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ENG CHEM > Chapter 8 Equilibra > Flashcards

Chapter 8 Equilibra Flashcards

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1
Q

Reversible reaction

A
  • Some reactions go to completion where the reactants are used up to form the product molecules and the reaction stops when all of the reactants are used up
  • In reversible reactions the products can react to reform the original reactants
  • To show a reversible reaction, two half arrows are used: ⇌
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2
Q

Dynamic equilibrium

A

In a dynamic equilibrium the reactants and products are dynamic (they are constantly moving)
-In a dynamic equilibrium the rate of the forward reaction is the same as the rate of the backward reaction in a closed system and the concentrations of the reactants and products is constant

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3
Q

A closed system

A

is one in which none of the reactants or products escape from the reaction mixture

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4
Q

In an open system

A

some matter is lost to the surroundings

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5
Q

Le Chatelier’s principle

A

says that if a change is made to a system at dynamic equilibrium, the position of the equilibrium moves to minimise this change
-The principle is used to predict changes to the position of equilibrium when there are changes in temperature, pressure or concentration

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6
Q

The position of the equilibrium

A

refers to the relative amounts of products and reactants in an equilibrium mixture.

  • When the position of equilibrium shifts to the left, it means the concentration of reactants increases
  • When the position of equilibrium shifts to the right, it means the concentration of products increases
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7
Q

Effects of concentration: increase in concentration of reactants and how does the equilibrium shifts

A

equilibrium shifts to the right to reduce the effect of increase in the concentration of a reactant

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8
Q

Effects of concentration: increase in concentration of products and how does the equilibrium shifts

A

equilibrium shifts to the left to reduce the effect of increase in the concentration of product

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9
Q

Effects of pressure: increase in pressure and how does the equilibrium shifts

A

equilibrium shifts in the direction that produces the smaller number of molecules of gas to decrease the pressure again

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10
Q

Effects of pressure: decrease in pressure and how does the equilibrium shifts

A

equilibrium shifts in the direction that produces the larger number of molecules of gas to increase the pressure again

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11
Q

Effects of catalysts: on equilibrium

A
  • A catalyst is a substance that increases the rate of a chemical reaction (they increase the rate of the forward and reverse reaction equally)
  • Catalysts only cause a reaction to reach its equilibrium faster
  • Catalysts therefore have no effect on the position of the equilibrium once this is reached
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12
Q

Effects of temperature: increase in temperature and effect on equilibrium

A

equilibrium moves in the endothermic direction to reverse the change

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13
Q

Effects of temperature: decrease in temperature and effect on equilibrium

A

equilibrium moves in the exothermic direction to reverse the change

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14
Q

Equilibrium expression

A

is an expression that links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account

  • Solids are ignored in equilibrium expressions
  • The Kc of a reaction is specific and only changes if the temperature of the reaction changes
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15
Q

Equilibrium expression: equation

A

K = ([C]c * [D]d)/([B]b * [A]a)

K = equilibrium constant
A, B,…= products
C, D,…= reactants
[A] = equilibrium concentration of A in moles

a=	number of moles of A
p=	number of molecules (or atoms) of R involved in the reaction
σ=	number of molecules (or atoms) of S involved in the reaction
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16
Q

Partial pressure of a gas

A

is the pressure that the gas would have if it was in the container all by itself
-The total pressure is the sum of the partial pressure

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17
Q

reactions involving mixtures of gases, the equilibrium constant Kp

A

is used as it is easier to measure the pressure than the concentration for gases

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18
Q

Equation to calculate the total pressure in a mixture of gases

A

P(total)= P(a) + P(b) + P(c)

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19
Q

Mole fraction=

A

no. moles of a particular gas / total no. of moles of all the gases in a mixture

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20
Q

To calculate the partial pressures of each gas the following relationship can be used:

A

Partial pressure = Mole fraction x total pressure

The sum of the mole fractions should add up to 1.00, while the sum of the partial pressures should add up to the total pressure.

