Chemistry HL/SL help

[dessskris]

Guys, I am doing a design on investigating the relationship between the concentration of a reactant and the rate of the reaction. Let's say I have 2 HCl solutions of the same volume, say 25 mL. The concentration of solution X is 1.0 M and that of solution Y is 3.

0 M. Will their masses be the same? They have the same volume but different concentration. How about the mass?

My thought: in solution X there will be 0.025 mol HCl. in solution Y 0.075 mol HCl. HCl's molecular mass is 36.46. So the mass of solution X would be 0.912 gr and that of solution Y would be 2.73 gr.

Is that correct?

I am reacting HCl and MgCO3 and measuring the mass loss. If my thought is correct, then can I do this design that way? By measuring the mass loss? Because I am thinking not to find the rate of reaction, but just drawing the graph and comparing the steepness of the graph. Is that feasible?

Thanks.

I'm hoping your math is right because it's too late for me to do that

You should be using c=n/v to find your moles of HCl in each solution and can find the mass from that. But for rates of reaction thats just added work you don't need. HCl and MgCO3 should release MgCl+H2O+CO2 (actually its HCO3 but at your temp and pressure it RAPDILY decomposes to water and CO2). Since CO2 is a gas that will be exiting the system you can record mass of the beaker+HCl+MgCO3 to find your initial mass, then after X seconds after beginning the reaction record the mass again. This should give you experimental data that you can find the rate of reaction from. Do this ON the scale so you can watch it drop mass and its just more precise in my opinion if you don't move it around everywhere (since that would actually be the equivalent of stirring which causes more collisions...you get it).

Reason I was avoiding the mass from the math is because of just human error, too much HCl or a little less MgCO3 etc etc, measuring isn't perfect

Ok I get that thanks, will put that as an explanation. Now on the mouth of the conical flask I added cotton wool to prevent any loss of acid due to acid spray. Then would I need to keep the mass of the cotton wool constant in the different repeats? Yes, stupid question but... yeah

Oh one more, is it so necessary to put a diagram to show the experimental setup?

EDIT: oh Drake I forgot, since you previously recommended me to also control temperature and pressure (a.k.a keeping them constant), I thought pressure only affects the speed of reaction when at least one of the reactants is in gaseous state? or... not? How to control a pressure though? by doing the experiment in the same place and using the same conical flask?

And for temperature, do I need to keep the temperature of the MIXTURE constant, or that of the SURROUNDINGS? Since the reaction might be an exothermic or endothermic reaction, the temperature cannot be constant right? I find it even more troublesome since I now have that cotton wool covering the flask. Where to put the thermometer? Using a water bath is not a good idea since I need to measure the mass and I think a water bath is too heavy to put on an electronic balance but if it is of the surroundings, can I just turn the AC off and close any window, to keep it constant?

[Hedron123]

Guys, I am doing a design on investigating the relationship between the concentration of a reactant and the rate of the reaction. Let's say I have 2 HCl solutions of the same volume, say 25 mL. The concentration of solution X is 1.0 M and that of solution Y is 3.

0 M. Will their masses be the same? They have the same volume but different concentration. How about the mass?

My thought: in solution X there will be 0.025 mol HCl. in solution Y 0.075 mol HCl. HCl's molecular mass is 36.46. So the mass of solution X would be 0.912 gr and that of solution Y would be 2.73 gr.

Is that correct?

I am reacting HCl and MgCO3 and measuring the mass loss. If my thought is correct, then can I do this design that way? By measuring the mass loss? Because I am thinking not to find the rate of reaction, but just drawing the graph and comparing the steepness of the graph. Is that feasible?

Thanks.

I'm hoping your math is right because it's too late for me to do that

You should be using c=n/v to find your moles of HCl in each solution and can find the mass from that. But for rates of reaction thats just added work you don't need. HCl and MgCO3 should release MgCl+H2O+CO2 (actually its HCO3 but at your temp and pressure it RAPDILY decomposes to water and CO2). Since CO2 is a gas that will be exiting the system you can record mass of the beaker+HCl+MgCO3 to find your initial mass, then after X seconds after beginning the reaction record the mass again. This should give you experimental data that you can find the rate of reaction from. Do this ON the scale so you can watch it drop mass and its just more precise in my opinion if you don't move it around everywhere (since that would actually be the equivalent of stirring which causes more collisions...you get it).

