Module 5.1 Flashcards
Meaning of rate of reaction
it is the change in concentration of reactant or a product per unit time .
what is the order
the order , with respect to a reactant is the power to which the concentration of the reactant is raised in the rate equation .
what is the rate constant k
is the constant that links the rate of equation with concentration of the reactants raised to power of their orders in the rae equation .
the rate equation
the rate equation for a reaction A+B–. C is given by ; rate =k[a[m[B]N , where m is the order of reaction with respect A and n is the order of reaction with respect to B .
what is the overall order
of a reaction is the sum of the individual orders m + n .
We learnt about collision theory in book 1 , how is this in relation to the rate equation
-reactions only proceed when successful collisions occur - that is particl emust collide with the correct orientaiton and with enough energy to overcoe the activation barrier 9i.e activation energy 0 . HOw frequently these successful collision soccur will determine the rate of a reaction .
what is the rate of reaction equaiton
rate of reaction = change in conc of reactant or product / time
How are rates measured
rates measured in moldm-3s-1 , (mol per dm3 per s 0 , but other units may sometimes , be more appropriate . If a reaction is very slow , a larger time scale (such as min 0 may be used . In this case , the units would e mol dm-3 min-1 . I
-If it is difficult to measure the concentration , you may use another measurement which allows you to monitor the amt of a product or reactant
. For example , if a gas is porduced in a reeaction , you could measure hte volume of gs porduced over a time period .
what is the order of reaction used for
-If more than one reactant is invovled in a reaction , each reactant can affect the rate of the reaction differently . The effect of the individul is described by stting an order with respect to each reactant . COnsider a reactant A , it is concentration affects the rate of a reaction . This can be expressed mathemticlly check page 10 .
learning tip about square brackets
square brackets are used to show a concentration
-the x denotes any power to which the concentration of a is raised .
fish sign means is proportional to .
meaning of zero order
zero order is if the order of 0 with respect to reactant A then rae is propotional to A to power of zero .
-THe rate is unaffected by changing the concentration of A .
-Note that any number to the power 0 is equal to 1 ..
meaning of first order
-If the order is 1 with respect to a reactant B then rate is proportional to [B]1 .
-If [B] , increased by 2 times , the rate also increases by 2 times .
-if [B] is increased by 3 times , the rate also increases by 3 times .
meaning of second order
If the order is 2 with respect to a reactant C , then rate is [c]2 . The change in rate will be equal to the change in concentration squared .
-If [C] increased by 2 times , the rate increased by 2squared = 4 times
-If [C] increases by 3 times , the rate increases by 32 squared , 9 times .
Rate reactions and overall orders
Chemist use rate equations to mathematically express the influence each reactant has on a reaction .
-Take the reaction A+B+C—. products .
-If the order for A , B and C were x y and z perspectively you could wirte the following expression for the rat .
rate is proportionla [A][B][C]
-The sign for proportional can be removed if a constant is addded into the equation . Chemists use the rate constant , k , he rate constant links the concentrations and orders of reactants to the rate . The expressiono then becomes a rate equation .
if any reactant is zero order , willit appear in the rate equation
no , that is because it does not affect the rate . For instance , in our example above , where the order are 0 , 1 , and 2 , with respect to A , B and C , the rate equation would be
rate=k[A]0[B]1[C]2
however , BECAUSE ANY NUMBER raised to the power of zero is one ,t he reactant can be remove d, powers of 2 can also e omited from outisde the brackets . but do not remove the reactant fromt he rate wquaiton .
rate ewution will become
RATE=K[B][C]2
what will the overall order of this reaction be
the overall order of a reaction is the sum of the individual orders .
-In this exampe abobe rate =k[B][C]2 and the overall order is +=
-the rate equation cn be determined only from experimental results . NOte that the orders are not the same as the numbers used to balance an equaiotn .
check calculating the value and units for rate constants
zero
first
second
third
do worked example 1 - page 12 - 13
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what is the half life
the half life of a reactant is the time taken for the concentration of the reactant to reduce by half
why do chemists use concentration time graphs
chemists can use concentration time graphs to fi dout about the rate and order of reactions . Theshapes of conentration time graphs depend on the order of the reaction .
finding rates from concentration - time graphs (1) - what can we measure for reactions involving cids or bases
pH changes by carrying out titrations .
pH changes using a pH meter
for reactions that produce gases , what can we measure
the change in volume or pressure .
-the loss in mass of reacants .
for reactions that produce visual changes what can we measure
the formation of a precipitiate
-colour change
how are visual changes monitoed
using a colorimeter , as the intensity of colour is directly related to the concentration of a coloured subtance .
example used - sulfur dichlordie dioxide decomposes t produce sulfur dioxide and chlorine
-was monitotired conc every 500 seconds , and hen plotted .
RATE OF REACTION found by taking a gradeint at the purple line , As the dat produces curve ,t angent
where does the tangent on page 1 shows lines at
t=0 and t=3000s
how is the rate calculated on page 14 (1)
after t=0 , (shown by the red , steeper tangent line ) .
Initial rate = change in concentration SO2CL2/ time for the change to take place .
0.500-0.00/3300-0
=1.5*104moldm-3s-1
how is the rate calculated on page 14 (2)
after t=3000s (shown by the green shallower tangent line )
rate = change in concentration of SO2CL2/ time for the change to take place
(0.38-0.14)/(4000-0)
=6.0*10-5moldm-3s-1
check the symbol for half life
what do the graphs look for zero order and first order reactions . in figure 2 and figue 3
how to workout the worked examples 1
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how do u use half lives to find the rate constant
because first order reactions have constant half-lives , the valule of the half-life , can be used to determine the value of the rate constant .
-The half-life and the rate constant are relaed bby the following equation .