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21
Q

Calculations involving Kc

A
  • In the equilibrium expression each figure within a square bracket represents the concentration in mol dm-3
  • The units of Kc therefore depend on the form of the equilibrium expression
  • Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
  • The concentrations of the reactants and products can then be calculated from the number of moles and total volume
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22
Q

concentration (mol dm^-3) =

A

no moles / volume (dm^-3)

  • Some questions give the initial and equilibrium concentrations of the reactants but products
  • An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation
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23
Q

Calculations involving Kp

A
  • In the equilibrium expression the p represent the partial pressure of the reactants and products in Pa
  • The units of Kp therefore depend on the form of the equilibrium expression
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24
Q

Some questions only give the number of moles of gases present and the total pressure

A

The number of moles of each gas should be used to first calculate the mole fractions
The mole fractions are then used to calculate the partial pressures
The values of the partial pressures are then substituted in the equilibrium expression

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25
Q

Changes that Affect the Equilibrium Constant: Changes in concentration

A

-If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in concentration of the reactants or product

-For example, the decomposition of hydrogen iodide:
2HI ⇌ H2 + I2
The equilibrium expression is:
kc = (H2)(I2)/(HI)^2 = 6.25 x 10^-3

-Adding more HI makes the ratio of [ products ] to [ reactants ] smaller
To restore equilibrium, [H2] and [I2] increases and [HI] decreases
Equilibrium is restored when the ratio is 6.25 x 10-3 again

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26
Q

Changes that Affect the Equilibrium Constant: Changes in pressure

A
  • A change in pressure only changes the position of the equilibrium (see Le Chatelier’s principle)
  • If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in pressure of the reactants and products
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27
Q

Changes that Affect the Equilibrium Constant: Changes in temperature

A
  • Changes in temperature change the equilibrium constant Kc
  • For an endothermic reaction such as:

2HI –>

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28
Q

Changes that Affect the Equilibrium Constant: Presence of a catalyst

A
  • If all other conditions stay the same, the equilibrium constant Kc is not affected by the presence of a catalyst
  • A catalyst speeds up both the forward and reverse reactions at the same rate so the ratio of [ products ] to [ reactants ] remains unchanged
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29
Q

Haber process

A

-The Haber process involves the synthesis of ammonia according to:

N­2(g) + 3H2(g) ⇌ 2NH3(g) ΔHr = -92 kJ mol-1

-Le Chatelier’s principle is used to get the best yield of ammonia

30
Q

Maximising the ammonia yield: pressure

A
  • An increase in pressure will result in the equilibrium shifting in the direction of the fewest molecules of gas formed to reduce the pressure
  • In this case, the equilibrium shifts towards the right so the yield of ammonia increases
  • An increase in pressure will cause the particles to be closer together and therefore increasing the number of successful collisions leading to an increased reaction rate
  • Very high pressures are expensive to produce therefore a compromise pressure of 200 atm is chosen
31
Q

Maximising the ammonia yield: temperature

A
  • To get the maximum yield of ammonia the position of equilibrium should be shifted as far as possible to the right as possible
  • Since the Haber process is an exothermic reaction, according to Le Chatelier’s principle the equilibrium will shift to the right if the temperature is lowered
  • A decrease in temperature will decrease the energy of the surroundings so the reaction will go in the direction in which energy is released to counteract this
  • Since the reaction is exothermic, the equilibrium shifts to the right
  • However, at a low temperature the gases won’t have enough kinetic energy to collide and react and therefore equilibrium would not be reached therefore compromise temperature of 400-450 oC is used in the Haber process
  • A heat exchanger warms the incoming gas mixture to give molecules more kinetic energy such that the gas molecules collide more frequently increasing the likelihood of a reaction
32
Q