Reason I was avoiding the mass from the math is because of just human error, too much HCl or a little less MgCO3 etc etc, measuring isn't perfect

Ok I get that thanks, will put that as an explanation. Now on the mouth of the conical flask I added cotton wool to prevent any loss of acid due to acid spray. Then would I need to keep the mass of the cotton wool constant in the different repeats? Yes, stupid question but... yeah

Oh one more, is it so necessary to put a diagram to show the experimental setup?

EDIT: oh Drake I forgot, since you previously recommended me to also control temperature and pressure (a.k.a keeping them constant), I thought pressure only affects the speed of reaction when at least one of the reactants is in gaseous state? or... not? How to control a pressure though? by doing the experiment in the same place and using the same conical flask?

And for temperature, do I need to keep the temperature of the MIXTURE constant, or that of the SURROUNDINGS? Since the reaction might be an exothermic or endothermic reaction, the temperature cannot be constant right? I find it even more troublesome since I now have that cotton wool covering the flask. Where to put the thermometer? Using a water bath is not a good idea since I need to measure the mass and I think a water bath is too heavy to put on an electronic balance but if it is of the surroundings, can I just turn the AC off and close any window, to keep it constant?

Hey, I never got to answer your post before. Pressure and temperature can be controlled by carrying out the experiment under the same weather conditions. This will assure constancy throughout the procedure (you don't need to use a water bath, don't worry).

As regards the experiment, you could use an indicator to see how much time does it take for all the reactant to be consumed. That might help you calculate the rate of the reaction.

P.S. You cannot calculate the mass of a solution by multiplying the number of moles and the molar mass, you need density values in order to do so.

Edited November 21, 2010 by Hedron123

[Drake Glau]

Guys, I am doing a design on investigating the relationship between the concentration of a reactant and the rate of the reaction. Let's say I have 2 HCl solutions of the same volume, say 25 mL. The concentration of solution X is 1.0 M and that of solution Y is 3.

0 M. Will their masses be the same? They have the same volume but different concentration. How about the mass?

My thought: in solution X there will be 0.025 mol HCl. in solution Y 0.075 mol HCl. HCl's molecular mass is 36.46. So the mass of solution X would be 0.912 gr and that of solution Y would be 2.73 gr.

Is that correct?

I am reacting HCl and MgCO3 and measuring the mass loss. If my thought is correct, then can I do this design that way? By measuring the mass loss? Because I am thinking not to find the rate of reaction, but just drawing the graph and comparing the steepness of the graph. Is that feasible?

Thanks.

I'm hoping your math is right because it's too late for me to do that

You should be using c=n/v to find your moles of HCl in each solution and can find the mass from that. But for rates of reaction thats just added work you don't need. HCl and MgCO3 should release MgCl+H2O+CO2 (actually its HCO3 but at your temp and pressure it RAPDILY decomposes to water and CO2). Since CO2 is a gas that will be exiting the system you can record mass of the beaker+HCl+MgCO3 to find your initial mass, then after X seconds after beginning the reaction record the mass again. This should give you experimental data that you can find the rate of reaction from. Do this ON the scale so you can watch it drop mass and its just more precise in my opinion if you don't move it around everywhere (since that would actually be the equivalent of stirring which causes more collisions...you get it).

Reason I was avoiding the mass from the math is because of just human error, too much HCl or a little less MgCO3 etc etc, measuring isn't perfect

Ok I get that thanks, will put that as an explanation. Now on the mouth of the conical flask I added cotton wool to prevent any loss of acid due to acid spray. Then would I need to keep the mass of the cotton wool constant in the different repeats? Yes, stupid question but... yeah

Oh one more, is it so necessary to put a diagram to show the experimental setup?

EDIT: oh Drake I forgot, since you previously recommended me to also control temperature and pressure (a.k.a keeping them constant), I thought pressure only affects the speed of reaction when at least one of the reactants is in gaseous state? or... not? How to control a pressure though? by doing the experiment in the same place and using the same conical flask?

And for temperature, do I need to keep the temperature of the MIXTURE constant, or that of the SURROUNDINGS? Since the reaction might be an exothermic or endothermic reaction, the temperature cannot be constant right? I find it even more troublesome since I now have that cotton wool covering the flask. Where to put the thermometer? Using a water bath is not a good idea since I need to measure the mass and I think a water bath is too heavy to put on an electronic balance but if it is of the surroundings, can I just turn the AC off and close any window, to keep it constant?