K=In2/tand a half this equation ONL APPLIES TO FIRST ODER REACTIONS . other order reactions have no half lives .
check worked example 2 p.g 15
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initial rates (1)
The rate at the very start of a reaction given the shorthand , t=0 , (which stands fo r time =0) , is known as the initial rate . Ifa tangent line is drawn at t=0 , on a concentration-itme graph, the gradient of this line is equal to the initial rate .
how to determine the order with respect , to each reactant , you carry out he reaction several time .
-to determine the order with respect of each reactant , to carry out the reaction sseveral times . Each time , you vary the concentration of one of the reactants . You can obtain the initial rte for eaach of these different concentrations and plot , the result , on a graph to produce a rate-concentration graph for each reactant ..
clock reactions (1)
Some reactionsNwill produce visible changes such as the formation of a ppt , or a dramatic colour change . Measuring the time taken for such changes to occur canbe used o find out how concenration affects the initial rate .
-Only the initial rate is consididered as the visile changes that occur are assumed to last occur as the reaction first happens , the rate is GENERALLY FASTEST .
-The time taken , for the visible event to occur is inversely proportional to the initial rate ;
the shorter the time taken for the change the faster the reaction must be . this cn be expressed mathematically proprotional 1/t
this can be expresed mathemastically as rate proportional 1/t
this means that when a graph is plotted 1/t is taken as a good approximation of the initial rate .
example of this rection between sodium thisosulfate , hcl
a cross is drawn on a piece of paper and placed under a beaker . The reactants are then added together in the beaker . The products , form a cloudy o solution as a ppt is ormed , so a stopwatch is used to time how long takes for the cross under the beaker to be obscured y solutoin.
prt 2 - odiumthiosulfate
the experiment , is then carried out using various concentrations of both sodium thiosulfate and hydrochloric acid to determine how thier concnerations affect the rate . graph 1/t , against concentation is plotted .
check f-1 , nd the iodine clock reaction(pagee 16)
What happens rate-concentration graphs - zero order
If the order is 0 , with respect to a given reactant , A , the rate-concentration graph will appear as in .
THE figure 2 , on 17 shows
-This graph , shows the rate proportional to A .
-Changes in the concentration of this rectant have no effect on the rate .
what happens in a first order reaction
-if the first order is 1 with respect to given reactant , B , the rate concentration graph will apper as in .
-This graph shows the rate proportional B .
-If the concentration of reactaaant B , is doubled , the rate will double .
-If the concentration of reactant B , is tripled the rate will triple .
-If the concenration of reactant B , is increased 100 fold the rate will increase 100 fold .
Determining , the rate constant from first order rate-concentration graphs
If a rate-concnetration , is first order , with respect , to a given reactant , it is possible to use the graph to determine the ate cosntant , k .
For the first order reactions ;
-Rate proportional [reactant]
-this means the rate=k[reactant]
If we arrange this to find k ;
k=rate/concentration
-Rate is ploted on he yaxi s, ont he rate-concentration graph ,
and conentration is plotted on the xacis so calculating , the rate is the samr as calcualting the gradient , as gradient =change in y divided by change in x .
-tje imot pf l , will depend on the raye equation for hthe givenr eaction.
what is a second order reaction
-if the order is 2 , with respect to a to a given reactant C , the rate cocnentration graph will apper as figure 4 .
–This graph shows the rate proportional [C]2
-if the concentraiton of reactant C is doubled the raye will incfrase by 2 squard .
-if the concenration b is tripled the rate will incrase by 3 squared .
if the concnetratio of rectant c is increased 100 fold , the rate will increase by 100 squared .
a reaction mechanism is what
a series of steps that together make up the overall reaction .
what is the rate determinising step
it is the slowest step is the slowest step in the reaction mechnism of a multistep reaction .
what is an inermediate
it is a species formed in one step of a multi-step reaction that is used , up in a subsequent step annd is not seen as either a reactant or a porduct or the overall eequation .
why is the rate determinig step useful for
reactions can occur in one step or in many steps Experimental resutls can be used to predict , how many steps will occur during reaction . The series of steps that occur during a reaction is called a reaction mechnaisms .
what will the slowest step in the reaction dictating
how quicklyt he reaction wil proceed . This slowest step is called the rate -d etermining step .
-when you measure the rate of any reaction , that has a multi-step reaction mechanism , you are effectively measuring the rate of this determinig step .
predicitng reaction mechanims from rate equaiton
-As you have learnt in previous topics , some reactants hvae no influence on the rate of reaction .
-Thee reactant have an order of zzero and do not appear in the rate equaiton .
-I fthey have no influence on the rate , they cannot be involved in the rate-determinig step .
check page 19 for hdyrolysis of the bromoethane
-What does this rate equation show .
-THIS SHOWS that only the concentraiton of CH3BR influences the rate . Therefore , only this moleucle is involed int he rate-determingi step .
step 2 - bromoethane
this suggests that the rate dereminig step must involve the c-BR bond breaking , breaking the OH- can take its place .
S1-CH3BR–CH3++BR-
STEP2;CH3+ +OH—-CH3OH
step 3
in this example , CH3+ must be involved in the second step , in the order to be used up as it does not apper in the overall equation .
-The sepecies invovled in the two steps , cancel one aonther to leav the overall equation .
Orders and the rate determini step -what doe the order show for the bromoethane hdyrolysis
-check the rate equaiton ofr htis
[CH3BR] is raised to the power of 1 , ( rememebr ]a 1 is not hown in the rate equaiton )
-This means tht the rate-determinig step invovles just one moelcule CH3br
consider the hydrolysisi of bromoethane - what if it was order is 2
-If the order is 2 with respect go 2e.g rate =k[x]2 , there will be wo moleucles of this reactant invovled in the rate-deerminign step . Doubling the concentration of each of the two
moelucle will double the rate so verall the rate is quadurpled .
what should be in place to be able to work out the reaction mechanism
a rate - determining step is given that involves the number of moleucles shown by the order for that reactant .
what should be in place to be able to work out the reaction mechanism (2)
subsequen steps are hown that eventually geenrate the products shown in the balanced equation
what should be in place to be able to work out the reaction mechanism (3)
any intermediate generated is not present int he overall equation (i.e it is used up within subsequent steps ).
