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reactivity Flashcards

1
Q

5 ways to measure rate of reaction

A
  • time taken for reactant to disappear
  • time taken for a colour change
  • loss of mass
  • rate of gas production
  • rate of precipitation
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2
Q

collision theory

A

substances must COLLIDE with the correct ORIENTATION with sufficient ENERGY TO BREAK THE BONDS

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3
Q

increasing reaction rate means;

A

increasing the number of successful collisions per seconds

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4
Q

3 ways to increase the reaction rate

A
  • increasing the frequency of collisions
  • increasing the number of collisions with sufficient energy to break the original bonds
  • increase the number of collisions with the correct orientation
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5
Q

increasing the frequency of collisions is:

A

temperature, surface area, concentration

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6
Q

increasing the number of collisions with sufficient energy to break the original bonds is:

A

catalyst, temperature

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7
Q

increase the number of collisions with the correct orientation is:

A

catalyst

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8
Q

concentration blurb:

A

increasing the concentration increases the number of particles available to react in a given volume.
so there will be more successful collisions per second.
resulting in an increased reaction rate

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9
Q

using the same amount of something means when comparing 2 different reaction rates:

A

both reactions will eventually produce the same volume of …. gas? as the same amount of … is used

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10
Q

surface area blurb

A

increasing surface area increases the number of particles available to collide.
so there will be more successful collisions per second.
resulting in an increased reaction rate

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11
Q

temperature blurb

A

increasing the temperature increases the kinetic energy of the particles.
this means the particles move faster, increasing the frequency of collisions.
in addition, the collisions are more likely to be successful because the kinetic energy of the particles has increased, so a greater proportion of particles have sufficient energy overcome the activation energy.
resulting in an increased reaction rate

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12
Q

catalysts

A

a catalyst provides an alternative alternative pathway for the reaction with a lower activation energy.
therefore, more reacting particles will collide with sufficient energy to overcome the activation energy.
so there will be more successful collisions per second.
resulting in an increase in the rate of reaction

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13
Q

energy diagrams: uncatalyzed reaction vs catalysed recation

A

refer to notes

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14
Q

equilibrium:

A

at equilibrium the forward reaction equals the rate of reverse reaction.
the concentrations of the reactants and the products are constant

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15
Q

equilibrium constant =

A

Kc

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16
Q

Kc calculation words

A

products over reactants

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17
Q

Kc calculation shown

A

refer to notes

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18
Q

what is not included in the Kc expression and why

A

solids, pure liquids and solvents as their concentrations do not change

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19
Q

Kc value:

A

the magnitude of Kc gives informations about the relative reactant and product concentrations at equilibrium

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20
Q

what does it mean if Kc is small

A

the reactant number is large.

there are more reactants than products at equilibrium

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21
Q

what does it mean if Kc is large

A

the product number is large.

there are more products than reactants at equilibrium

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22
Q

what is the value of Kc not affected by

A

a change in concentration of reactants or products
a change in pressure
the presence of a catalyst

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23
Q

what is the value of Kc affected by

A

A change in temperature

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24
Q

what does the reactant quotient Q mean?

A

The ratio of products to reactants at any point other than at equilibrium is represented by the Reaction Quotient, Q.