Hey, I never got to answer your post before. Pressure and temperature can be controlled by carrying out the experiment under the same weather conditions. This will assure constancy throughout the procedure (you don't need to use a water bath, don't worry).

As regards the experiment, you could use an indicator to see how much time does it take for all the reactant to be consumed. That might help you calculate the rate of the reaction.

P.S. You cannot calculate the mass of a solution by multiplying the number of moles and the molar mass, you need density values in order to do so.

Yes the mass of cotton wool needs to stay constant

Diagram of setup, why not, makes you look better and its easier to read which is a point in your design if I remember right.

And yea, by constant I meant just do all your trial in the same room on the same day so it won't vary a ton Not have EXACTLY 25 degrees Celsius for both reactions. One or 2 degrees won't change it enough in my opinion.

P.S She can calculate the mass without density. You use density to calculate mass when you only know the volume. She knows the concentration and volume and thus can find her moles and then its just a mol->mass problem. C=n/v is density for solutions in this case so your n is your "mass" in mole form.

[dessskris]

Sorry I am not quoting your posts because it's going to be too long and confusing for me

It is just a design so can I not care about the mass?

Calculate the rate of reaction? I think it would be rather confusing to find the no. of mol of CO2 lost and divide it by the time taken :/ is it possible to just divide the loss in mass by the Mr of CO2 to find the n of CO2? So rate of reaction can be found easily

I was actually referring to an O Level practical book, they have a very similar experiment with my design. In there they ask us to draw graphs of mass against time, and then based on the steepness of the graphs, we are asked to conclude the relationship between concentration of reactant and rate of reaction. Is this too simple for IB Chem SL design?

-------------------------------

Btw, this is another experiment about enthalpy change. What is the formula of ΔH?

Is it

ΔH= mc ΔT / n

OR

ΔH= -mc ΔT / n

??

The one with the minus sign was given by my teacher but it does not make sense since a negative value of ΔT (or a drop in temperature) would result in a positive value of ΔH. Vice versa. But I did some googling and some sources did mention the one with minus sign. Which one is correct?

[Drake Glau]

+delta H should be a drop on your temperature, its endothermic

-delta H should increase temp, its exothermic.

Negative delta H means that the energy of your reactants is greater than the energy of your products so during the reaction the level of H decreased (negative) and released energy. Opposite for the other. And as far as I know it's always benn q=mc deltaT for our class...

WTB post button for insert symbol...since we have smilies and everything else...

And finding the moles of CO2 lost would simply be taking the mass difference since it was the only gas being evolved and dividing it by the molar mass of CO2. So lets say your solution and beaker are 100g, at the end of the reaction its 95g, 5g of CO2 was released.

5g/44.01=moles of CO2

You have your time already from the stopwatch

moles/time=rate of reaction.

This does not work for every reaction though, this one only has one gas leaving the system which is why it works so well. If two gases are evolved from a reaction you can't tell the mass of each and so this does not work.

Another thing you can do is opposite. But it's a bit harder to do in this case...since there's a gas leaving...

[dessskris]

+delta H should be a drop on your temperature, its endothermic

-delta H should increase temp, its exothermic.

I still don't get this! Be patient please, sorry.

A temperature drop, the solution gets cooler, less hot, so it loses heat. When it loses heat, enthalpy must be negative

Vice versa. Is that correct?

[Drake Glau]

+delta H should be a drop on your temperature, its endothermic

-delta H should increase temp, its exothermic.

I still don't get this! Be patient please, sorry.

A temperature drop, the solution gets cooler, less hot, so it loses heat. When it loses heat, enthalpy must be negative

Temperature drops, means there is less energy in the solution by the time its done. Enthalpy is the total energy of a system. So if your temp drops meaning an energy drop, your initial energy will be greater than your end energy causing it to be negative when you do initialH-FinalH

It feels cold because the energy in the solution is much smaller than, let's say your hand, so the solution begins to suck energy from your hand making your hand feel cold. Or with a thermometer, the solution is absorbing energy/heat from the thermometer so your temp drops.

Vice versa. Is that correct?

Gah how I hate energetics now...finally got it over with last week!

[genepeer]

+delta H should be a drop on your temperature, its endothermic

-delta H should increase temp, its exothermic.

I still don't get this! Be patient please, sorry.

A temperature drop, the solution gets cooler, less hot, so it loses heat. When it loses heat, enthalpy must be negative

Vice versa. Is that correct?

