer 9

• Only for reactions that are homogeneous in the gas phase
• Little of no effect for solids/liquids

In general:
- Look at which has more moles (in either products/reactants)
- If u increase pressure, will favour the side with the least moles
- Opposite for decreasing pressure, favour side with most moles
- However, if moles on both sides equal, no change in equilibrium. Could increase rate tho

Question 10

Le Chatelier’s principle in terms of concentration change

Answer 10

• If you increase the concentration on the left hand side
• Equilibrium will move to the right hand side to try minimise the effect
• More product will be made
  U should know what that means now

Question 11

Le Chatelier’s principle in terms of catalysts

Answer 11

• Actually, they don’t affect the position of equilibrium
• Employed solely to increase the rate at which the equilibrium is established
• Speeds up both forward and reverse reactions equally

Question 12

How to tell if it’s exothermic or endothermic based on change in heat?

Answer 12

Negative heat = exothermic
Positive heat = endothermic

Question 13

In endothermic reaction, more heat = more product = more profit, which is desirable. What’s the downside?

Answer 13

Higher temperature = higher price ∴ may not be profitable

Question 14

In the effect of pressure change, may be desirable to increase pressure to increase products if there is more moles on the reactants side. What’s the downside?

Answer 14

• Higher pressures may be harder to control/maintain
• May possibly cause an explosion

Question 15

Tell me about the Contact process in terms of dynamic equilibrium

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) △H = -196 kJmol^-1

Answer 15

The Contact process:
The production of sulfuric acid
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) △H = -196 kJmol^-1
Several ways to make this profitable:
1. Catalyst = vanadium oxide catalyst at 450°C w/ 1-2atm pressure
2. High pressure = more product
- May be harder to control/maintain
3. Low temperature = more product
- May also be harder to control/maintain
4. Remove SO₃ as it is produced
- Causes equilibrium to move towards the right hand side by forming more product to replace the removed one

Question 16

Tell me about the Haber process in terms of dynamic equilibrium

3H₂(g) + N₂(g) ⇌ 2NH₃(g) △H = -92 kJmol^-1

Answer 16

The Haber process:
The production of ammonia (NH₃) used for fertilizers
3H₂(g) + N₂(g) ⇌ 2NH₃(g) △H = -92 kJmol^-1
Several ways to make this profitable:
1. Catalyst = iron catalyst at 400-450°C around 250 atms
2. Low temperature = more product
- May be harder to control/maintain
3. Higher pressure = more product
- Harder to control/maintain
4. Perhaps increasing concentration of 3H₂ or N₂ = more product
- May be pricey affecting profit
5. Remove 2NH₃ as it is produced
- Causes equilibrium to move towards the right hand side by forming more product to replace the removed one
- In addition to this, unused N₂ & H₂ recycled

Question 17

Tell me about equilibrium constant, Kc

Answer 17

• Used to find the unit for certain reactions
• Which the size of that can determine the position of the equilibrium

For the general reaction:
aA + bB ⇌ cC + dD
Then:
K_c = [C]^c[D]^d/[A]^a[B]^b <- divided by
Large Kc values = more products = pos. of equilibrium will lie to the right
Kc less than 1 = more reactants = pos. of equilibrium will lie to the left
- IN ADDITION, [X] = concentration of X
- Power = moles

First 4 examples:
1. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]²/[SO₂]² + [O₂] = 1/moldm^-3 = dm³mol^-1

2. 2HI(g) ⇌ H₂(g) + I₂(g)
   Kc = [H₂] [I₂]/[HI]² = no units
3. 2NH₃(g) ⇌ N₂(g) + 3H₂(g)
   Kc = [N₂] [H₂]³/[NH₃]² = moldm^-3moldm^-3/1 = mol²dm^-6
4. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
   Kc = [NH₃]²/[N₂] [H₂]³ = 1/moldm^-3moldm^-3 = dm⁶mol^-2

Well my issue is, how am I supposed to know which side it favours based on this???

In terms of a set of results:
- Well there’s gonna be some equation given
- And results of “equilibrium conc”
- As [X] = conc of X
- And power = moles
- U can just find out the Kc
- Find its units first, so without the conc. first

My best bet for this:
Just do a few questions on this ¬.¬

Question 18

What is the difference between acids and bases?

Answer 18

Acid - any hydrogen-containing substance that is capable of donating a proton (hydrogen ion) to another substance
Base - a molecule or ion able to accept a hydrogen ion from an acid

Question 19

How to tell whether an acid is strong/weak?

Answer 19

• Based on the amount it dissociates in water
  e.g.
  Strong acid = HCl(aq) -> Cl^-(aq) + H⁺(aq)
• Equilibrium lies almost completely to RHS
• There is no ⇌ symbol within equation
  Weak acid = CH₃COOH(aq) ⇌ CH₃COO^-(aq) + H⁺(aq)
• Clearly see H₃ still within compound
• Hasn’t fully dissociated
• ⇌ symbol used in the equation

In addition, applies to alkali’s as well, incase u get confused

Question 20

Difference between strength and concentration in acids

Answer 20

Strength = amount an acid is dissociated
Concentration = quantity of ions in solution

• Concentrated acid has a high number of ions (in a small n° of water molecules)
• Dilute acid has a low n° of ions in a large quantity
  Therefore in this example:
  0.0005moldm^-3 of H₂SO₄
  Still a strong acid - it’s just very dilute

Question 21

Tell me about the measurement of acidity

Answer 21

pH - a measure of the acidity of a solution

• When considering acidity of different substances
• Usually measured in water
• Hence convenient to line up acids according to their ability to donate protons to water molecules
  HA(aq) <===> H⁺(aq) + A^-(aq)
• Greater the ability of acid to donate protons = greater the concentration of H⁺ ions
• Greater the ability of the acid then the stronger the acid

Question 22

How to actually gain the pH of an acid?

Answer 22

pH = -log₁₀[H⁺]
e.g.
[H⁺] = 1.0 x 10^-pH moldm^-3
pH = -log₁₀([H⁺] /moldm^-3)

Buncha questions to find pH:
a.) 0.1 moldm^-3
= -log(0.1) = 1
b) 3.0 x 10^-4 moldm^-3
= -log(3 x 10^-4) = 3.52 (2dp)
c) 6.51 x 10^-2 moldm^-3
= -log(6.51 x 10^-2) = 1.18

Buncha questions to find concentration:
- If pH = -log₁₀[H⁺]
Then [H⁺] = 10^-pH (antilog on calculator)

Concentration of [H⁺] if pH is:
a) 3.42
= 10^-3.42 = 3.80 x 10^-4
b) 13
= 10^-13 = 1 x 10^-13
c) 2.5
= 10^-10 = 3.16 x 10^-3

Another best bet with this:
Just do a few questions on these too

Question 23

Tell me all about titration

Answer 23

Basic equipment u’d def need:
- Read the book D: