# Chemistry HL/SL help

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I can try and help as well, maybe not as well, but I can do my best. For those are in a different timezone by several hours and can't really wait for an answer (don't procrastinate people! )

But, on topic sorta I do have a question

I'm doing a lab about the enthalpy of neutralizing HCl of NaHCO3. I have my mass of NaHCO3, volume of HCl (reaming constant), my temperature change, and from all that I found my q from q=mc deltaT equation. But that's not the enthalpy change. All I know is that enthalpy change would be my kJ (q) per mol but I don't know which substance to use for my moles? Also do I need to figure all this out really? The sheet she gave us simply said Enthalpy lab and i kind of feel like stopping with q and how much energy my reaction released should be enough...

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Well if you write out the balanced chemical equation you should use whichever reactant you want, it's easier to just use the one that only requires 1 mol. Also, think about the sources of error in your experiment and how they would affect your enthalpy change...let's just say you should see if you could have lost any heat and not been able to factor that into your calculations, and, if there is some way, figure out how to include it in your calculations.

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Yea I figured it out after I left, thanks though, and I just found my limiting reactant and took my q divided by the moles of the limiting reactant.

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Yea I figured it out after I left, thanks though, and I just found my limiting reactant and took my q divided by the moles of the limiting reactant.

Hey Drake, sorry for not answering before. When you are working at a constant pressure Qp (meaning Q at a constant pressure) = ∆H.

Demonstration)

H = E + PV (2nd law of T.D)

so:

∆H = ∆E + ∆PV (I)

E is the internal energy of the system which is equal to: Q + W (heat + work)

∆E = Qp - P∆V (II) (we are assuming pressure is constant)

∆H = ∆E + P∆V (I took P as a common factor)

By replacing ∆E from eq. II in eq I:

∆H = Qp - P∆V + P∆V

=> ∆H = Qp

This is derived from the 2nd law of Thermodynamics. I'm not sure if I have cleared your thoughts, be sure to ask again if you didn't understand. The units of ∆H will depend upon the heat capacity you use: specific heat capacity will result in joules per units of mass while molar heat capacity will result in joules per moles.

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Yea I figured it out after I left, thanks though, and I just found my limiting reactant and took my q divided by the moles of the limiting reactant.

Hey Drake, sorry for not answering before. When you are working at a constant pressure Qp (meaning Q at a constant pressure) = ∆H.

Demonstration)

H = E + PV (2nd law of T.D)

so:

∆H = ∆E + ∆PV (I)

E is the internal energy of the system which is equal to: Q + W (heat + work)

∆E = Qp - P∆V (II) (we are assuming pressure is constant)

∆H = ∆E + P∆V (I took P as a common factor)

By replacing ∆E from eq. II in eq I:

∆H = Qp - P∆V + P∆V

=> ∆H = Qp

This is derived from the 2nd law of Thermodynamics. I'm not sure if I have cleared your thoughts, be sure to ask again if you didn't understand. The units of ∆H will depend upon the heat capacity you use: specific heat capacity will result in joules per units of mass while molar heat capacity will result in joules per moles.

We haven't done really anything with the laws. We are still in the review stage of thermodynamics from the SL curriculum so we just kind of stop at q=mc deltaT and from that jut find the limiting reactant inside the calorimeter and find the moles of it and then its just q/mol to find kJ per mol.

That's how I worked it out at least, it worked out quite nicely when you look at the data, the trends follow exactly how they should according to everything I've learned lol

I understand how your equations work though, probably make more sense when we go over though. Thanks

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

I found out i was using 4.18kJ/g/C for my specific heat of water, so i fixed all my values to joules because its actual 4.18J/g/C. But now i have things like 37 joules per mol for a delta enthalpy but my lit value is 46kJ/mol, 46000 joules, thats a GIANT error....

Edited by Drake

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

HCl + NaHCO3 ===> NaCl + H2O + CO2

Okay, if you have delta H literature values the calculation would be:

∆H = ∆H products - ∆H reactants

(∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3)

(-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol)

∆H = -50.32kJ/mol

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

HCl + NaHCO3 ===> NaCl + H2O + CO2

Okay, if you have delta H literature values the calculation would be:

∆H = ∆H products - ∆H reactants

(∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3)

(-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol)

∆H = -50.32kJ/mol

K cool i got 46kJ/mol probably just because we have diff values from diff sources.

BUT. my experimental values are like 37 JOULES, with a 96% error...

The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J.....

Edit: I can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right?

Edited by Drake

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

HCl + NaHCO3 ===> NaCl + H2O + CO2

Okay, if you have delta H literature values the calculation would be:

∆H = ∆H products - ∆H reactants

(∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3)

(-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol)

∆H = -50.32kJ/mol

K cool i got 46kJ/mol probably just because we have diff values from diff sources.

BUT. my experimental values are like 37 JOULES, with a 96% error...

The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J.....

Edit: I can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right?

It's not surprising that the results diverge from the real value. There are two factors that are determining. First, the literature values for delta H you're using are for the reaction at STP (this is not true under lab conditions). Second, the reactions are done inside a calorimeter which prevents most (if not all) of the heat loss to the surroundings. This a MAJOR error of the experiment since despite the accuracy of the method you're using, there will always be heat loss and so the result for delta H will not be 100% certain.

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

HCl + NaHCO3 ===> NaCl + H2O + CO2

Okay, if you have delta H literature values the calculation would be:

∆H = ∆H products - ∆H reactants

(∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3)

(-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol)

∆H = -50.32kJ/mol

K cool i got 46kJ/mol probably just because we have diff values from diff sources.

BUT. my experimental values are like 37 JOULES, with a 96% error...

The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J.....

Edit: I can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right?

It's not surprising that the results diverge from the real value. There are two factors that are determining. First, the literature values for delta H you're using are for the reaction at STP (this is not true under lab conditions). Second, the reactions are done inside a calorimeter which prevents most (if not all) of the heat loss to the surroundings. This a MAJOR error of the experiment since despite the accuracy of the method you're using, there will always be heat loss and so the result for delta H will not be 100% certain.

But a 96% error? I mentioned how the calorimeter could be a terrible insulator and let energy out of the system. I also mentioned that the CO2 could also escape through the lid and take energy with it. I've also mentioned that for my last 2 x values, the mass i used includes excess NaHCO3 and not just the solution so its q value is actually kind of high.

Oh wait...ok I see what you mean, my reaction is releasing energy (endothermic right?) and the calorimeters insulation is preventing it from releasing and its going back into the solution thus skewing my temperature readings? So scratch the bad insulator thing, my calorimeter's insulation being good was actually an error? o.O

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Ugh, K, I found all my own experimental values by finding q based off all my data, then dividing it by my limiting reactant's moles so i can get a kJ/mol number. And now i found these literature values for the delta enthalpy of my NaHCO3, HCl, H2O, CO2, and NaCl. Now I THINK I'm supposed to take the enthalpies of my reactants, add them together, then subtract the sum of the enthalpies of my products from it and that should be kJ/mol yes?

It just doesn't feel right because what i got was like 47kJ/mol but what i got was well over 1000 kJ/mol. And MY enthalpy values actually go up and i dont see how my lit values would move =/

HCl + NaHCO3 ===> NaCl + H2O + CO2

Okay, if you have delta H literature values the calculation would be:

∆H = ∆H products - ∆H reactants

(∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3)

(-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol)

∆H = -50.32kJ/mol

K cool i got 46kJ/mol probably just because we have diff values from diff sources.

BUT. my experimental values are like 37 JOULES, with a 96% error...

The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J.....

Edit: I can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right?

It's not surprising that the results diverge from the real value. There are two factors that are determining. First, the literature values for delta H you're using are for the reaction at STP (this is not true under lab conditions). Second, the reactions are done inside a calorimeter which prevents most (if not all) of the heat loss to the surroundings. This a MAJOR error of the experiment since despite the accuracy of the method you're using, there will always be heat loss and so the result for delta H will not be 100% certain.

But a 96% error? I mentioned how the calorimeter could be a terrible insulator and let energy out of the system. I also mentioned that the CO2 could also escape through the lid and take energy with it. I've also mentioned that for my last 2 x values, the mass i used includes excess NaHCO3 and not just the solution so its q value is actually kind of high.

Oh wait...ok I see what you mean, my reaction is releasing energy (endothermic right?) and the calorimeters insulation is preventing it from releasing and its going back into the solution thus skewing my temperature readings? So scratch the bad insulator thing, my calorimeter's insulation being good was actually an error? o.O

I'm not sure about what method you used. What I mean is the literature value you are comparing your result to is defined at STP (standard conditions of temperature and pressure - 273 K and 1 ATM) which are not the conditions you are under in the school's lab. This is one of the reasons for it being different.

According to your literature value, the reaction is endothermic (∆H > 0 = endothermic) meaning that is absorbs energy from the surroundings but there isn't much energy to absorb because the calorimeter acts as an insulator and the reaction is (theoretically) not in contact with the surroundings.

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Ok, but the temperature of my solution went down, meaning it released energy not absorbed it o.O

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Ok, but the temperature of my solution went down, meaning it released energy not absorbed it o.O

Then you should use my delta H result which is negative, meaning that the reaction is exothermic. If you have to state where you took the values from:

Chang, Raymond (2007). Chemistry. China: McGraw-Hill. Appendix 3.

Edited by Hedron123

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Ok, but the temperature of my solution went down, meaning it released energy not absorbed it o.O

Then you should use my delta H result which is negative, meaning that the reaction is exothermic. If you have to state where you took the values from:

Chang, Raymond (2007). Chemistry. China: McGraw-Hill. Appendix 3.

Too late now cause it's turned in but, all my calculations gave me positive results, endothermic, yet the temp of solution went down. This is kinda where it all blew up in my mind =/ Thanks for the help though.

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Hey, I was just wondering what and how exactly Titrations work, and if there are any sample problems/notes on this topic, as unfortunately I did not understand my teacher's explanation.

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Hey, I was just wondering what and how exactly Titrations work, and if there are any sample problems/notes on this topic, as unfortunately I did not understand my teacher's explanation.

Ok so with a titration the main goal is to discern the unknown concentration of a liquid with a known volume, using a liquid of known volume and concentration.

The basic method is to use a large titration set.

For example:

Lets say you have NaOH of twenty ml (unknown conc.) and HCl of, lets say, 0.1molar.

You know the reaction equation.

So you use a titration funnel to, drop by drop, set the HCl into the NaOH. Once the NaOH (which is naturally purple) turns clear (meaning that it is neutral) the reaction has occured. This can be aided by indicator solutions.

Thus, you can measure the amount of HCl needed for the reaction, and then you can use stoichiometry to discern the unknown concentration.

Sorry that was a round about way of discussing it-I hope I helped

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just to add to centre field's very good explanation, and continuing to use the example of adding HCl to NaOH:

Just to clarify, most of the complicated stuff your teacher was talking about was probably due to rinsing equipment, using pipettes correctly etc. This is just to be precise. In essence, a titration is: I have X amount of NaOH. I can find out exactly how much by seeing how much HCl I need to neutralise it (in this case, if I used 0.5 mol of HCl to get it to pH 7, that means there must have been 0.5 mol of NaOH present).

And NaOH is naturally clear. You will choose an indicator, that, in this case, is e.g. purple when the solution is basic (i.e. when you haven't yet added enough HCl to neutralise all the NaOH so you've still got unreacted NaOH), but will turn clear when the solution reaches pH 7. So you'll have been adding the HCl VERY slowly, drop by drop, and at the precise moment it changes, you look at the burette to see how much HCl was needed. As you know the concentration of the HCl, and now you know what volume you needed to add, you know precisely how many mol of HCl you used, and therefore how many mol of NaOH were there.

If you need equivalence points explained, here's my attempt:

The equivalence point is when the reaction is finished: all the NaOH has reacted with all the HCl so there is none of either left (only the salt they produce and water. Actually that's not true. It's an equilibrium reaction, so the reaction doesn't go to completion. However, there will come a point when most of the NaOH has reacted, and there is an equal amount of unreacted HCl and NaOH present, which will mean you still get pH 7). However, often the indicator only changes once you get just past that point: once the solution actually becomes acidic. This is called the end-point. However, the idea is you go just beyond pH 7, i.e. making the end-point as close as possible to the actual equivalence point, so you get a solution that is just acidic in order to just get the colour change (the end-point) and then stop. So it's so close it's practically the real equivalence point (ok, sorry for the repetition and italics )

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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.

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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

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