OCR A A Level Chemistry Module 5

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Last updated 9:19 AM on 2/5/26
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287 Terms

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Equivalence point

The volume of one solution that reacts excatly with the volume of the other solution

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End point of a titration

When [HIn] = [In-] (colour is a mixture of two colours in indicator equation)

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Nitric acid formula

HNO3

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Nitrous acid formula

HNO2

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Zero order

When the concentration of a reactant has no effect on the rate

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First order

When the rate depends on its concentration raised to the power of one (if x doubles, y doubles)

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Second order

When the rate depends on its concentration raised to the power of two (if x tripled, y increases by nine) x2

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Rate equation

Rate = k [A]^m [B]^n

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rate constant (k)

Number that mathematically converts between rate of reaction and concentrations

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Overall order of reaction

Overall effect of the concentrations of all reactants on the rate of reaction (sum of powers in equation)

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Units for overall order = 0

Mol dm-3 s-1

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Units for overall order = 1

S-1

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Units for overall order = 2

Dm3 mol-1 s-1

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Units for rate

Mol dm-3 s-1

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Initial rate of reaction

Rate when t = 0 (draw tangent to curve at t = 0)

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Two methods for continuous monitoring of reactions

Gas collection

Mass loss

Colour change

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concentration time graph gradient

rate of reaction

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zero order line on conc time graph

straight line, negative gradient

value of gradient = rate constant k

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first order line on a conc time graph

downward curve with decreasing gradient over time

time taken for conc of reactant to half is constant (confirm first order using half lives)

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second order line on a conc time graph

downward curve with a steeper gradient at start which tails off more slowly

hard to distinguish between first and second order so use value of half lives

if constant its first order if they're not constant its second order

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half life (t1/2)

time taken for half of a reactant to be used up

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exponential decay

when first order reactions have a constant half life with the concentration of reactant halving every half life

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[x]0

initial concentration of x when t = 0

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use can use a tangent on a conc-time graph to measure...

rate

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how to calculate rate constant k from using tangents on a conc-time graph

rearrange equation

substitute value for rate and the conc at which this rate was measured

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k rate constant units

s-1

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calculating k using half life equations

k = ln2/t1/2

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why are rate conc graphs so important

offer a route into the direct link between rate and conc in the rate equation

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zero order line on a rate conc graph

horizontal straight line with zero gradient

rate = k

intercept on y axis gives k

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first order line on a rate conc graph

straight line graph through origin

rate directly proportional to conc

rate constant = gradient of straight line

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second order line on a rate conc graph

upward curve with increasing gradient

rate constant cannot be obtained directly from graph

plot second graph of rate against concentration squared - straight line through orgin

gradient of this second graph is equal to rate constant k

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initial rate

instantaneous rate at the start of a reaction when time = 0

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clock reaction

more convenient way of obtaining initial rate of a reaction by taking single measurement

time (t) from start of experiment is measured for a visual change to be observed e.g. colour/precipitate

assumed that average rate of reaction over this time is equal to initial rate

initial rate proportional to 1/t

repeated several times at different concentrations and values of 1/t calculated

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iodine clock

formation of iodine

starch usually added as gives more distinct colour change from orange to black

sodium thiosulfate Na2S2O3 removes iodine as it forms then when used up iodine forms

graph of 1/t against conc plotted

thiosulfate ions react with iodine forming I- ions which are not coloured when all thiosulphate reactions I- will build up in solution cauing colour change as i2 is formed

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accuracy of clock reactions

in these reations you are measuring average rate during first part of the reaction

over this time you can assume that average rate of reaction is constant and thus same as initial rate

the shorter the period of time measured the less the rate changes and hence more accurate

reasonably accurate providing less than 15% of reaction has taken place

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reaction mechanism

series of steps that make up overall reaction

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rate determining step

slowest step in a multi step reaction

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rules for rate equation/rate determining step

rate equation only involves species involved in the rate determining step

orders in the rate equation match the number of species involved in the rate determining step

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effect of temperature on rate constant

as temp increases, rate increases and so k will increase

each 10 degrees rise in temp rate constant doubles

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two factors which contribute to higher rate constant at higher temps

Boltzmann distribution shifted to the right increasing proportion of particles that exceed Ea

particles move faster and collide more frequently (higher ke) this factor is comparatively small compared with right shift so change in rate mainly governed by Ea

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the Arrhenius equation

k = Ae-eA/RT

<p>k = Ae-eA/RT</p>
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exponential factor e-Ea/RT represents ...

the proportion of molecules that exceed Ea and have sufficient energy for a reaction to take place

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pre exponential term A

takes into account the frequency of collisions with the correct orientation gives rate if there was no activation energy

