You are Here: Home

# enthalpy change working out question...confused? Tweet

Chemistry discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. enthalpy change working out question...confused?
POST DELETED...sorry
Last edited by krazzyraji; 08-09-2012 at 17:06.
2. Re: enthalpy change working out question...confused?
(Original post by krazzyraji)
This is where i've got so far:
q=m.c ΔT
q=75 x 4.18 x 8.5= 2664.75/1000 = 2.66
n=m/mr:.......confused on here onwards (wat would the mass be??)

thanks
You don't need to worry about the mass of each reactant used -- since you know the concentration and volume used of each, you can just use n = V.(conc) to work out how many moles there were.
3. Re: enthalpy change working out question...confused?
(Original post by BJack)
You don't need to worry about the mass of each reactant used -- since you know the concentration and volume used of each, you can just use n = V.(conc) to work out how many moles there were.
would it be:
2.00 x what??
4. Re: enthalpy change working out question...confused?
(Original post by BJack)
You don't need to worry about the mass of each reactant used -- since you know the concentration and volume used of each, you can just use n = V.(conc) to work out how many moles there were.
as in 2.00 x 25 or 50...how would you know which one?
5. Re: enthalpy change working out question...confused?
ok so you've worked out your q, and you've got it in kilojoules. Now to work out the mass, you know your density of the final solution to be 1gcm^-3
Density=mass/volume so rearrange that to make mass= density x volume (volume has to be the volume of the final solution which is 25 +50)

Now you also know that the reactants by the equation so you get the Mr of that from the values on your periodic table. Then you can work out the amount of moles total.

Now you have q and moles. Use the equation enthalpy change= q/moles and thats your final answer.

HOWEVER:

Because the reaction is exothermic you have to put your final answer as a negative value, otherwise you'll lose most of the marks for the question.

Hope that makes sense
Last edited by m1a1tank; 04-02-2012 at 16:41. Reason: Made a mistake lol
6. Re: enthalpy change working out question...confused?
(Original post by m1a1tank)
ok so you've worked out your q, and you've got it in kilojoules. Now to work out the mass, you know your density of the final solution to be 1gcm^-3
Density=mass/volume so rearrange that to make mass= density x volume (volume has to be the volume of the final solution which is 25 +50)

Now you also know that the reactants by the equation so you get the Mr of that from the values on your periodic table. Then you can work out the amount of moles total.

Now you have q and moles. Use the equation enthalpy change= q/moles and thats your final answer.

HOWEVER:

Because the reaction is exothermic you have to put your final answer as a negative value, otherwise you'll lose most of the marks for the question.

Hope that makes sense
Thanks a lot...so would the answer be 2.00 x (75/1000) = 0.15
ΔH = q/n = 2.66/0.15 = 4.83...??
7. Re: enthalpy change working out question...confused?
i haven't done the question. but your final answer, make sure to change it to a negative because the reaction is exothermic. do you know why its exothermic or would you like me to explain it?
8. Re: enthalpy change working out question...confused?
(Original post by m1a1tank)
i haven't done the question. but your final answer, make sure to change it to a negative because the reaction is exothermic. do you know why its exothermic or would you like me to explain it?
ok,,,,i get the exothermic bit and answer would be 17.7..i did wrong calc above...
but thanks alot, i get it now
9. Re: enthalpy change working out question...confused?
ok kl. thats good then just keep on doing loads of practice questions on it so you can do it with anything that they give you.
10. Re: enthalpy change working out question...confused?
(Original post by krazzyraji)
kk, also would you know how to answer the question below by any chance:

"Look up the standard enthalpy change of combustion figures (ΔH^[circle with line in middle]) for all three alcohols and compare them with your calculations. Are your results accurate? If not, why not?

I dont get this question above...

previously just before this question, we had to answer the enthalpy change for each, which are as follows:
-ethanol: -350.8 kjmol-1
-propanol: -716.6 kjmol-1
-hexanol: -1550.2 kjmol-1

So what do we do next?? I dont get what it means by the combustion figures?

Thanks
It means that you must look up in a data book the standard enthalpy of combustion values for the same alcohols and COMPARE the values with the values that you obtained by experiment (or calculation)
11. Re: enthalpy change working out question...confused?
^^imagine that!!!