learning tip on page 20
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worked example page 20
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what is the arnhenius plot
is a graph of ln k=lnA -Ea/R
*1/T where ln k is plotted against 1/T
rate constant k
reacttion rates depend on both the rate constant and the concentrations of the reactants presents in the rate equatioN
check the rate ewution
-The larger the value of K , the faster the reaction .
the effect of temperature on the rate constant k (1)
an increase in temperature , gives more energy to the molecules . This means that collisions are more frequent , and more of the collisions exceed the activation energy of the reaction . EXPLAINED IN BOLTZMANN DISTRIBUTION BOOK 3.2.
the effect of temperature on the rate constant k (2)
the key factor affecting the reaction rate is the number of collision that exceed the activtion energy . This mean that rate increases with temperature by much more thn can be expallined soley from any increased frequency of collisons .
what happens in this reaction , if the rate increases with increasing temperature , when the concentrations are the same , then the rate constant must increase with temperature .
-raising the temperature speeds up the rate of most reactions by incresin the rate constant k .
-for many reactions , the rate doubles increse in 10 degrees in tmp . This reflect the getaer number of recting particles that exceed the acctivation energy .
proportionality of doubling the rate , what does this mean
-typically , doubling the rate will double the value of the rate constant k
-check figure 1
what is the ARRHENIUS equation showing
write equation
-what does the equation mean
-arrhenius equation is used to describe , mathematiclly , the exponential relationship between the rate constant and the temperature .
k-rate constant
Ea -activation energy
T - temperature (in kelving )
e=mahematicl cosntant value with value 2.71828 -log
A - preexpoentnial factor
R-gas constant .
what can the arrhenius equation tell us (1)
to be able to react , moleucles have to collide with enough energy , to overcome the activation energy . It has been found that at any given emperature , T, rate constant , k is proprotional to e-EA/RT , this is expressues amthetmitcallyas .
what can the arrhenuius euation tell us (2)
this can be turned into an equation by adding a constant , the ‘pre-expoenntial facotr’ , A giving k==Ae -Ea/R . This is the Arrhenuius ewqution
What does the Arrhenius equation tell us (3)
-Temperature , T, and the rate constant , k , are related exponentially .
-As temperture increases , the rate cosntant increases .
what does adding a catalyst do to the rate constant , k .
-adding a catalst provides an alternative reaction path , with a lower activiation enery . If ou follow this through maethematiclaly , a lower value of Ea increases the rate constant K .
Taking logarithims of thr ARRHENIUS EUATION
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check the end of page 22 .
figure 2 , and worked example
im being lazyyy
what does the equillibrium law state
state that for the equilibirum look a tpage 23
what dos a homogeneous equlibirum show
it is an equlibirum in which ALL THE SPECIES , making up the reactants and products , are in the same physical state .
what does a heterogenous equlibrium show
it is an equlibirum in which species making up the reactants and products are in different physical states .
dynamic equlibrium , what is it and Kc
-not all reactions go to completion . An equiliibirum ma be established , this is when the raes of the forward and reverse reactions are equal .
in a dynamic equlibirum 0 does equilibrium occur straight away
no , the reverse reaction cannot take place until the forward reaction has produced a high enough concentration of the product .
example of decompostion of n204 a dynamic equlibrium ,w hat happens (1)
-at first the cocnentration of N204 is high and the concentration of N02 , is zero .
example of decompostion of n204 a dynamic equlibrium ,w hat happens (2)
as the reaction proceeds , the concentration of N204 , falls and the cocnenration of N02 , rises . The reaction mixture , appers , colourless intiially then a pale brownn colour will appear .
example of decompostion of n204 a dynamic equlibrium ,w hat happens (3)
once equliibrium is reached he concentration of each subsance remians unchanged - the reaction has a brown appearnce .
example of decompostion of n204 a dynamic equlibrium ,w hat happens (4)
if conditions change , the equilibrium will need to be re-establsihed . A colour change will be observed until the equilibrium is resored .
According to to the equlibrium law - for the reaction .
-equilibirum consant Kc , can be calcualed as follwos .
-once an equilibrium is established , kc remians constant unless the emp of reaction changes .
units of Kc dh3dk page 23
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how to dertermine concentrations at equlilrium - what are the two common ways
-in order to determine a value for Kc , cocnentration of reactans and prodcuts at equlibirum need ot be known .
titrtion using a colorimeter
how to determine the concentrations at equilibirum - TITRATION
titrating one of the reactants or products agaisnt a subance , with a known concetration will enable us to find out how much of the reactant / product is present .
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how to determine the concentrations at equilibirum - TITRATION DRAWBACKS
-Reactant or product has to be removed from the reaction mxiture , and this will alter the position of the equilibirum .
-Also other subtnces may be present that can affect he results of titrations
how to determine the concentrations at equilibirum - TITRATION DRAWBACKS (2)
e,g - if an alakli is used to titrate agaisnt the equlibirum subances in question and an acid catalsyt is also used , both equilibirum subtance and atalyst will react with the alakli .
-The amount o any such sucbances must be carefullyd etermined when the titration results are used to determine equiilbrium cocnetations .
equilibirum investiagion and worked example 1
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calcualting unknown equlibirum concentraitons - for the balanced equation , what are the two things you can find (stochiometric)
-the reactaitng quntities neeedto prepare a requried quanitty of a product .
-the quantities of product formed by reching togeher known qunatities of reactnt .
what does kc help us to determine
helps us to deterime the equilibrium concentrations of the components in a equilibirum mixtrue .
worked example 2
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learning tip on page 25
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if a heterogenous equilibrium is present then what happens to the molar concentration and volumes
-molar concenraions for solids and pure liquids do not change BECAUSE THEIR VOLUME REMAINS CONSNA .
how does expression of rkc change with heteroegenous equilibrium
-the concentraions of solid substances are not included in the expression .
