Data obtained from lab (neutralization):
H2SO4:
C =1.0mol/l
V = 30 mL
Initial temp. = 24.1
NaOH
c= 1.0mol/L
v= 50 mL
initial temp. = 23.9
Final temp. of H2SO4 & NaOH = 32.9
Assume specific heat capacity of both these substance is equal to water's = 4.19
Calculate the molar heat of reaction for the NaOH and find the percentage difference of the value you obtained from your experiment with the widely accepted value for the molar heat of neutralization of sodium hydroxide, -57 kJ/mol
My solution:
nH=mc(∆t)
H = (0.03 + 0.05 kg) x 4.19 KJ/ kg c x ( 32.9) - (23.9 + 24.1 /2) / 0.05 mol
H = 60 Kj/mol
% difference = -57 - (60) / -57 x 100
i get -205 % ..... is that correct?? can % difference be negative and more than 100??
Yes, you may get negative numbers. Yes, the error can be more than 100%; however, I think you have made an assumption that isn't true AND I think there is an error in your percent calculation. I think your answer is a lot closer than that.
Three problems: Please check my reasoning carefully. The easy two first.
1. This is an exothermic reaction; therefore, your answer of 60 should be -60(but see #3 below).
2. The percent error is (experimental-accepted/accepted)*100 = [-57-(-60)/-57] = about +5% but you can do it exactly after looking at #3 below.
3. I don't think your "averaging of the two different temperatures is correct. You have averaged the 24.1 and the 23.9 as 24.0 (23.9+24.1)/2 = 48.00/2 = 24.00 BUT the 23.9 degrees is for 50 mL and the 24.1 degrees is for 30 mL; therefore, the average can't be split down the middle since the volumes were not the same. I think what you need to do to calculate the heat of neutralization is
[0.05 x 4.19 x (23.9-23.9)] + [0.03 x 4.19 x (32.9-24.1)] = 2.991 kJ or something like that. You can do it exactly, then divide by 0.05 mol = 59.8 kJ and recalculate the percent error from that. I get about +4.9%.
Again, check my thinking.
The calculation of the molar heat of reaction appears to be correct. However, the percentage difference you have calculated is incorrect.
To calculate the percentage difference, you need to use the formula:
Percentage Difference = (Experimental Value - Accepted Value) / Accepted Value x 100
In this case, the experimental value is 60 kJ/mol and the accepted value is -57 kJ/mol.
Percentage Difference = (60 - (-57)) / (-57) x 100
Percentage Difference = 117 / (-57) x 100
Percentage Difference ≈ -205.3%
The negative sign indicates that the experimental value is lower than the accepted value. The magnitude of the percentage difference can exceed 100%, but it is typically reported as a positive value. Therefore, in this case, the percentage difference is approximately 205.3%.
It's worth noting that a negative percentage difference for the molar heat of neutralization suggests that the experimental value obtained is higher than the widely accepted value, but it may indicate some experimental error or systematic deviation.