Removing ammonia

A
  • by condensing it to a liquid causes the equilibrium position to shift to the right to replace the ammonia causing more ammonia to be formed from hydrogen and nitrogen
  • The removed ammonia is stored at very low temperatures and there is no catalyst present with the stored ammonia so the decomposition reaction of ammonia to decompose back into hydrogen and nitrogen will be too slow
33
Q

Maximising the ammonia yield: Catalysts

A

in the absence of a catalyst the reaction is so slow that hardly anything happens in a reasonable time!
Adding an iron catalyst speeds up the rate of reaction

34
Q

Maximising the sulfuric acid yield: Pressure

A
  • An increase in pressure will result in the equilibrium shifting in the direction of the fewest molecules of gas formed to reduce the pressure
  • In this case, the equilibrium shifts towards the right so the yield of sulfuric acid increases
  • In practice, the reaction is carried out at only 1 atm
  • This is because Kp for this reaction is already very high meaning that the position of the equilibrium is already far over to the right
  • Higher pressures than 1 atm will be unnecessary and expensive
35
Q

Maximising the sulfuric acid yield: Temperature

A
  • The same principle applies to increasing the temperature in the Contact process as in the Haber process
  • A compromise temperature of 450 oC is used
36
Q

Maximising the sulfuric acid yield: Catalysts

A

The Contact process uses vanadium(V) oxide as a catalyst to increase the rate of reaction

37
Q

Common Acids

A

An acid is a substance that neutralises a base forming a salt and water:
2HCl(aq) + CaO(s) ⇌ CaCl2(aq) + H2O(l)

  • Acids are also substances that release hydrogen ions when they dissolve in water
  • Acids dissociate in water to release a hydrogen ion
  • In organic acids (such as carboxylic acids) only some of the hydrogen atoms can form ions when the acid dissociates
38
Q

Common Alkalis

A

-A base is a compound that neutralises an acid forming a salt and water

2HCl(aq) + CaO(s) ⇌ CaCl2(aq) + H2O(l)

base salt

  • A base is a substance that accepts hydrogen ions or a compound that contains oxide or hydroxide ions
  • A base that is soluble in water is called an alkali
39
Q

Brønsted–Lowry Theory

A

defines acids and bases in terms of proton transfer between chemical compounds

  • A Brønsted-Lowry acid is a species that gives away a proton (H+)
  • A Brønsted-Lowry base is a species that accepts a proton (H+) using its lone pair of electrons
40
Q

The Brønsted-Lowry Theory is not limited to aqueous solutions only and can also be applied to reactions that occur in the

A

gas phase

41
Q

Strong acids

A
  • A strong acid is an acid that dissociates almost completely in aqueous solutions
  • –HCl (hydrochloric acid), HNO3 (nitric acid) and H2SO4 (sulfuric acid)

-The position of the equilibrium is so far over to the right that you can represent the reaction as an irreversible reaction

42
Q

The solution formed is highly acidic due to the

A

high concentration of the H+/H3O+ ions
Since the pH depends on the concentration of H+/H3O+ ions, the pH can be calculated if the concentration of the strong acid is known

43
Q

pH of a solution calculation

A

pH = -log(H+(aq)

-log(H+(aq) = concentration of H+/H3O+ ions

44
Q

Weak acids

A
  • A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions
  • —Eg. most organic acids (ethanoic acid), HCN (hydrocyanic acid), H2S (hydrogen sulfide) and H2CO3 (carbonic acid)
  • -The position of the equilibrium is more over to the left and an equilibrium is established
45
Q

The solution is less acidic due

A

to the lower concentration of H+/H3O+ ions

  • Finding the pH of a weak acid is a bit more complicated as now the concentration of H+ ions is not equal to the concentration of acid
  • To find the concentration of H+ ions, the acid dissociation constant (Ka) should be used
46
Q

Strong bases

A

strong base is a base that dissociates almost completely in aqueous solutions
-The position of the equilibrium is so far over to the right that you can represent the reaction as an irreversible reaction

-The solution formed is highly basic due to the high concentration of the OH– ions