-the concentrations of pure liquids are not included in hte xpression .
e.g – when phosphorus reacts with oxygen , following hetereogeneous equillibrium is esbalished check p.g 25
example of a heterogenous catalyst - when phosphorous reacts with oxygen , the following equilibrium is established .
P4(s)+502P4O1-
What is a mole fraction ?
The mole fraction of a substance is a measure of how much of a given substance , is present in a reaction mixture .
What is the partial pressure ?
The partial pressure of a substance is the pressure an individual gaseous substance would exert if it occupied a whole reaction vessel on its own.
How is the mole fraction of a substance calculated ?
mole fraction Xa = number of moles of substance A / total number of moles of all substances .
How can we calculate the partial pressure if we know the total pressure ?
For substance A ; partial pressure , PA = mole fraction * total pressure
The concentration of a substance is proportional to its partial pressure .
round brackets are used for …
partial pressures
square brackets are used for …
concentrations
check end of page 26
equilibrium expressions can be written using partial pressures instead of concentrations .
Check how to write kp .
go through the worked example on page 27
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what is the expression for kp when a reaction is heterogenous it will alter as dollows
-solids will not be included in the expression .
-pure liquids will not be included in the expression .
what does the magnitude of the equilibrium constants kc and kp indicate ?
they indicate the extent of a chemical reaction .
what would an equilibrium constant with a value of 1 indicate
that the position of equilibrium is halfway between reactants
when k is greater than 1 what does it suggest in an equilibrium constant
the reaction is product-favoured .
-the products on the right hand side predominate at equilibrium .
when k is less than 1 what does it suggest in an equilibrium constant
the reaction is reactant favoured .
the reactants on the left hand side predominate at equilibrium .
how does an increase in temperature shift the position of k .
an increase in temperature shifts the position of equilibrium in the endothermic directions .
how does a decrease in temperature shift the position k ?
a decrease in temperature shifts the position of the equilibrium in the exothermic direction .
what does the ^_ H values get affected in an equilibrium
they have the same magnitude but have opposite signs .
what is the equilibrium constant , controlled by ?
rate constant , which changes its vale only with changes in temperature . When any other changes occur to conditions k will remain constant .
what will happen to k , if the forward reaction is endothermic .
the equilibrium , yield of the products on the RHS , increases .
The equilibrium yield , of the reactants on the LHS , decreases .
what will happen to k , if the forward reaction is exothermic
-the equilibrium , yield of the products on the RHS , decreases .
-equilibrium yield of the reactants on the LHS , increases .
k increases when
k decreases when
k increases as the temperature rises if the forward reaction is endothermic .
-k decreases as the temperature rises if the forward reaction is exothermic /
changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (1)
-work out the kp expression which now gives the ration x .
-the system is no longer in equilibrium .
changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (2)
the equilibrium position must shift to restore th ratio to the Kc value of x . The system must ;
-Increase x - on the top expression for Kc .
-Decrease x on the bottom of the expression of kc .
changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (3)
-This will cause a shift in the equilibrium , position from the left to right , the same will be true , for the rate constants Kp , as t his will also remain constant when conc or the equivalent partial pressures are changed .
what will changes in pressure do to kc - if the pressure is doubled , what will happen to the concentrations of both gases
it will also effectively double , .
learning tip for pressure and volume - being inversely proportional
if the pressure is increasing , and there are no changes in temperature , or concentration therefore then the volume must be decreasing .
-concentration and volume can be linked by n=cv .
-If the number of moles stay the same , and volume decreases , concentration must increase .
-The molar volume of any gas is proportional to its partial pressure , so increases concentration = increases partial pressure .
what if the value of kc is now 25.6
the system is no longer in equilibrium so the equilibrium positions must shift , to restore the ratio to the kc so the value of 12.8 , the system must ;
-Decrease of NO2 on the top
Increase in N204 on the bottom .
-Causing shift in the equilibrium position form right to the left , this will also be true for the rate constant Kp.
how does the presence of catalsyt affect k
it dosent for kc and kp ,
-it affects the chemical reaction but not the position of equilibrium .
-Catalyst speeds up both the forward and reverse reactions in the equilibrium by the same factor . Equilibrium , is reached more quickly , but the equilibrium position and hence the value of the equilibrium constant is unchanged , by the action of a catalyst.
What is a Bronsted lowry acid
it is a proton H+ donor .
what is a Bronsted lowry base
it is a portion H+ acceptor
what does the Bronsted lowry base model show
-the reactions between acids and bases states that ;
-reactions between acids and bases involves the transfer of H+ ions (known as protons ) .
-What is a Bronsted lowry acid - is any substance that can donate a proton .
what is a Bronsted lowry base - it is any substance that can accept a proton .
Models for cids and bases over time (1)1770S
-In the 170s many chemists were investigating air and gases within it . Joseph Pruestky and Jarl Scheele both reported findings that suggested the existence of the gas we know now as oxygen .
-Another chmist , Antoine - Laurnet , de Lavoiser , found among other things , tht this gas was important int he rusting of metl and in 1778 , he proposed that it was the source of acidity . Lavoiser was wrong , but it was an early step towards , the understadnig of aids .
in 1815 What did Humphtry Davy show
that some acidicacuduc substances such as HCL did not actuall contain oxygen .
what did justus liebig do in 1832
he defined an acid as a substance containing hydrogen , that could be repalced by a metal - close to todas definiton .
