4.3 How fast? — rates
Posted by: intechemistry on: September 23, 2010
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50 Responses to "4.3 How fast? — rates"
October 31, 2010 at 7:46 pm
Salam. I have links to two Chemistry Dictionaries in my links menu 
Here’s what dictionary #1 says:
Stoichiometry
Description of the quantitative relationships among elements and compounds as they undergo chemical changes.
http://home.nas.net/~dbc/cic_hamilton/dictionary/a.html
In simple terms, it’s the related/interconnected/balanced-ratio of coefficients in a balanced chemical equation. 
The statement is incorrect in the sense that it doesn’t always apply. A reaction can have a low activation energy but the concentration of the reactants might be so low that the rate is very low.
& Thanks for posting this Q in the correct place
October 31, 2010 at 8:14 pm
how about,
-rate-determining step=slow step, correct?
-n why in multi step reactions rate-determining step has higher activation energy?
October 31, 2010 at 8:29 pm
rate-determining step=slow step, correct?……YES!
The slowest step is the one that determines the rate of the reaction and this slow step has the greatest energy requirement. The highest energy requirement of ANY reaction is the one that has E >= Ea. Once you get over Ea, all steps in the reaction have lower energy states.
October 31, 2010 at 9:08 pm
thanks a lot sir
later if i’ve got any problem, i’ll i ask you more….:D
October 31, 2010 at 10:18 pm
salam sir.
when gas react with solid, the rate is independent of the pressure of the gas. is this case the same as the metal-catalysed gaseous reactions?
November 4, 2010 at 10:02 am
Salam Dzatil.
Actually I feel like a bit ‘unsure’ about this question :s
If we apply a simple part of collision theory a collision has to happen in order for a reaction to occur to this question, then as P increases, the number of collisions per unit time will increase, so we can imagine that as pressure increased, the rate would increase.
rate = k [reactant1]^a [reactant2]^2 etc..
and
k= A e^(Ea/RT)
The term ‘A‘ is sometimes called the ‘collision frequency factor’ (or certainly, that’s what I sometimes call it!), and as the name implies, it will be affected by the pressure, therefore the rate will also be affected. {I don’t know the mathematical description of the variable ‘A’}
If you read somewhere that in this case, (g) + (s) reaction that rate wasn’t dependent on P , then I think what it really means is the effect is not very significant. I am quite sure (because of the mean velocity of gas particles is generally fast hence already yields a large number of collisions p.u.t.) that the surface area of the solid would have a far greater effect on the reaction, hence be the rate determining/limiting factor here.
A metal-catalysed reaction has a number of other differences compared to a gas-solid reaction, so one cannot really use one system to say what definitely happens in the other system. In the metal-cat gas phase reaction, there is no longer the limiting factor of surface area, so the emphasis shifts from solid surface area to pressure as a means of increasing collisions, but most importantly, the catalyst provides a second (alternative) way for the reaction to happen which is much faster than the gas phase reaction, hence reducing the significance of pressure.
That’s how I look at it anyway :s
So now you’re going to ask me about ammonia, yes?
“why is high P used there if it’s a metal-catalysed gas phase reaction? To which I will offer the answer that the yield of ammonia is very low for the conditions used (necessary high temp for increased rate but reduces yield as rxn is exo) that almost every possible way to increase yield must be used and this includes the application of high pressure.
Your 2 line Q is pretty tough.
For exams, you will most likely only have to state the effect of pressure on both systems (or perhaps, just the latter system) so you could do it in a sentence.
Is my attempted answer satisfactory?
November 4, 2010 at 12:44 pm
Salam sir,
can i have the answer for past year papers unit 5.3: redox and chemistry of transition metal?
thank you very much..
November 4, 2010 at 2:14 pm
ohh i see, now i have the idea on how are the reactions take place. from your explanation sir, i can conclude that :
~ for (g)+(s) reaction, the rate is limited by the surface area of the solid, hence we can say the pressure of gas has no effect on the reaction.
~ for metal-catalysed reaction, the metal catalyst provide an alternative pathway for the reaction, hence increasing the proportion of effective collision and rate of reaction. this means, any increase in pressure of gaseous reactants will not alter the rate. but in ammonia case, high pressure is used to increase the yield, as high temperature will reduce it due to the reaction is being exothermic.
is it correct sir? correct me if i’m wrong
November 4, 2010 at 3:00 pm
assalmualaikum sir.
Can you explain about the reaction between propanone & iodine? why need quenching & also, what’s the significance of each step? Is this considered also as clock reaction?
And, how about da reaction between hydrogen peroxide & iodine? Are these two experiments basically the same?