A temperature drop means that the system is absorbing heat from the environment (including solution). This heat is being stored as chemical energy in the bond of the products. So the system (products) has absorbed heat and therefore the enthalpy change is positive.

[dessskris]

A temperature drop means that the system is absorbing heat from the environment (including solution). This heat is being stored as chemical energy in the bond of the products. So the system (products) has absorbed heat and therefore the enthalpy change is positive.

OOOOOHHH I GET IT NOW!!!

Oh my God.. So as the solution absorbs heat, the surroundings would be cold?

Wait, but the temperature sensor (we were using data logger) was put into the solution, not put on the surface of the polystyrene cup

[genepeer]

Well you might say the chemicals are actually absorbing the heat from the water (the majority of the particles in the solution) and surrounding air. So the solution (water in particular) is losing heat, but the chemicals that we are actually interested in are absorbing the heat energy. So the solution (water in particular) is part of the environment.

[Drake Glau]

A temperature drop means that the system is absorbing heat from the environment (including solution). This heat is being stored as chemical energy in the bond of the products. So the system (products) has absorbed heat and therefore the enthalpy change is positive.

OOOOOHHH I GET IT NOW!!!

Oh my God.. So as the solution absorbs heat, the surroundings would be cold?

Wait, but the temperature sensor (we were using data logger) was put into the solution, not put on the surface of the polystyrene cup

Yea, the temp probe is the environment too The solution is taking energy from it too which is why it picks up cold temps, that's how thermometers work

[dessskris]

In a lab, I am finding the enthalpy of neutralisation of different acids with NaOH solution. The acids used were HCl, HNO₃, CH₃COOH and H₂SO₄. It happens that the acids from the greatest enthalpy of neutralisation to the lowest are in the following order:

H₂SO₄ > HCl > CH₃COOH > HNO₃.

Why is that so?

I guess it's because of the individual average bond enthalpies? is that correct?

If yes, do I then need to calculate the ΔH bonds broken and ΔH bonds formed?

If not, why is it then?

[caro0049]

hi people =)

does one of you know a link to download the chemistry HL paper 1 and 2 from 2009?????

i would be very thankful

[dessskris]

hi people =)

does one of you know a link to download the chemistry HL paper 1 and 2 from 2009?????

i would be very thankful

Sorry but IBS is against distributing past IB papers. We do not have it in the Files section and we do not give links where you can get them from. We suggest you to ask your teacher to give you a copy, or if they are not willing nor able to give, google it by yourself.

[dessskris]

is anybody familiar with spectrophotometer? can I just say SPM instead? thanks. I had an experiment on measuring rates of reactions from the light absorption.

I do not really get how SPM works. it measures the absorption of the substance put in the cuvette.. but how does it absorb light? does the substance have to be coloured to be able to absorb? or any solid can just absorb light?

I got 4 questions to be answered but I do not even understand how SPM works so I apparently cannot think of the answer to those. anybody could help? thanks!

[Drake Glau]

Guess I'll copy over my reply to your message for this

Umm I've only used one of these once and it was for biology when we were messing with chlorophyll. From what I remember it simply measures the amount of light that gets reflected and thus records how much was absorbed right?

If that's so your graphs make perfect sense to me. Lets say you react two very dark chemicals (I don't know what yours look like except HCl is clear I know this). This means they are reflecting most of the light, because you can see it lol so the level absorbed is really low. Now lets say those reactants react and being forming a clear solution. Now they are absorbing all the light because it's clear and not reflecting the light so the absorbancy level is now high. At the beginning of a reaction the rate is at it's highest so it will display the absorbancy level rising with the highest gradient and as the reaction continues the rate decreases because they is less and less reactants as they are used up and less reactant means less frequency of particle collision and thus lowering your rate of reaction. So as the rate decreases the absorbancy level would decrease too.

This will change depending on the reaction of course because it can either be clear reactants forming colored or the other way. What kind of polynomial is this btw? I THINK you should have a sigmoid curve which I think is 3rd degree, but I'm not sure lol. Should be exponential, then kind of linear, then plateau off near the end of the time/reaction.

Course it looks out of context sort of because I mention your graphs but whatever. Hope it helps some other people

[dessskris]

Mind if I reply the PM in here too? Forget about the PM and let's just use this thread

If my classmates are smart enough to google and find IBS, this would also be useful for them

Yeees that is what I read in my handout, too.