What happened int he alte 1880s ,
Swaante Arhenius proposed that acids dissociated in water to form the hydrogen ios H+ . and that bases dissocited in water to form hydroxide ions , OH- this is true , fo rmany acids and bases and they can be referred as Arhenius acicdsa nd bases , However , the model breaks down when the acids and bases are not solution sin water or when bases are not soluble hydroxides .
what happened in 1929
the Bronsted Llowry definiton discussed , above was proposed by the Danish chemist , Johannes Bronsred , and Brithish chemist thomas lowry .
exampleof bronsted lowry of acid in acrion ,
bronsted lowry acid hcl
for hydrogen chloridw HCL
HCL — H+ CL-
hcl has donated a H+ ion it i a bronsted lowry svif .
exampleof bronsted lowry of acid in acrion ,
bronsted lowry acid h2s04
sor sulfuric acid
H2s04– H+—HS04-
-H2s04 has donated an H+ ion , it i a bronsted lowr aicd , h2s04+ can actually go on to donte another H+ ion , (sulfuric acid is dibasic ) this will e covered alter .
bases accept protons - give an example o a bronsted lowr base
For ammonia , NH3- , NH3+H+— NH4+
-The ammonia has accepted (h+) , IT IS A BRONSTED WORY BASE .
Why does a bronsed acid base invovle ionic equations
because it involve the transger of H+ ions , as they re usuall epresnted by ionic equations .
what do ionic equations show (1)
-the H+ ions being donted
-tje ions tht is acepting the H+ ion
what do ionic euqations show (2)
-the product is formed
-all changes correctly balanec (they cancel out )
-state sumbolds for all substances inclduing ions .
what is the ionic equation for the rection between hydrochloric aicd and sodium hydroxide
H+ +OH- – H2-
what are monobasic aicds
HCL IS A MONOBASIC AICD BECAUE EACH MOLEUCLE CAN RELEASE ONE PROTON .
hcl– h+-CL-
what are dibasic acids
H2S04 , is a dibasic acid becsue each moelcule can release two protons . This is done in two stages
-H2sO4 – H+ (aq) +Hs04
-HSO3 – H+ +SO42-
What are tribasic acids
Tribasic acids are where each moleucle can release three protons this is done in three syages
H3PO4 – H+ + H2PO4
-H2PO4– H+ +HPO42-
-HPO42- — H+ + PO42-
What are conjugate acid bases
an acid base pair is a set of two species that transform into each other by gain or loss of a proton .
-F1 - gives an example of conjugate aicd-base pair , for the dissociation of a nitrous acid HN02 . It also shows how the acid and bases in acid - base pair are linked by H+
how many acid - base pairs do ewuilibria inovvle ,
-give an example reaction
acid - base equilibria invovle two acid - base apirs . The equilibbrium below shows the dissociationf of ntiorus aicd HN02 in water .
HN02 +H20 -_ , H20+ + NO2-
What does the frowad reaction show for acid base equilibria
-The acid HN02 releases a proton to form its conjugate bases H2- .
-tHE BASES nO2- accepts the proton to form its conjugate acid HN02 -
What does the frowad reaction show for acid base equilibria (2)
-The acid H30 + reeleases a proton to form its conjugate ase H20 .
HN02 and NO2- accepts the proton , to form its conjugate acid HNO2- .
pair one
HN02 AND NO2- differ by H+ and make up one aicd - base pair
pair two
H3-P+ AND H20 differ by H+ , and make up a second acid - base pair .
What is an alkali
it is a vase that dissolves in water forming OH ions .
What is neutrakisation
It is a cgenucak reacuib ub which an acud abd a vase react together to produce a salt and water .
What is a strong acid
is an acid that completely dissociates in solution .
what is a weak acid
is an acid that partiall dissociation in solution .
What is the acid dissociation constant Kp
check sheet it is of an acid Ha is defined …
what -pKa
-log 10 Kp
What is Kp
10-pkA
typical acid base reactions
-aqueous takes part in typical acid base reactions with carbonates bases and alkalis .
-In all of these reactions a neutrlisaiton reaction occurs and water is formed as one of the products .
Learning tip one - figure 1 - when acids release protons in water
-when hcl is added to water for example whaat reaction occurs
HCL+H2O– h20 +CL-
-H+ ABD H20+ can be used interchangeaky when describing aqueous acid solutions and H+ is usually used in ionic equations fro acid -nase reactuibs wgucg us wgt tge a vive reaction can be simpliied
HCL– H+CL-
Reactions with carbonates
-In the reactions of acids , that follow ionic equations are used to show the important role of the H+ ions .
-The ionic equation show only H+ . from the acid . This H+ , could be provided by hdyrochloric acid HCL , ntrici aicd HNO3 , or sulfuric acid H2S04 .
-ThESE ACIDS have the same ionic equaiton for each type of reaction becasue they ll release H+ ions , whent hey disovl in wter .
Reactions with carbonate part two - auqesous acids can react with solid carbonaed to form - a salt , carbon dioxide and water .
2HCL – aCO –> Cacl2 + Co2 ,, +H20 .
all ions ;
-2H+2CL- check the PAGEEE UHH H FOR CANCELATIONS
THen the ionic euation for that
2H+– CACI2 –CA3+ +CO2+H20
what should be the sum of charges on either side of an equation
-the sum of the charges on either side of an ionic equaiton need notbe zero , but must be the same on the two sides . In this xample , the two H+ ions are balanced by one Ca+ ion - there is a charge of ++ one ach side of the ewuation .
If the carbnate is in solution , what does the final ionic euaion simplife too when the carbonate is dissociated
check to see the full euation an then wehn all the ions cancel out , the ionic equation .
should there be state symbols in ionic ewuations
yes , this is especially important when an acid reacts with a solid as solids will not be dissociated into ions .
-Notice , in the ions , euaation the species which remian unchanegd are being called out , there are spectator ions .
aqueous cids reacting with bases - forming salt and waer
-dheck the full equation
-and then the ionic equaiton
-full equaiton .
-ionid equation .
aqueous acids recting with alkali forming a salt and water
-Ful Equation .
-Ionic equationnn check page
the reaction of a metal with an acid is what kinda rection
-it is redox resction , so does not fit in with acid-base model . Howebeer , theya re a coommon , acid reaction .
-The generl equation ;
acid+metal– salt+hydrogen
aqueous acids recting with metlas to form show in the equation
ful equatin and ionic equation , but hwat is the probelm
although most acids react with metals in this way , you have o be carefuls .