November 4, 2010 at 4:43 pm
8, diyanahM7, November 4, 2010 at 12:44 pm
Salam. I will try and upload the answers tonight, God willing.
November 4, 2010 at 4:53 pm
#9, dzatil, November 4, 2010 at 2:14 pm
~ for (g)+(s) reaction, the rate is limited by the surface area of the solid, hence we can say the pressure of gas has no effect on the reaction.
Intechemistry replies:I’m sticking with the answer that there is a small, perhaps not very significant, effectin rate with pressure.
~ for (g) + (g) with metal-cat, the inc in P will affect the rate but again only slightly and not very significantly, as the high temp and cat, account for the greatest increase in rate. High P is used for yield purposes… more moles of gas on the reactants side etc…
November 4, 2010 at 6:14 pm
10, (:Nurul Afiqah:), November 4, 2010 at 3:00 pm
Salam.
Can you explain about the reaction between propanone & iodine? Intechemistry replies: Not unless you tell me what it is about the reacton you are having peoblems with. It would be very long to discuss it here. Besides, the books should already give it good coverage as it’s something mentioned numerous times in the specification. What point are you stuck on?
Quenching is done to stop the reaction, so the concentration of the reactants and/or products can be determined at a specific time. If we didn’t quench the reaction, it would continue and we could not get accurate time/concentration data.
As for the signifance of each step, I’m not sure what steps you mean. Could you write them out and tell me what you think the significance is? It’s helpful to know how you are percieving these things.
Is it a clock reaction? No, but doubtless it could be modified into a clock reaction but to do so would be an unnecessary burdeon as the reaction can be well followed using titration.
I think the H2O2 reaction with I- is done as a clock reaction. Check these out:
a) http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA3/MAIN/CLOCKRX/PAGE1.HTM
b) http://en.wikipedia.org/wiki/Iodine_clock_reaction#Hydrogen_peroxide_variation
c) http://www.practicalchemistry.org/experiments/iodine-clock-reaction,55,EX.html
November 4, 2010 at 8:40 pm
oh ok, thank you sir…
don’t worry sir, take your time..ngee =)
November 4, 2010 at 11:41 pm
Alhamdulilllah, I think now I get it, mb the weather’s too hot dis evening dat i could not look at the book clearly:P so, thank you sir:)
November 4, 2010 at 11:55 pm
If it’s of any help, when it comes to this propanone, H+ and I2 mechanism, I think Hodder’s 4 step mechanism is nice and simple, lots of short cuts 
. I don’t think curly arrows are required for this particular mechanism, it’s a bit too long.
Strangely, of the three three edexcel endorsed A2 revision books I have, only 1 gives the mechanism (similar to what appears in Hodder). Creation of the rate equation from data and identification of what must be the slow step (involving 1 molecule of propanone and H+ only) is perhaps the most important thing here.
November 11, 2010 at 2:06 pm
salam sir,
in iodine clock reaction, what is the purposes of sodium thiosulfate being added into potassium iodide but it is not shown in equation?and why we have to change the concentration of hydrogen peroxide and water, not KI instead?
November 11, 2010 at 10:34 pm
Salam. In the iodide solution, I- will be oxidised to I2. The thiosulphate is there to remove the I2 once it’s formed. When all the thiosulphate has been used up, the build up of I2 will turn the starch blue-black.
the H2O2 is an oxidising agent. It oxidises the I- into the I2. Changing it’s concentration changes the rate of reaction (the rate of I2 production) The amount of water is usually added to keep the total volume constant hence some of the concentrations of certain species will also be constant and NOT affect the rate.
If you wanted to, you could vary KI only. Then the rate of reaction (assuming the order of reaction w.r.t. I- is not zero) would be related to the concentration of KI.
There are lots of variations that can be done in this reaction.
You could (if you wanted to) vary the concentration of H+ ions to see what effect acidity had on the reaction.
In the final exams, they would be pretty mean to ask you to describe how to do an iodine clock reaction. (but I guess they could do it!) They are more likely to give you a particular set-up of such a reaction and ask questions about it.
P.S. I think they would give an equation between I2 and (S2O3)2- somewhere, especially if they take the opportunity to ask about the REDOX aspect of the reaction. Not so significant perhaps when it comes to the Kinetics, which, if you are only varying the H2O2 concentration, then the fixed (S2O3)2- is of little consequence.
November 19, 2010 at 12:07 am
salam sir…
what are the typical steps of doing the iodine clock reaction? (if the question asks us to describe one)
November 24, 2010 at 3:02 am
Salam Amirul.
