Question
I have solved the following question but i am not sure whether it is right or wrong..
Calculate the molar heat of reaction for the NaOH(aq) using the data obtained from this experiment. (this is a neutralization question)
H2so4
Volume= 30ml
Concentration=1.0mol/l
Initial temp. = 29.9
NAOH
Volume= 50ml
C = 1.0 mol/L
Initial temp. = 29
Final temp of the two contents = 32.9
This is my solution:
El = Eg
NaOH H2O
nH = mc∆t
0.05mol x H = 0.08L x 4.19 x 3.9
H = 26.15
The standard molar heat of neutralization of sodium hydroxide,
is -57 kJ/mol.
so shouldn't my calculated value be close to the standard value? Please tell me whether my solution is right or wrong!
Thank you in advance =)
Calculate the molar heat of reaction for the NaOH(aq) using the data obtained from this experiment. (this is a neutralization question)
H2so4
Volume= 30ml
Concentration=1.0mol/l
Initial temp. = 29.9
NAOH
Volume= 50ml
C = 1.0 mol/L
Initial temp. = 29
Final temp of the two contents = 32.9
This is my solution:
El = Eg
NaOH H2O
nH = mc∆t
0.05mol x H = 0.08L x 4.19 x 3.9
H = 26.15
The standard molar heat of neutralization of sodium hydroxide,
is -57 kJ/mol.
so shouldn't my calculated value be close to the standard value? Please tell me whether my solution is right or wrong!
Thank you in advance =)
Answers
DrBob222
1. Yes, I think the value should be closer to the standard listed.
2. I think you should take into account that the temperature of the NaOH and the H2SO4 were not the same initially.
[50 x 4.184 x (32.9-29.0)] + [30 x 4.184 x (32.9-29.9)] = 1,192 joules or 1.192 kJ.
1.192 kJ/0.05 mol = 23.8 kJ/mol but that isn't much improvement.
2. I think you should take into account that the temperature of the NaOH and the H2SO4 were not the same initially.
[50 x 4.184 x (32.9-29.0)] + [30 x 4.184 x (32.9-29.9)] = 1,192 joules or 1.192 kJ.
1.192 kJ/0.05 mol = 23.8 kJ/mol but that isn't much improvement.
Marium
could it be that H needs to be negative??