-Some acids such as sulfuric acids and nitric acids are powerful , oxidising agents , and othe reactions ma also take place especiall when the acids are cconcentrated .
-CLEAR AWAY FROM H2S04 AND HNO4 when giving examples of cid reactions iwth emtals .
strong acids n weak acids
-check page 34
-in aqueous solution , acid dissociate and an equilibrium , is set up . The equilibrim shows the dissociation of an acid HA in water .
-the strength of an acid HA is the extent of its disscoaition into H+ and A- ions .
what is the difference between strength and concentration - LEARNING TIP
-Make sure youa re clar about the differences between strength and concentration .
-STRENGTH is the extent , an acid dissociate into H+ and A- .
-Concentration , of an acid is how many moles of the acid are present in a given vlume .
-It is perfectly possible to hae a dilute soltuion of strong acid and a concentrated soltuion of a weak acid .
SA1
-HCL - hdrochloric acid
SA 2
HN03 nitric acid
SA 4
H2S04 - SULFURIC AICDS
HI HYDROIC ACID
HCLO4 - CHLROIC (VII) PECHLORIC ACID
how do weak acids dissociate
a weaj acud, only partially dissociates in an aqueous solutions , man naturally occueing cids are weak .
what is the equilibrium et up when ehtnaoic aicd reacts with water
CH3COOH ____ H+ CH3COO-
-The equilibrium position lies well over to the elft .
-There are onlys mll concentrations of dissocaited ions , H+ and CH3COO- compared witht he cocnentration of undissociated CHCOOH .
-We can actually say that CHCOO- is a very good base , ti i er good at accepting the dissociated H+ back .
what is the acid dissociation contant Kka
-the actul extent of aicd dissociatio is measured by an equilibrium constant called the acid dissociation constnat Ka .
- weak acid Ha , has the following equilibriumin aquesous soltuoin ,
chck pahe for more .
what dies a large ka value indiciate
a large extent of acid dissociationthe acid is stong .
what does a small ka value indiicate
a small extent of dissociation , the acid is wek .
pka is ore manageable than ka , a ka is vsats so what are the two values of ka
pka=-logK
-kA =-pKa
-whatd oes a low value of Ka mean
high vaue for pKa
what does a high value of Ka match with
a low value fo rpka
the smaller the pKa value …
the stronger the acid .
-check table 2 .
how to work out pH with hdrogen ions
-log (H+)
how o work out ph with hdyroge ions
-Pj
skim page 36 and worked example and 2 .
…
How to calculate the Ph of strong acids
-A strong monobasic aicd HA , has virtually complete dissociaiton inw ater .
Meaning hydrogen of h+ of a strong acid is approimatelye wwul to the cocnentration of the acid HA
-hdyrogen ion = HA .
U CALCUATE PH JUST BY -LOG check worked example 4
calculating the Ph of weak acids is slightly mroe difficult why
as a weak monobasic aicd only partialll dissociaites thereofre , WHEN SETTING UP THE EQUILIBRIUM .
-We can no longer assume hdyrogen+ is ewqual tot hte concentration of the aicd , we needto work out the vle of K for the reaction - to find the extent of disscoiationt hat has occured to dictate the value of H+ .
Ka= (h+)(A-)/HA
things to know about calculaitng weak acids (1)
for weak acids (H+) is much less than HA because the extent of dissociation is so SMALL .
-When HA moleucles dissociate , H+ and A- ions are formed in equal quantities and it is iassumed that any water present will l hae dissocited to such a negligible maount htat it will not affect the cocnentration of H+
thigns to note about calcualting Ph of weak acid (2)
thereofre h+ and a - can e consdiered equal . -This means the expression H+ and A- in the Ka , cslculation becomes the H+)2/
as the ha moelcules have dissocited (HA) WILL HAEVE REDUCES SLIGHT SO WHAT DOES THE EQUILIBRIUM COCNENTRATION BECOME
CHECK PAHGE 37 FO RHIGY
and hen worked exampel 4.
check
the leanign tip on pGW
38
What is the limiition of the scientifc approximation
-when calclaitng the Ph ofor a weak acid , it is assumed that so ie , of the originl cids , has dissocitedtha tthe cocnenration of the acid a the equilibrium is ffectiely the ame as the cocnentration of the oiginal amount o aciid .
-THIS CAN BE EXPRESSED AS HA EQUILBRIUM = HA undissociated
What is the limiition of the scientifc approximation (2)
it is actually the case that there will be some acid , whidh does dissociate even if only for a smll amount . If this is the amoutn less than 5% of the total for Hha IS UNDISSOCIATED it , then it deemd for safe ,t ouse the abov approximation this iwll be the case for very weka acids iwll very low Ka vlaues .
what isntuemnts do you have to use to calcuualt ek for a weaka cid
use a ph meter - also need the cocnentraiton o the weak acid - see worked example 5 .
What is the ionic product of water
Kw is defined as Kw = H+OH- at 25 degrees = 1.00*10-14
is water an acid or a base
water cn act as a n acid by donating a proton
water can act as a base bya ccepting a proton .
-This back and fourth transfer of porotns happens conitnously in water . In fact ,w ater exists t euilibrium ,,, known as ionisation of water .
-EUWILIBRIUM LIES well to the elft , and onlya n extremel tinya mt of wate ris dissociated at aigven time .
chekc forkc expression .
as the amount of dissociaiton , ionisaiton of water is so small what does thi mean
it is consdiered a contant like kc , we rearrange the equation as follows to combine hte two cosntants
Kc*(H20) = (H+)(OH-)
-This can be furhter simplified by reerring to the expressio kc *H20 as a new cosntant KW .
Check sheet fo rhte implfiyin gof kw , what is kw
the ionic product of eate r.
-the units of kw are always that ebcuae
-25d egress the measured ph of water is 7 and hydrogen =10-7mol .