Have you read the links I posted above? You should be able to determine the steps there, especially if you correlate it with what your books say
January 8, 2011 at 10:21 am
salam sir,
if the question asks to describe the chemical and physical method by which the progress of the reaction could be followed, what am I suppose to answer? is it the process of conducting the experiment?
the question is about this equation ;
2MnO4- (+) 6H+ (+) 5(COOH)2 = 2Mn2+ (+) 8H20 (+) 10CO2
January 8, 2011 at 11:08 am
Salam.
For the benefit of all reading this, then in general, when you do these Q’s, try and spot
a) if the colour will change, then use colorimetry [note: not calorimetry!
b) If there is acid or base consumed (or produced) then use titrimetry (using standardised acid or base in the burette)
c) If there is a species known to undergo REDOX reactions
(e.g. I-, (S2O3)2-, and of course the transition metals Fe2+Cu2+ etc) then use redox titration.
d) If the reaction involves a change in the ions as productsor reactants then you can use a conductometric experiment using a conductivity probe
e) If a gas is produced we can use volumetric methods employing a frictionless graduated syringe or an inverted graduated measuring cylinder full of water sitting in a trough of water [Note: can also do if similar techniques if a a gas is consumed but would be a lot more problematic to set up the experiment]
Your question is great because there are many methods you can use.
1) Colorimetry is the easiest and gives giid accuracy if you use a calibrated colorimeter. That covers the physical method (you are not doing any ‘new’ cheical reactions – simply monitoring the reaction you gave above). You could have used gas/volumentric measurment or conductance measurment also but more tricky to do experimentally.
2) As an A-Level answer ,I would say for the chemical method, do acid/base titrimetry. Using standardised NaOH(aq) solution {i.e. NaOH(aq) whose concentration is accurately known) and calculate the moles of H+ present [NOTE: NaOH will react with the H+ AND the (COOH)2 and the Mn2+ (forming Mn(OH)2 ) so that must be remembered in the calculation step! The other possible complication that the NaOH added might make change the REDOX potential of (MnO4)- causing different/side reactions is not likely to be a significant factor as you should have QUENCHED the sample of the reacting solution, so it should still be quenched when you add the NaOH(aq) AND the conditions are still, in effect, acidic - until the equivalence point is exceeded]
Hope that helps?
January 8, 2011 at 3:10 pm
salam..
i want to ask about the quenching reagent(sodium hydrogencarbonate)..it is used to stop the reaction for example reaction with iodine and propanone..do the sodium hydrogencarbonate react with the mixture if it is added?it will not affect the concentration of the reacting mixture?
January 8, 2011 at 6:50 pm
Salam zophie.
Na2CO3(aq) or NaHCO3(aq) can be used to quench a specfic type of reaction.
Obviously it will react with any H+ so you don’t use it as a quenching agent for acid/base titrations, because there would be no more H+ left for you to analyse!, but you do (usually)use it for redox titrations.
The propanone/iodine/acid reaction can be followed by redox reaction as the conversion between I2/I- is very easy to achieve and is a redox process.
January 8, 2011 at 7:06 pm
P.S. Titrations involve balancing moles, and not concentrations. this is a common misperception. It’s why one can wash-down the neck of the conical flask when close to the end point, making sure all the reactants are in contact with each other.
Yes, when you add the Na2CO3, the concentration of the reactants in the conical flask (e.g. I2(aq)) will become lower, but it doesn’t matter because the same number of moles of that reactant is present in the conical flask., meaning you will add the same number of moles of the liquid from the burette {sometimes called the titrant}.
There are some reasons why students make this error, but I will try and avoid discussing it as it might be confusing.
February 18, 2011 at 2:39 pm
Salam, sir. Is sampling method (using quenching & titration) is the same as titrimetic analysis (which is used to measure the fall in acid concentration during esterification)?
February 18, 2011 at 2:40 pm
Salam, sir. Is sampling method (using quenching & titration) is the same as titrimetric analysis (which is used to measure the fall in acid concentration during esterification)?
February 21, 2011 at 12:31 am
Sampling can involve a number of methods of analysis. Titrimetry (or just ‘titration’) is one such method within the sampling category, and when the titration method is done, the process of quenching.is used. Quenching is done because titration isn’t an instantaneous method, i.e. it takes some time to perform the titration, one must therfore ‘quench’ to stop the reaction to allow an accurate time period and concentration value withing that time period.
Quenching is titration are separate things, but are done together when titration is to be used to determine the rate of reaction.
May 1, 2011 at 11:29 am
Salam sir…
may i know why order of reaction cannot be deduced from the stoichiometric equation of a reaction?
