2004 - Written Paper - Q8

Posted 25 May 2006 - 03:07 PM

Hi,
could someone please explain to me how to get the answer to Q8, the enthalpy of neutralisation question, i got the answer to come out as 65 which is wrong !

Michael

Posted 25 May 2006 - 03:37 PM

The Enthalpy of neutralisation is the energy required to produce one mole of water.

NaOH + HCl ---> NaCl + H2O
1mole NaOH ----> 1mole H2O

E=cmΔT
=4.18x0.04x6.5 temp change = 25.5 - average of the other solns=6.5
=1.0868

n=cv
n=1x0.02
n=0.02 moles

0.02moles ------->1.0868
1 mole -----------> -54.34 kjmol-1

Nathan

Posted 25 May 2006 - 03:42 PM

NaOH + HCl ---> NaCl + H2O

*remember that enthalpy of neutralisation is the energy required to produce 1 mole of water*

moles of NaOH = 0.020 x 1 = 0.02 moles
moles of HCl = 0.020 x 1 = 0.02 moles

avg start temp = (18+20)/2 = 19
H = 25.5-19 = 6.5

H = cm T
= -4.18 x 0.04 x 6.5
= -1.0868 for 0.02 moles water

1/0.02 = 50 multiply H by 50

H = -1.0868 x 50
= -54.34 kj mol-1

(rounded to -54 in answer scheme)