HCl is clear I know this

Na2S2O3 (aq) + 2HCl (aq) --> 2NaCl (aq) + SO2 (g) + S (s) + H2O (l)

HCl is colourless and clear. Na2S2O3 is colourless and clear. The products are white (I guess it's because of the sulfur solid) and smelly because of the SO2 gas.

I get it now btw, all coloured substances reflect/transmit the light, that's why the %absorbance is low, right? As time goes by, what happen? The reaction should produce more S solid (if this is a slow reaction) and so the solution should be even whiter. Thus absorbance should decrease..?

What does SPM do to the solution in the cuvette? Nothing, right? So I thought absorbance should be constant? But why does it increase?? This does not make sense now..

In your example, the reactants are coloured and the products are clear, yeah? So the absorbance increases over time. My reaction is the other way round, shouldn't absorbance decrease over time??

And how can I find initial rate of reaction?

Thank youuu Drake..

I would thank any other people who could help me, too.

EDIT: my graphs were 4th degree polynomial graphs but I guess I should change it to 3rd?

[Drake Glau]

Mind if I reply the PM in here too? Forget about the PM and let's just use this thread

If my classmates are smart enough to google and find IBS, this would also be useful for them

Yeees that is what I read in my handout, too.

HCl is clear I know this

Na2S2O3 (aq) + 2HCl (aq) --> 2NaCl (aq) + SO2 (g) + S (s) + H2O (l)

HCl is colourless and clear. Na2S2O3 is colourless and clear. The products are white (I guess it's because of the sulfur solid) and smelly because of the SO2 gas.

I get it now btw, all coloured substances reflect/transmit the light, that's why the %absorbance is low, right? As time goes by, what happen? The reaction should produce more S solid (if this is a slow reaction) and so the solution should be even whiter. Thus absorbance should decrease..? Umm yes, absorbance would decrease because white is the reflection of all color and you started with clear solutions.

What does SPM do to the solution in the cuvette? Nothing, right? So I thought absorbance should be constant? But why does it increase?? This does not make sense now..

The SPM does nothing to the reaction, absorbance doesn't stay constant because the reaction is going on inside and as the reaction continues more and more of that white substance is being made. Absorbance increasing...umm...I don't know, I might have something messed up on how this SPM works

In your example, the reactants are coloured and the products are clear, yeah? So the absorbance increases over time. My reaction is the other way round, shouldn't absorbance decrease over time??

Again I could be messing up on how this SPM is working and so my example's increasing absorbancy could possibly be backwards making your lab make more sense

And how can I find initial rate of reaction?

I don't really know to be honest, we never spent this much detail on kinetics, or equilibrium for that matter, sorry

Thank youuu Drake..

I would thank any other people who could help me, too.

EDIT: my graphs were 4th degree polynomial graphs but I guess I should change it to 3rd?

Can you send me your graphs by chance?

Comments on red so I can address each question easier

Edit - I love the panic emoticon

Edited December 19, 2010 by Drake

[Tilia]

In a lab, I am finding the enthalpy of neutralisation of different acids with NaOH solution. The acids used were HCl, HNO₃, CH₃COOH and H₂SO₄. It happens that the acids from the greatest enthalpy of neutralisation to the lowest are in the following order:

H₂SO₄ > HCl > CH₃COOH > HNO₃.

Why is that so?

I guess it's because of the individual average bond enthalpies? is that correct?

If yes, do I then need to calculate the ΔH bonds broken and ΔH bonds formed?

If not, why is it then?

Think in terms of strong/weak acid and bases. Was it a negative or positive enthalpy change?

[dessskris]

@Drake haha thank you, I will try to find out more about this transmittance and absorbance and SPM thingy, my CE is messed up!

Think in terms of strong/weak acid and bases. Was it a negative or positive enthalpy change?

Negative.

H2SO4: -61.9 kJ/mol

HCl: -59.8 kJ/mol

CH3COOH: -57.3 kJ/mol

HNO3: -56.8 kJ/mol

CH3COOH is a weak acid and the rest are strong acids, right?

How come HNO3's neutralisation releases the least energy, though? Btw I reacted them with NaOH.

---

EDIT:

Never mind, that order was in accordance with my experimental values. When I looked at the lit values, the order should be H2SO4 > HCl > HNO3 > CH3COOH, so yeah that's because of the strengths of the acids. Thank you!