May 1, 2011 at 2:12 pm
Salam.
The rate equation tells you the number of molecules of reactants used up to and including the slow step.
The chemical equation tells you the moles relationships of reactants and products involved in the reaction.
^^take note of the moles / molecules issue also
After the slow step there may be more steps consuming even more reactants which would therefore increase the coefficient with the reactants in the chemical reaction but would not be present in the rate equation.
May 31, 2011 at 5:53 pm
Hi sir,
Br2 + HCOOH —> 2Br- + CO2 + 2H+
Facer says use colorimetry. You mentioned that colorimetry isn’t a suitable method as CO2 released may affect its accuracy? How can we follow the rate of the rx then?
From Facer A2 page 13, the Examiner’s Tips on the bottom right mention that ‘Following the concentration of one of the reactants give one of the overall order of reaction, not the partial order w.r.t. the substance whose concentration is being measured.’ I don’t understand why it is so. If the graph of [A] against t, why doesn’t it just give the partial order wrt to A?
Thanks!
May 31, 2011 at 6:52 pm
if temperature increases, there will be more successful collision.
is this correct?
but why george facer marking scheme wrote cannot write there will be more successful collision if temperature increase because over time there will be the same number of successful collision
June 6, 2011 at 5:22 am
@ 31, chocoviolicious1808, May 31, 2011 at 5:53 pm
I have done that reaction with ALM2 (and ALM3 I think). The CO2 formed caused problems. The gas formed, stuck to the inside of the cuvette in the form of bubbles, which grew in size and eventually floated up to the surface {while new bubbles slowly grew instead}. The path length of the light continually changed. The results were pretty meaningless. We really tried to make sure the inside of the cuvette was clean too!, but the problem persisted. We even tried playing with the concentration so that the bobbles would be less etc.. but still had problems.
However, that technical problem is probably going to be overlooked by Edexcel as the answer use colourimetry is pretty standard here. The problem described above above “the gas produced in the solution may interfere with the transmission of light in the sample holder” may be present in a follow-up Q where some problems of the this technique may be asked about.
As for the ‘Following the concentration of one of the reactants give one of the overall order of reaction, not the partial order w.r.t. the substance whose concentration is being measured.’ I;ll be honest, to try and rationalise that, I have to jump through a number of cosmic mental leaps, and even then I don’t feel the answer is legitimate. This question was addressed here:
https://intechemistry.wordpress.com/2011/05/05/some-awaited-comments-on-equilibrium-and-kinetics/ In preparing that post, I tried to get a solid answer/explanation and failed 
and in this case, I feel that such a statement is very unlikely to be given in simple error given it’s assertion yet I lack the ability to explain why. My mind is ‘fixed’ towards the opposite.
If you plot [A] vs time it will give you the order w.r.t. A
If you plot [B] vs time it will give you the order w.r.t. B
If order of A was 1 and order of B was 2 how could the individual graphs, in all cases, give the same third order shape?.
I am afraid I have been unable to penetrate this issue in terms of being able to explain it to my students (or to myself!), yet I don’t want to say GF is wrong. My advice is stick with GF. But I will try (time permitting) to re-examine this issue. Sadly however it may come a bit late.
June 6, 2011 at 5:31 am
@32, emrys ong May 31, 2011 at 6:52 pm
It seems a bit strange. Can you give the exact reference as to where this appears?
In the mean time perhaps it means this:
The successful collision is the one that makes (or leads towards) the formation of products.
Say 1 mole of product is formed. The 1 mole can form at a lower temp over a longer period of time than 1 mole at a higher temp in a shorter amount of time, both had the same number of successful collisions overall as the same amount of product was formed.
Certainly however, the initial number of successful collisions at the higher temp will occur, which is why it is essential when saying more ‘successful collisions’ at a higher temp that you add PER UNIT TIME, i.e. hot and cold reactions considered over the same chosen period of time. That would be the correct and probably what you were thinking of.
June 6, 2011 at 1:46 pm
is on pg26 george facer A2 question 2b. but i understand more from your explanation. thanks a lot
June 6, 2011 at 9:19 pm
Re: #32
Thanks and noted!:)
As for the partial order issue, i don’t understand how to ‘stick to GF’. I mean… if we’re given a graph with constant half-life, we can only say that the order w.r.t. the substance A is 1. We can’t say that the overall order is 1, can we?
June 12, 2011 at 12:09 pm
@ chocoviolicious1808, June 6, 2011 at 9:19 pm
But if GF is right, why can’t we say that? {and actually we’d have no choice but to say this!}
If we get the constant half-life graph, then GF is telling us to say overall order is one. The order w.r.t. A can also be one as the the other reactants would (have to) have zero order.
i.e. rate = k [A]^1 [B]^0 <– Order w.r.t. A is one and overall order is 1 also.
