what would be the difference when calculating the enthalpy change of neutralisation for these two reactions: H2SO4 + 2NaOH ------ Na2SO4 + 2H2O HCl + NaOH ------ NaCl + H2O
hye, i already do this experiment and i got -66.9 degree celcius. what is the justification of the enthalpy change? why the enthalpy change that i get not similar to enthalpy change in the book?
Are enthalpy changes of neutralisation the same for strong acids with different concentrations? e.g. is the enthalpy change of neutralisation the same for 1moldm-3 of HCl as it would be for 0.1moldm-3 HCl?
Hi there. The overall ionic equation is still the same i.e. H+ + OH- -> H2O since the 3H+ from the acid require 3 moles of OH- and make 3 moles of water. Hope that helps :)
Ellie P NaOH and HCL always react in a 1:1 ratio. I think what you're looking for is the enthalpy change when 1 mole of water is produced, so for other acids and bases with different ratios, the amount of H20 produced would be different. So if you're reacting 1 mole H2SO4 with 2 moles of base, you get 2 moles of water. So if you have the moles of acid, so to find enthalpy change you divide your energy change (in kJ) by twice this value for amount of acid (as this is how many moles of water you have), as the ratio of acid to water is 1:2, while the ratio of base to water is 1:1.
Ellie P I nearly put that in, should have. In that case you work out both of the moles and the least will dictate the number of moles of water formed. You use them to scale up. Remember that your total volume is the sum of the two individual vols and your mass will be whatever that is in g, unless the density is not 1 g/cm3. I've seen that once before.
Aim to have equal moles of acid and alkali so there's no limiting reagent to worry about. Make sure the concentrations of your acid and alkali aren't too low otherwise you'll get a very low temp rise
thank you sir, in terms of the molar ratio of acid and water, if its 1:2 then do i divide the joules by /2 to get one mole of product of water?? thank you for the videos really helpful :)
I have a question... if you are given 2 acids. the one that requires the most NaOH to be neutralised is the strongest acid. can you please explain why that is wrong
Thank you for taking the time to do this! Know that your effort is not under-appreciated in the least! =)
Love being taught chemistry by Ross Noble
How do you work out the moles of water if it was not a one to one ratio and the concentrations and stuff were different
Hello, sir. Approximately how many minutes are needed to reach the maximum temperature?
1 hour until my 2024 chem paper 3 chat am i cooked
what would be the difference when calculating the enthalpy change of neutralisation for these two reactions:
H2SO4 + 2NaOH ------ Na2SO4 + 2H2O
HCl + NaOH ------ NaCl + H2O
Where is the 2.5 x 10^-3 mol from?
hye, i already do this experiment and i got -66.9 degree celcius. what is the justification of the enthalpy change? why the enthalpy change that i get not similar to enthalpy change in the book?
Are enthalpy changes of neutralisation the same for strong acids with different concentrations? e.g. is the enthalpy change of neutralisation the same for 1moldm-3 of HCl as it would be for 0.1moldm-3 HCl?
Yup because you always scale to the production of 1 mole of water
How do you work out the moles of water formed when a diprotic acid is used and this is the limiting reagent?
Hi there. The overall ionic equation is still the same i.e. H+ + OH- -> H2O since the 3H+ from the acid require 3 moles of OH- and make 3 moles of water. Hope that helps :)
Oh right makes sense! Thank you :)
what would you do if it wasn't a 1:1 ratio of NaOH and HCl reacting?
Ellie P NaOH and HCL always react in a 1:1 ratio. I think what you're looking for is the enthalpy change when 1 mole of water is produced, so for other acids and bases with different ratios, the amount of H20 produced would be different. So if you're reacting 1 mole H2SO4 with 2 moles of base, you get 2 moles of water. So if you have the moles of acid, so to find enthalpy change you divide your energy change (in kJ) by twice this value for amount of acid (as this is how many moles of water you have), as the ratio of acid to water is 1:2, while the ratio of base to water is 1:1.
TheSteelEcho666 ahh okay, thank you!
Ellie P No worries.
Ellie P I nearly put that in, should have. In that case you work out both of the moles and the least will dictate the number of moles of water formed. You use them to scale up. Remember that your total volume is the sum of the two individual vols and your mass will be whatever that is in g, unless the density is not 1 g/cm3. I've seen that once before.
MaChemGuy so for example if it was 0.020mol and 0.040mol, then you would divide ____kj by 0.020mol to give you the final answer?
sir what would be some tips when you carry out this experiment??
thank you
Aim to have equal moles of acid and alkali so there's no limiting reagent to worry about. Make sure the concentrations of your acid and alkali aren't too low otherwise you'll get a very low temp rise
thank you sir, in terms of the molar ratio of acid and water, if its 1:2
then do i divide the joules by /2 to get one mole of product of water??
thank you for the videos
really helpful :)
I have a question... if you are given 2 acids. the one that requires the most NaOH to be neutralised is the strongest acid. can you please explain why that is wrong
Sounds like you're confusing strength with concentration
+MaChemGuy I don't understand
+MaChemGuy I think I worded it wrong
I mean why do weak acids need more naoh than strong acids to get neutralised
+MaChemGuy btw the question is not related to the video I'm just confused about the question my teacher asked I'm class
+MaChemGuy so I'm guessing what your saying is that it is dependant on the concentration not the strength of the acid?
no
wait a second, why can't it be C(H3)COOH+4(NaOH) = CC(Na4)OO+4(H2O)?