47
Q

Weak bases

A
  • A weak base is a base that partially (or incompletely) dissociates in aqueous solutions
  • NH3 (ammonia), amines and some hydroxides of transition metals
  • The position of the equilibrium is more to the left and an equilibrium is established

-The solution is less basic due to the lower concentration of OH– ions

48
Q

Hydrogen ions in aqueous solutions can be written as either as

A

-H3O+ or as H+ however, if H3O+ is used, H2O should be included in the chemical equation:

HCl(g) → H+(aq) + Cl–(aq)

or

HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq)

49
Q

The pH Scale

A

is a numerical scale that shows how acidic or alkaline a solution is

  • The values on the pH scale go from 1-14 (extremely acidic substances have values of below 1)
  • All acids have pH values of below 7, all alkalis have pH values above 7
  • The lower the pH then the more acidic the solution is
  • The higher the pH then the more alkaline the solution is
50
Q

pH of water

A

-An equilibrium exists in water where few water molecules dissociate into proton and hydroxide ions
H2O(l) ⇌ H+(aq) + OH–(aq)

-The equilibrium constant for this reaction is:

Kc= (H+)(OH-)/(H2O)

Kc x (H2O)=(H+)(OH-)

51
Q

pH of acids

A
  • Acidic solutions (strong or weak) always have more H+ than OH– ions
  • Since the concentration of H+ is always greater than the concentration of OH– ions, [H+] is always greater than 10-7 mol dm-3
  • Using the pH formula, this means that the pH of acidic solutions is always below 7
  • The higher the [H+] of the acid, the lower the pH
52
Q

pH of bases

A
  • Basic solutions (strong or weak) always have more H+ than OH– ions
  • Since the concentration of OH– is always greater than the concentration of H+ ions, [H+] is always smaller than 10-7 mol dm-3
  • Using the pH formula, this means that the pH of basic solutions is always above 7
  • The higher the [OH–] of the base, the higher the pH
53
Q

Strong and weak acids can be distinguished from each other by their

A
  • pH value (using a pH meter or universal indicator)
  • Electrical conductivity
  • Reactivity
54
Q

The pH meter

A
  • is connected to the pH electrode which shows the pH value of the solution
  • The most accurate way to determine the pH is by reading it off a pH meter
55
Q

Strong and weak acids can be distinguished from each other by their: electrical conductivity

A
  • Since a stronger acid has a higher concentration of H+ it conducts electricity better
  • Stronger acids therefore have a greater electrical conductivity
  • The electrical conductivity can be determined by using a conductivity meter
56
Q

conductivity conductivity

A
  • Like the pH meter, the conductivity meter is connected to an electrode
  • The conductivity of the solution can be read off the meter
57
Q

Strong and weak acids can be distinguished from each other by their: Reactivity

A
  • Strong and weak acids of the same concentrations react differently with reactive metals
  • This is because the concentration of H+ is greater in strong acids compared to weak acids
  • The greater H+ concentration means that more H2 gas is produced
58
Q

pH value of a strong acid

A

pH of 0.1 mol dm^-3 solution:
HCL(strong) = 1
CH3COOH(weak) = 2.9

59
Q

A neutralisation reaction

A

is one in which an acid (pH <7) and a base/alkali (pH >7) react together to form water (pH = 7) and a salt

  • The proton of the acid reacts with the hydroxide of the base to form water
  • The spectator ions which are not involved in the formation of water, form the salt
60
Q

titration

A

is a technique used in neutralisation reactions between acids and alkalis to determine the concentration of the unknown solution

61
Q

Equivalence point

A

→ moles of alkali = moles of acid

-this is where neutralization takes place

62
Q

What are pH titration curves

A
  • Titration is a technique used in neutralisation reactions between acids and alkalis to determine the concentration of the unknown solution
  • It involves adding a titrant of known concentration from a burette into a conical flask containing the analyte of unknown concentration
  • An indicator is added which will change colour at the endpoint of the titration
  • The endpoint is the point at which equal number of moles of titrant and analyte react with each other
  • The equivalence point is halfway the vertical region of the curve
63
Q