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25
how is Q calculated
substituting given concentrations into the expression. This number is then compared to the Kc value.
26
if Q < Kc
the product concentration is too low
27
if Q > Kc
the product concentration is too high
28
which will be favoured if Q > Kc
the reverse reaction will be favoured to reach equilibrium
29
which will be favoured if Q < Kc
the forward reaction will be favoured to reach equilibrium
30
reaction quotient Q
low. forward reaction. reverse reaction. high | Q ----------------------------> Kc
31
Le Chatelier’s Principle
If a change is made to a system at equilibrium, the system will adjust to minimise that change and re-establish equilibrium.
32
what is the equilibrium position affected by, relating to Le Chatelier’s Principle
A change in concentration (reactant or product). A change in pressure. A change in temperature. The equilibrium system will minimise these changes by favouring either the forward or reverse reaction.
33
concentration (equilibrium)
If a reactant or product is added, the system will oppose the change by favouring the reaction (forward or reverse) that uses up the added chemical, and re-establishes equilibrium.
34
if concentration of reactants increases (Le Chatelier’s Principle)
rate of reaction increases. | favours forward reaction
35
if concentration of products decreases (Le Chatelier’s Principle)
rate N/A | favours forward reaction
36
if pressure increases (Le Chatelier’s Principle)
rate of reaction increases | favours side with least gas mole
37
if pressure decreases (Le Chatelier’s Principle)
rate of reaction decreases | favours side with most gas mole
38
if temperature increases (Le Chatelier’s Principle)
rate of reaction increases | favours endo / ΔH +ve
39
if temperature decreases (Le Chatelier’s Principle)
rate of reaction decreases | favours exo / ΔH - ve
40
if there is a catalyst (Le Chatelier’s Principle)
rate of reaction increases | no change in yield
41
increasing pressure (equilibrium)
Increasing the pressure results in the system opposing the change by shifting in the direction of the lowest number of gas moles. This will decrease the pressure and re-establish Kc.
42
decreasing pressure (equilibrium)
Decreasing the pressure results in the system opposing the change by shifting in the direction of the greatest number of gas moles. This will increase the pressure and re-establish Kc.
43
decreasing temperature (equlibrium)
A decrease in temperature causes the system to oppose the change, by shifting in the exothermic direction to replace the lost heat energy.
44
increasing temperature (equilibrium)
An increase in temperature causes the system to oppose the change, by shifting in the endothermic direction to absorb the added heat energy.
45
change in temperature affect on Kc
A change in temperature changes the ratio of products and reactants, and therefore the value of Kc. If the temperature change shifts the equilibrium towards the products, the value of Kc is increased. If the temperature change shifts the equilibrium towards the reactants, the value of Kc is decreased.
46
addition of a catalyst (equilibrium)
A catalyst speeds up the rate of both the forward and reverse reaction equally. The ratios of products to reactants remains the same. (Kc is unchanged) Equilibrium is reached faster.
47
does an acid donate or accept protons
donate
48
does a base donate or accept protons
accept
49
conjugate base pairs
When an acid donates a proton, it becomes a conjugate base. | When a base accepts a proton, it becomes a conjugate acid.
50
ionisation of water
An ionisation reaction in pure water or in an aqueous solution, in which a water molecule loses a proton to become OH−. The proton is accepted by another water molecule to form an hydronium ion, H3O+
51
equilibrium expression for ionisation of water
Kw = (H30+) (OH-)
52
Whether a solution is ‘acidic’, ‘basic’ or ‘neutral’ depends on the relative concentrations of H3O+ and OH-.
Acidic: [H3O+] > [OH-] Basic: [H3O+] < [OH-] Neutral: [H3O+] = [OH-]
53
the pH scale
pH is used to express the acidity or basicity of an aqueous solution.
54
calculating pH
The pH of an acid or a base can be calculated from its concentration. (1) Write the equation for the acid or base. (2) Multiply the concentration by the number of moles of H3O+ or OH- in the reaction. This will give [H3O+] or [OH-]. (3) Calculate the pH from [H3O+] or [OH-].
55
calculating concentration
The concentration of an acid or a base can be calculated from its pH. (1) Calculate the [H3O+] from the pH. [H3O+] = 10-pH (2) For a base, convert the [H3O+] into [OH-]. (3) Write the equation for the acid or base. (4) Divide the [H3O+] or [OH-] by the number of moles of H3O+ or OH- to give the concentration.
56
strength of acids and bases
The strength of an acid or base depends on its ability to ionise/dissociate.
57
what happens to a strong acid
completely ionises
58
what happens to a weak acid
partially ionises
59
what happens to a strong base
complete dissociates
60
what happens to a weak base
partially ionises
61
strength and reactivity
A strong acid produces a higher [H3O+] than a weak acid of the same concentration. Therefore, there are more H3O+ in the same volume, meaning a greater number of effective collisions per second. Resulting in a higher reaction rate.
62
strength and conductivity
Conductivity depends on the concentration of ions in solution. A strong acid or base produces more ions in solution than a weak acid or base of the same concentration, and are therefore better conductors.
63
strength and pH
A strong acid will produce a higher [H3O+] than a weak acid of the same concentration, resulting in a lower pH. A strong base will produce a higher [OH-] than a weak base of the same concentration, resulting in a higher pH.
64
neutral solutions
If a salt dissolves in water and neither of the dissociated ions react with water, the resulting solution is neutral.
65
example of neutral solutions using NaCl
NaCl → Na+ + Cl- Neither the Na+ ions or the Cl- ions react with water, therefore no H3O+ or OH- are produced. ∴ [H3O+] = [OH-] ∴ pH = 7
66
hydrolosis
When a salt dissolves in water and one of the dissociated ions reacts with water, this is called hydrolysis. The resulting solution is either acidic or basic.
67
acidic solutions
If a salt dissolves in water and one of the dissociated ions reacts with water to produce H3O+, the resulting solution is acidic.
68
acidic solution example using NH4Cl
NH4Cl → NH4+ + Cl- NH4+ + H2O ⇌ NH3 + H3O+ ∴ [H3O+] > [OH-] ∴ pH < 7
69
basic solutions
If a salt dissolves in water and one of the dissociated ions reacts with water to produce OH-, the resulting solution is basic.
70
basic solutions example using NaCH3COO
NaCH3COO → Na+ + CH3COO- CH3COO- + H2O ⇌ CH3COOH + OH- ∴ [H3O+] < [OH-] ∴ pH > 7

chemistry (3 decks)

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