As Chemistry

phishfood

How would the molarity of the acids affect it?

Reply 3

phishfood

i think you were correct in supposing 1M HCl to give the same enthalpy as 0.5M H2SO4, since full ionisation means that both would give the same [H+]

Reply 4

anchemis

yeah.. as HCl is monobasic... one mole H+ is dessociated...
H2SO4 is diabasic... two moles of H+ is dessociated

Reply 5

results for 3 different acids

question says use your results to calculate the enthalpy change per mole of acid for each neutralisation. Express your values in units of kJ/mol

E =mc*delta theta
HCl - E = (50.5*4.2*6.6)/(25.5*1/1000) = 547896J/mol
For bottom paRT, no of moles of acid = alkali at neutralisation therefore
no of moles of acid = volume*conc = volume of NaOH*conc
= 25.5*1/1000 = 0.0255

HNO3 - E = (48.5*4.2*7.1)/(23.5*1/1000) = 61543J/mol

H2SO4 - E = (51*4.2*8.15)/(26*0.5/1000) = 134287J/mol

HCl 1 molar 54.8kJ/mol
HNO3 1 molar 61.5kJ/mol
H2S04 0.5 molar 134.3 kJ/mol

Are these results okay?

Reply 6

'correct' value is about 57kJmol-1

however as heat is often lost to the surroundings you would always expect a lower value. Other inaccuaracies creep in such as:

solution concentration (unless the acids are standarised they are unlikely to be exactly the molarity stated)
instrumental tolerance (for example a pipette has a tolerance of +/- 0.04ml even when being used 100% correctly (also rather unlikely)
Ease of reaing small changes in temperature (if the thermometer can be read to within 0.2º there is an error of +/- 0.2 in the first reading and +/- 0.2 in the second reading i.e. a total possible error of +/- 0.4. If you are taking a temperature difference of 10ºC then this amounts to a possible 4% error in the thermometer readings alone.

If you have values higher than 57kJ then either the design of your experiment is wrong or the sum of your inaccuracies has taken the reading above 57 - I repeat that values higher than 57 are unlikely simply because of heat losses to the surroundings

In your suggested values sulphuric acid is clearly erroneous.

Reply 7

OK I see - you have calculated for mole of acid used not mole of water formed

Enthalpy of neutralisation definition is per mole of water formed.

If calculating per mole of acid used, then as sulphuric acid has two hydrogen per formula unit it will produce twice the energy.

However your results suggest that the HCl is 2M, HNO3 is 2M and the H2SO4 is 1M

?

Reply 8

out of curiosity did u use a temperature probe to get such 'accurate' temperature readings?

Reply 9

Well, I don't understand because I showed my teacher afterwards and she said that these results were fine.

Reply 10

charco

the HCl is 1 molar, the HNO3 is 1 molar, the H2SO4 is 0.5 molar.

Reply 11

help me! I have to hand in my write up on monday.

Reply 12

I don't think there is anything wrong with my results because i compared them with friends' after practical and they had very similar results.

Reply 13

Widowmaker

I don't think there is anything wrong with my results because i compared them with friends' after practical and they had very similar results.

I'm thinking that perhaps my school made the concentration of H2SO4 wrong.

Reply 14

charco

out of curiosity did u use a temperature probe to get such 'accurate' temperature readings?

I drew a graph and found the temperature distance from this.

Reply 15

Here ae the questions.
d) Use your results to calculate the enthalpy change per mole of acid for each neutralisation. Express your values in units of kJmol^-1.
e) Compare the three values of enthalpy change of neutralisation and comment on the suggestion made in the introduction. The enthalpy of neutralisation of an acid with an alkali does not depend on the particular acid or alkali providing both a strong?
f) As part of your evalaution you will need to assess the sources of error and decide if the results are the same of significantly different.

If it helps, the concentration of NaOH was 1 moldm^-3.

Reply 16

I really think that the H2SO4 is 1 molar, because I really couldn't be 60kJ/mol out!

Reply 17