Strong acid + strong alkali pH titration curve

A
  • Initially there are only H+ ions present in solution from the dissociation of the strong acid (HCl) (initial pH about 1-2)
  • As the volume of strong alkali (NaOH) added increases, the pH of the HCl solution slightly increases too as more and more H+ ions react with the OH– from the NaOH to form water
  • The change in pH is not that much until the volume added gets close to the equivalence point
  • The pH surges upwards very steeply
  • The equivalence point is the point at which all H+ ions have been neutralised (therefore pH is 7 at equivalence point)
  • Adding more NaOH will increase the pH as now there is an excess in OH– ions (final pH about 13-14)

-The equivalence point is still 7

64
Q

Strong acid + weak alkali pH titration curve

A
  • Initially, there are only H+ ions present in solution from the dissociation of the strong acid (HCl) (initial pH about 1-2)
  • As the volume of weak alkali (NH3) added increases, the pH of the analyte solution slightly increases too as more and more H+ ions react with the NH3
  • The change in pH is not that much until the volume added gets close to the equivalence point
  • The equivalence point is the point at which all H+ ions have been neutralised by the NH3 however the equivalence point is not neutral, but the solution is still acidic (pH about 5.5)
  • This is because all H+ have reacted with NH3 to form NH4+ which is a relatively strong acid, causing the solution to be acidic
  • As more of the NH3 is added, the pH increases to above 7 but below that of a strong alkali as NH3 is a weak alkali

The equivalence point is still about 5.5

65
Q

Weak acid + strong alkali pH titration curve

A
  • Initially there are only H+ ions present in solution from the dissociation of the weak acid (CH3COOH, ethanoic acid) (initial pH about 3-4)
  • As the volume of strong alkali (NaOH) added increases, the pH of the ethanoic acid solution slightly increases too as more and more H+ ions react with the OH– from the NaOH to form water
  • The change in pH is not that much until the volume added gets close to the equivalence point
  • The pH surges upwards very steeply
  • The equivalence point is the point at which all H+ ions have been neutralised by the OH– ions however the equivalence points is not neutral, but the solution is slightly basic (pH about 9)
  • This is because all H+ in CH3COOH have reacted with OH– however, CH3COO– is a relatively strong base, causing the solution to be basic
  • As more of the NaOH is added, the pH increases to about 13-14

-The equivalence point is still about 9

66
Q

Weak acid + weak alkali pH titration curve

A
  • Initially there are only H+ ions present in solution from the dissociation of the weak acid (CH3COOH, ethanoic acid) (initial pH about 3-4)
  • In these pH titration curves, there is no vertical region
  • There is a ‘point of inflexion’ at the equivalence point
  • The curve does not provide much other information
67
Q

Indicators

A

are substances that change colour when they are added to acidic or alkaline solutions

68
Q

Choosing indicators for titrations: Strong acid and strong alkali

A
  • The colour change for both indicators takes place at a pH range that falls within the vertical region of the curve
  • Therefore, both indicators can be used
69
Q

Choosing indicators for titrations: Strong acid and weak alkali

A

-Only methyl orange will change colour at a pH close to the equivalence point and within the vertical region of the curve

70
Q

Choosing indicators for titrations: Weak acid and strong alkali

A
  • Now, only phenolphthalein will change colour at a pH close to the equivalence point and within the vertical region of the curve
  • The pH range at which methyl orange changes colour falls below the curve
71
Q

Choosing indicators for titrations: Weak acid and weak alkali

A

Neither indicator is useful, and a different method should be considered

72
Q

Indicator & pH range (methyl Orange, Phenolphthalein)

A
  • Methyl Orange= 3.1-4.4

- Phenolphthalein=8.3-10

ENG CHEM (9 decks)

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