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Logarithmic form of the Arrhenius equation

lnk = -Ea/RT + lnA

<p>lnk = -Ea/RT + lnA</p>
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Arrhenius equation graph

plot ln k against 1/t

straight line graph

gradient = -Ea/R

intercept = ln A

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equation for Kc

conc of products/conc of reactants

raised to the power of their stoichiometric numbers

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value of kc and position of equilibrium

larger the value, the further the position of equilibrium to the products

smaller the value, the further the position of equilibrium towards the reactants

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units of kc

substitute units into expression for kc

cancel common units

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homogenous equilibria

equilibrium species that all have the same phase e.g. gas

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heterogeneous equilibria

equilibrium species that have different phases

any species that are solid or aqueous are omitted from the Kc expression

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calculating equilibrium concentrations

equilibrium amounts in mol / total volume

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Kp used in

equilibria involving gases

easier to measure pressure than conc of gases

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mole fraction

number of moles of gas x/total number of moles in gas mixture

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sum of mole fractions always =...

1

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partial pressure

mole fraction x total pressure

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sum of partial pressures always = ...

total pressure in question

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what values are used in Kp expression?

small p(x) is equilibrium partial pressure

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calculating Kp steps

calculate total number of gas moles

calculate mole fractions of each gas

calculate partial pressure of each gas

substitute into kp expression use units given in question

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Value of equilibrium constant K gives ...

exact position of equilibrium

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at a set temp... does conc, pressure or catalyst change value of K?

no

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what is the only condition which changes value of k

temperature change

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effect of temp on k explained

if forwards reaction is exothermic an increase in temp will decrease value of k as raising temp decreases eqm yield of products/pos of e shifts to the left

if the fwd reaction is endothermic the eqm constant k increases with increasing temp as raising temp increases eqm yield of products as partial pressure of products must increase to match kp

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explain the non change to kp/kc with changing conc/pressure

kp/kc stays constant but ratio doesn't match so must change to restore eqm

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effect of catalyst on p.o.e

catalysts affect/alter the rate of a chemical reaction but not the p.o.e

catalysts speed up both forward and backward reactions by same factor to reach eqm quicker

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bronsted lowry acid

proton donor

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bronsted lowry base

proton acceptor

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alkali

soluble base that releases OH- ions in aqueous solution

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Strong base

Alkali that completely dissociates in aq solution

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conjugate acid base pair

two species interconverted by transfer of a proton

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hydronium ion

H30+

conjugate acid of H2O

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Monobasic acid

an acid that has only one hydrogen ion to donate to a base in an acid-base reaction. Therefore, a monobasic molecule has only one replaceable hydrogen atom

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Dibasic acid

2x h+ ions can be replaced

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how to tell if an acid is mono etc

count number of hydrogen atoms in formula not including organic carbon chains

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writing equations for complete neutralisation

determine whether mono di or tri

write equation using as many base units needed to neutralise number of h

these can be used to calculate volume needed to neutralise a set volume of acid

if mono then same

if di then double

if tri then triple

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spectator ions

ions that do not change over course of reaction they are removed when constructing ionic equations

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redox reactions between acids and metals

produces salts and hydrogen gas

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neutralisation of acids with carbonates

forms salt water and carbon dioxide

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neutralisation of acids with metal oxides

forms salt and water only

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neutralisation of acids with alkalis

same for metal oxides forms salt and water only just in solution

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why is a negative logarithm used when measuring ph

makes scale more manageable

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pH =

-log[H+]

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[H+] =

10^-pH

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ph of one has ... the conc of h+ ions than a solution of ph 2

10x

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strong acid ph calculation

[H+] = [HA]

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Stong acids

Nitric (HNO3)

HCL

H2SO4

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Stong acid

completely dissociated in aqueous solution

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weak acid

partially dissociates in aqueous solution

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acid dissociation constant Ka

[H+][A-]/[HA]

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Factors affecting Ka

Temperature

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pKa =

-log(Ka)

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Ka =

10^-pKa

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low pka =

stronger acid

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Weak acid equilibrium

ha <-> h+ + a-

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[H+] depends on ..

concentration of acid

acid dissociation constant ka

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When HA dissociates, H+ and A- are formed in ...

Equal quantities

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Approximations for calc Ph of weak acids

1) H+ conc from dissociation of water very small compared to H+ from HA and so neglected

2) HA start much much greater than H+ at eqm due to partial dissociation so neglect any decrease in conc of HA so HA start = HA eqm

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Simplified Ka expression after approximations

Ka = [H+]2/[HA]start

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Finding H+ /ph of weak acid (expression)

[H+] = square root of Ka x [HA]

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Experimentally determining Ka for weak acid

Prepare standard solution of weak acid at a known concentration

Measure ph of standard solution using ph meter

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Breakdown of approximations involving weak acids

1) if pH is greater than 6 then dissociation of water will be significant compared with dissociation of weak acid so approximation breaks down for very weak acids or very dilute solutions

2) when [H+] becomes significant with stronger weak acids you cannot assume that [HA] is much much greater than [H+] as there is more dissociation (approx doesn't hold for acids with Ka over 10-2 or dilute solutions