June 15, 2011 at 11:00 am
Hi sir!
In SAM, #18(c)(vii),
‘If ethanoic acid of the same concentration and at the same temperature is used instead of hydrochloric acid, explain how the rate would differ.’
It’s an experiment to find the activation energy of the reaction between Mg(s) ribbon and HCl. Temperature is altered to get 1/time. A graph is drawn and Ea is calculated from the gradient. The question used HCl. Now it suggests using ethanoic acid.
The answer gives: The rate should be lower, since ethanoic acid is a weaker acid (compared to hydrochloric acid) and so there will be a lower concentration of hydrogen ions present.
But CH3COOH + NaOH —> CH3COONa + H2O. It doesnt matter whether the acid is strong or weak, because as the H+ is consumed the equilibrium of CH3COOH = CH3COO- + H+ is shifted to the right, producing more H+. So, 1 mol of ethanoic acid reacts with 1 mol of NaOH, and strength of acid does not matter.
Why is it that in the reaction between an acid & Mg (as in the question), the strength of the acid matters? I’m confused. :S
June 15, 2011 at 11:24 am
Rate is = k [A]^x [B]^y etc.
i.e. rate is proportional to concentration. Weak acids only produce a low concentration of H+ in solution hence the rate would be lower.
Yes because of eqm the weak acid eventually gives off all it’s H+ in the end, but along the way to getting to the end of the reaction, the weak acid is only ionising to a very small extent all the way through it’s reaction with Mg. So the concentration all the way through the reaction is permanently low.
For HCl it’s much higher throughout compared to the weak acid, hence a faster reaction with Mg.
February 6, 2012 at 11:05 pm
salam sir,
Based on GF A2(pg 14)i don’t really understand question(b)..what is the relationship between the gradient of the graph and the order with respect to oxygen?how the order with respect to oxygen can be determined by using gradient of the graph of other reactant’s concentration?
February 7, 2012 at 12:07 am
Salam.
i don’t have GF with me now. Any chance you could take a pic of it and e-mail it to
a l l a n @ s a l a m dot u i t m dot edu dot my
There is one graph in GF in the Kinetics section that I am not able to figure out. I think this could be the one (but x remember)
February 7, 2012 at 1:42 pm
ok..never mind sir..i’ll ask you in class tomorrow..i’m Nurul Nabihah from 11m1..
April 28, 2012 at 10:46 am
salam sir Allan,
i have a question about the enthalpy level diagram for SN1 reaction. there is one intermediate form in the mechanism am i right? so, the energy level of this ‘intermediate’ is higher or lower than the energy level for the reactants? is the question clear?
April 28, 2012 at 11:37 am
Salam Ezza.
Yes, you are right. , For A-Level, SN1 mechanisms do have one intermediate & the most commonly encountered one here is a carbocation.
The intermediate is at a higher enthalpy than the reactants.
(A bond had to be broken to make the intermediate, hence required energy)
May 2, 2012 at 7:06 am
but sir, in facer AS page 270, the diagram shows that the intermediate is at a lower enthalpy level..does this mean the intermediate is more stable with respect to the reactant?
May 2, 2012 at 12:54 pm
@46 | Ezza, May 2, 2012 at 7:06 am
What appears there is not the usual illustration.. I know for a fact that A-level authors sometimes say/depict something in a different way just because they don’t students to think that the common encountered explanations are the only way.
You are right about the interpretation of the diagram. It would mean the intermediate is more stable than the reactant. For most molecules occurring naturally at room temp, if you break one of their bonds (a radical or ionic species would likely be produced) then the intermediate is going to be LESS stable than the molecule it came from
May 2, 2012 at 7:11 pm
so sir, this means the intermediate is at HIGHER energy level than the reactants so LESS stable than the reactants. because energy is NEEDED to break a bond to form intermediate. is this right? so, in exam, i must draw intermediate at higher level than reactant… not follow facer..
May 2, 2012 at 9:24 pm
Yes. What you say is by far the most common case, hence usually more correct.
Actually, I’m seriously struggling to think of an intermediate that is of lower enthalpy than the reactant it came from.
Overall, I don’t think this point has importance for Edexcel Chemistry, so even if you were to follow Facer, you are likely to still get full marks…. or so I believe.
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October 31, 2010 at 5:52 pm
salam alayk,
sir,
-what is stoichiometry??
- the slower the reaction the higher the energy needed for the reaction. is this statement correct??