-when water ionises the samennmber of h and oh ions are relaed
meanong of that the cocnentraio of OH- ions are also equalt o 10-7 mol
check sheet for equation .
the values o h+ and oh - are only eqal if the water is pure and neutral .
how can the balance beteee nH and OH be afected
byt he addition of extr H+ or Oh- ions that cause the equilibrium to shift . Remember that as with all euqilibrium follow le chateiller the equilibrium wills hift to restore value of kw , and changes only if TEMP IS ALTERED .
KW CONTROLS BALNCE BETTWEEN H+ AND OH- IN AQUEOUS SOLTUIOSN .
at 25 degrees wgt is the ph value of 7 in terms of h and oh-
it iss the neutral point at which h+ and oh- cocnwntrations are the same equal 10-7 , moldm-3 applying to water and also to neturals oltuions .
relaativve cocnetration sof h+ and oh- are determined by kw .
at 25 degrees is rain wwater neutral
-at 25 dehreees rainater is actually acidic , carbon dioxide dissovles inw ater to forma weak acid H2C03 , which is able to dissociate and release H+ ions into the water .
in rain water when H+ has icnreas OH ions will fall until
H+ and OH- = 1.00x10-14 .
-The overall efect is that Ph decreases ,r ainwter ph is abt 56 ,
mienral water fo rexmaple contains dissovled ions suhc as carbonate co32- , these ions reat with acid and thus remoe H+ ions at a lower (h+) .
(h+) , has creases (oh-) , must increaseuntil
(h+)(oh-( =1.00*10-14mol-14mol2dm6 at (25 degrees )
-The overalle fecct is that Ph increase . tHE ph OF MINTERL wate ris 7-8 abd tgat nebs nuebrak water us akjakube ,
check table 1 on page 40
…
strong bases tend to do what give example
weak bases tend to what give an example
strong ases tend to be hdyroxies of the emtals in group1 and 2
ammonia nh3 is a weak base , in aqueous soltuion n equilibriu is set up and the equilirium postion lies well to the LHS .
How do you use Kw , to calculate the Ph of strong abses
-to workout the pH of stong bses we need to know H+ and this depnds on
-the cocnentriton of the base
-the ionic product of water Kw = 1.00*10-14 (because th is links with (h+) AND (Oh-)
a strong monobasic alkali - give an example of this
NAOH is comeptlelu dissocited in aqueous solution . This means that (oh-) of a strong base is equal to the cocnentratio of the basw
NAOH —NA+ +OH-
SO (Oh-)=(NAOH)
we can find (H+) from kw and (Oh-)
kw=(H+) and (oh-).
Kw =(h+)(0h-)
(h+) =kw/(oh-)
the pH canbe calcualted using pH=-log10(H+)
see worked exmaple of 41 ., also check learning tip .
What is a buffer solution
it is a mixture that minimised pH changes on addition of small amounts of acid or base . The word minimises is essential to this defnition .
a buffer solution cannot prevent the pH from changing slightly but …
pPH changes are minimised , at least for as long as the buffer solution reamins .
give an example of how a buffer can be made
-it can be made from a weak acid and a salt of the weak acid , for example ehtnaoic acid (ch3cooh) and sodium ehtanoate (ch3coona)
what is a buffer solution a mixture of
a weak acid HA
and its conjugate base A-
whta sissociates fully and whagt dissociaes partially in he ch3cooh and ch3coona uffer sytem
check page 42 .
-the equilibrium mixture formed contains a high concenraiton of the undissociaed weak aci CH3COOH , and its conjugate abse CH3COO- . The high concenration of the conjugate base pushes the equilibrium ot he elt , so the concentraion of H+ ion sis ver small .
The RESULTING buffer soltuion contains alrge resvoirs of the weak acid and its conjugate base . seen in figure 1 .
can buffers be made from a weak acid and a strong alkali
-In this situation , a solution containnig a mixture of the salt and excess of weak acid is formed . For exampe , a weka acid sucha s methanoic aic d, HCOOH ,can be partially neutralised by an aqueous alkali such as NaOH .
How does a bufffer act to cotnrol pH
In a buffer solution ,t he weak acid HA ,a nd the conjugate base , A- , are both responsible for controlling pH . The buffer solution minimises pH changes b ussing the equilibrium
HA —- H+ + A-
OVERALL PRINCIPLE BEING
-the weak acid , HA removes added alkali
-the conjuage bas , A- removes added acid .
In a buffer , what does the addition of an acid H+ do to a buffer solution
(H+) IS INCREASES
THE CONJUAGTE BASE a- REACTS WITH H+ ions .
-the equilibrium shifts to the elft removing msot of the added h+ ions .
tthe euiqlibirum shifts to the elft HA— H+ + A-
-On addition of an alakali , OH- to buffer mixture
-OH- is increases small cocnentraion of H+ ions react witht he OH- ions .
Ha dissociates shifting the equilibrium tot hre right to resotre m ost of the H+ ions that have reacted
equilibriu shifts to the right
HA— H+ + A- .
These actions of a buffer are summarised in figure 2
HA — H+ +A-
left added acid right added alakli
true orfalse , a buffer cannot remove all of anya cid or alkali that is added
TRUE IT JUS MINIMISES THE pH changes .
why do we need to check the concentrations of HA and A -
as only a very small proprotion o HA dissocites , so , as disucssed in topic 5.1.11 onn calcualitng hte pHo f weka acids we can ssume that Ha equilibrium = HAA undissociated .
lculaitng the phH OF BUFFER SOLUTIONS
THE pH OF A BUFFER SOLTUIOND EPEDNS ON
-tHE ACID DISSOCIATION CONSTNAT kA , OF THE BUFFER SYTEM .
-THE COCNENTRATION RATIO FO THE WEak acid and its conjugate base .
-gor a buffer consiting of a weak aci HA and its conjuagete base A-
Ka =H+ A-/HA
menaing h+=ka* HA/A
what does the salt of the weak acid do
it is ionic , and sissociates completel in aqueous solution show example pn page 43 .
check learning tip and workd example on page 43
…
what is the equivalence point
it is the point ina titration at which the volume of one solution has reached exxlt the point of volume in the second solution .
