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AQA Chemistry Unit 2 Help

Hey been struggling with a few questions not knowing how the answers have been arrived at, if any of you could please help, please do!
heres the first Q!
from a june 2012 paper

A student carried out an experiment to determine the enthalpy change when a sample
of methanol was burned.
The student found that the temperature of 140 g of water increased by 7.5o
C when
0.011 mol of methanol was burned in air and the heat produced was used to warm the
water.
Use the student’s results to calculate a value, in kJ mol–1, for the enthalpy change when
one mole of methanol was burned.
(The specific heat capacity of water is 4.18 J K–1 g
–1).
Reply 1
Original post by SannahKhan
Hey been struggling with a few questions not knowing how the answers have been arrived at, if any of you could please help, please do!
heres the first Q!
from a june 2012 paper

A student carried out an experiment to determine the enthalpy change when a sample
of methanol was burned.
The student found that the temperature of 140 g of water increased by 7.5o
C when
0.011 mol of methanol was burned in air and the heat produced was used to warm the
water.
Use the student’s results to calculate a value, in kJ mol–1, for the enthalpy change when
one mole of methanol was burned.
(The specific heat capacity of water is 4.18 J K–1 g
–1).


a) Aim - to determine enthalpy change for burning 1 mol of methanol

b) calculation, heat change = m * c * delta temperature (make sure units cancel out so that you have heat change in J)

then how many moles of methanol is used, hence how can you get the value of heat change/enthalpy change for 1 mol of methanol burned?
Reply 2
Original post by shengoc
a) Aim - to determine enthalpy change for burning 1 mol of methanol

b) calculation, heat change = m * c * delta temperature (make sure units cancel out so that you have heat change in J)

then how many moles of methanol is used, hence how can you get the value of heat change/enthalpy change for 1 mol of methanol burned?


Thanks for that, yeah i tried that in the beginning.. let me paste the mark scheme
"M1 q = m c ΔT OR q =140 x 4.18 x 7.5
M2 = 4389 (J) OR 4.389 (kJ) OR 4.39 (kJ) OR 4.4 (kJ)
(also scores M1)
M3 Using 0.0110 mol
therefore ΔH = 399 (kJmol-1
)
OR 400"

thats the mark scheme, i done the q=mcT bit, and got the answer of 4389, but that moles bit is a little confusing, how did they reach to -399 / -400 ?
Original post by SannahKhan
Thanks for that, yeah i tried that in the beginning.. let me paste the mark scheme
"M1 q = m c ΔT OR q =140 x 4.18 x 7.5
M2 = 4389 (J) OR 4.389 (kJ) OR 4.39 (kJ) OR 4.4 (kJ)
(also scores M1)
M3 Using 0.0110 mol
therefore ΔH = 399 (kJmol-1
)
OR 400"

thats the mark scheme, i done the q=mcT bit, and got the answer of 4389, but that moles bit is a little confusing, how did they reach to -399 / -400 ?


Moles = q/0.0110 = -4389/0.0110 = -399000J mol-¹

Then you divide by 1000 to give the answer in kilojoules so -399000/1000 = -399Kj mol-¹

(you change the sign to a minus because its an exothermic reaction - COMBUSTION is exothermic)

Hope that helps! :smile:

Posted from TSR Mobile
(edited 10 years ago)
Reply 4
Original post by SannahKhan
Thanks for that, yeah i tried that in the beginning.. let me paste the mark scheme
"M1 q = m c ΔT OR q =140 x 4.18 x 7.5
M2 = 4389 (J) OR 4.389 (kJ) OR 4.39 (kJ) OR 4.4 (kJ)
(also scores M1)
M3 Using 0.0110 mol
therefore ΔH = 399 (kJmol-1
)
OR 400"

thats the mark scheme, i done the q=mcT bit, and got the answer of 4389, but that moles bit is a little confusing, how did they reach to -399 / -400 ?


sometimes, it is not about just following the mark scheme. the mark scheme is very simplified to suit markers' taste, not you as a student who should work it out with what you know and then cross-check with the method and/or answer on mark scheme (it is not the same as model answer, it is a mark scheme)
Reply 5
Original post by namelocscott
Moles = q/0.0110 = -4389/0.0110 = -399000J mol-¹

Then you divide by 1000 to give the answer in kilojoules so -399000/1000 = -399Kj mol-¹

(you change the sign to a minus because its an exothermic reaction - COMBUSTION is exothermic)

Hope that helps! :smile:

Posted from TSR Mobile

thanks for the help you're a life saver!

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