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# Edexcel Chem 3 question watch

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1. The question with heat of neutralisation of NAHCO3.....

5g of NAHCO3......... added to 50cm3 of acid....
when calculating the mass.....do we take it as 50g (density 1g/cm3).... or as 55g (50g +5g of NAHCO3)?
2. omg good question. I used 50cm^3.

Didn't we find out it was 5.0g after that question?
3. its meant to be 55. i just checked it out.
4. Evem i used 50.... but i was just wonderin.. in the textbooks...the experiments all say something like.... 100 cm3 of NAOH to 100cm3 of HCL...... mass becomes 200g...... how much naoh or hcl is dissolved in the water is not considered... even though we know the concentration and volume.....we can find the no. of moles and therefore the mass.......and we can add that to 200..... but its not done...
confusing?!?!
5. I wrote in the assumption that the volume of the solution does not change when the solid is added. So can the mass be assumed to be 50g?
6. It seems that the temperature decreases from 22 deg to 15.5 deg and so it is endothermic? But it is a neutralisation reaction...

I have got another probelm. Look at the equation:
HCl + NaHCO3 ---> NaCl + H2O + CO2

CO2 is evolved and there is a constant mass loss. Also, the temperatures of CO2 molecules when they leave the solution are unknown, so the heat carried away by CO2 is not known. It depends on the instantaneous reaction rate and the thermal conductivity of CO2. So the effective mass in the calculation of heat change lies between 50g and 55g.
7. (Original post by Hash)
its meant to be 55. i just checked it out.
wicked- thats what i did.. but then when u had to work out delta H of neut for it in KJmol-1 what number of moles needed to be used in order to scale up???
i did my answer x 1/.12
but i got an answer quite low, around 10-20 KJmol-1, cant rememberr exactly what. can anyone shed some light?
8. (Original post by misty)
wicked- thats what i did.. but then when u had to work out delta H of neut for it in KJmol-1 what number of moles needed to be used in order to scale up???
i did my answer x 1/.12
but i got an answer quite low, around 10-20 KJmol-1, cant rememberr exactly what. can anyone shed some light?
it should be positive

no of moles = 5/molar mass of NaHCO3
9. um for all those of u who thinks its 55 it is 50 i think cos its the mass of water which provides the energy
10. i got +22.9 kj/mol (using 50 grams)......... but i was stupid enough to put a -ve sign infront of it... thinkin that heat of neutralisation is always exothermic.lol.......

moles of nahco3 = 5/84 =0.0595238 moles

heat = (50)(4.2)(22-15.5)=1365J

heat of neutralisation = 1365/0.0595238 = 22932J = +22.9 KJ/mol
11. i got 22.x, positive
12. (Original post by [_Z_])
i got +22.9 kj/mol (using 50 grams)......... but i was stupid enough to put a -ve sign infront of it... thinkin that heat of neutralisation is always exothermic.lol.......

moles of nahco3 = 5/84 =0.0595238 moles

heat = (50)(4.2)(22-15.5)=1365J

heat of neutralisation = 1365/0.0595238 = 22932J = +22.9 KJ/mol
woohoo, totally agree
13. y is it +ve?
14. (Original post by [_Z_])
Evem i used 50.... but i was just wonderin.. in the textbooks...the experiments all say something like.... 100 cm3 of NAOH to 100cm3 of HCL...... mass becomes 200g...... how much naoh or hcl is dissolved in the water is not considered... even though we know the concentration and volume.....we can find the no. of moles and therefore the mass.......and we can add that to 200..... but its not done...
confusing?!?!

Thats true, coz the liquid is absorbing the heat, so its the mass of the liquid which concerns us, i.e. 50 grams...
15. (Original post by misty)
y is it +ve?
oops nm!
16. (Original post by misty)
y is it +ve?
actually it is not a neutralisation which is defined as:
H+ + OH- --> H2O

In this case, CO2 is produced. It absorbs the heat and carries it away, just as those in coke do so as to make it cooler.
17. how did you find paper 3 as a whole? i thought it was pretty hard compared to unit 1 and 2

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