This matches the stochiometry of the reaction taking pace .
what is the end point in a titration
the end point in a titration at which there are equal concentrations of the weak acid and conjugatebase forms of the indicagotr . The colour at the end point is midway beween the colours of the acid and the conjugate base forms .
when you carry out a titration , what are you determining
you are determining the volume of one solution that reacts exactly with a known volume of another solutoin knowna s the equivalnece point of the titration . At this point ,t he solution in the concical flask has exactly rected with the solution int he burette .
what waycan titratios be carrried out aicd to base terms
acid to base or base to acid - same pricniples apply to both routes .
check figure 1 section 1 on page 45 what is it showing
A slight increase in pH occurs as base is added , because the acid is in such excess . It is importaant to note that the section of the curve is not horizonral , pH is rising slightly .
check figure 1 section 1 on page 45 what is it showing
check figure 1 section 2 on page 45 what is it showing
the sharp rise in PH occurs the acid is no longe rin excess so any base added has a large impact on the pH .
check figure 1 section3 on page 45 what is it showing check figure 1 section 1 on page 45 what is it showing
a slight increase in pH occurs as a further base i added . The increase is only slightly because the base is in excess , now and extra base has had little impact on the pH .
-It is importanat to note however that this section is not horizontl - pH is rising slightly .
check figure 1 section equivalence point on page 45 what is it showing
the equivalence point is the centre of the vertical section of the titration curve
ttitration curces
pH meters or data loggers can be used to mesure the pH of the reactionmixture , as solution from the burette is added over time .
-The vsarying pH values can then be plotted as grpah , nown as an acid-base pH titration curve . THe cure mabe plotted by hand or computer pH data logger .
Typical cid -base titration
pH curve is shown in figure 1 for the reaction between a strong acid and strong base .
CURVE HAS THREE DISTINCT AREAS .
1-A slight increase in pH
2- a sharp increase in pH
3 - a slight increase in pH .
wh at i an acid base indicator
an acid base indicator is a weak acid often represented as Hln . An indivaotrhas one colour in it acid formHln and a different colour in its conjugate base form IN-
Investigation - purpose of a pH meter
-an indicator such as universal indicator can only give a general indication of a pH solution where as Ph meter can be accurate two 2 d.p
pH meter stage one
calibrate so the results are accurate .
pH meter stage two
-the probe is removed from its storage solution and rinsed with deionised water .
Ph meter stage three
-The probe is blotted dry and then placed into a solution of a known pH such solutions are usually manufactured professionally .
Ph is allowed to settle before checking that the pH has registered .
-process is repeated for other known Ph’s .
what colour is Hln
H+ and In-
what happens when there are equal amounts of weak acid and conjugate base
the indicator is at its end point .
how do you know that a titration has reached its end point
end points are accompanied by a visual colour change that allows for a titration to be monitored easily . For example ,t he colour of the end point for methyl orange is orange , at which point HIN (red) and (In-) are eqqual .
most indicators change colour over what range
of about two pH units .
-The pH ranges for the indicators shown in figue 2 on page 46 .
How to use the correct indicator for a titration ?
-Choose a pH value close to the end point as possible to the pH value of the titrations equivalence point .
-In practice , a suitable indicator , changes colour within the pH range in the vertical section of the titration curve .
For the following curves , the ranges of the indicators are not actually on the equivalence point , so how can you show that the reaction has reached this point ?
As the equivalence point , occurs halfway through an incredibly sharp rise or fall in pH , the moment at which the indicator passes through its end point is effectively the same as the moment that the equivalence point takes place .
STRONG ACID STRONG BASE TITRATION - check figure 3 on page 46
-What points can we notice on the curve
-The vertical section of the graph covers a large change in pH , starting around pH 3 , and ending around pH 11 , with an equivalence point at pH 7 .
-Both the indicators , methyl orange and phenolphthalein have end points that fall within this pH range .
-Either indicator would be suitable to use in a titration between strong acid and a strong base .
STRONG ACID - WEAK BASE TITRATION - figure 4 on page 47
-What points can we notice on the curve .
-The vertical section of the graph , covers a small change in Ph starting around pH 3 , and ending around pH 7.5 , with the equivalence point occurring at more acidic value , i.e. at a Ph lower than 7 .
-Methyl orange has an end point that galls within this Ph range .
STRONG ACID - WEAK BASE TITRATION - figure 4 on page 47
-What points can we notice on the curve . (2)
-Phenolphthalein does not have an end point that falls within this pH range .
-One methyl orange would be a suitable indicator for a titration between strong acid and a weak base .
WEAK ACID - STRONG BASE TITRATION - figure 5 on page 47 (1)
-The vertical section of the graph covers a smaller change in pH , and occurs further up toward the higher pH values , starting around pH 6.5 , and ending around pH 11.5 .
-The equivalence point occurs at a more basic Ph i,e pH value above 7 .
WEAK ACID - STRONG BASE TITRATION - figure 5 on page 47 (2)
-Methyl orange has an end point that falls outside this pH range .
-Phenolphthalein has an end point that falls within this pH range .
-Only phenolphthalein would be a suitable indicator for titration between a strong acid and a weak base .
WEAK ACIDS - WEAK BASE TITRATIONS (1) FIGURE 6 47
-There is no real vertical section .
-Neither indicator has an end point near the equivalence point .
-Neither indicator suitable .
WEAK ACIDS - WEAK BASE TITRATIONS (2) FIGURE 6 47
-An indicator would change colour gradually , over a few cm3 , of base added .
-No indicators , are really suitable for a weak acid - weak base titration .
Investigation for titrations summary (1)
-Set up titrations between acids and alkalis of different strengths . You should select some weak acids and bases and some strong acids and bases .
Investigations for titrations summary (2)
Carry out titrations using various combinations of weak and strong acids and bases . Use a pH meter to record the pH throughout the titrations . Use your results to plot acid-base titrations curve . Identify the equivalence points for each titration .
