a neutralization reaction occurs when 120ml of 1.00mol/l lioh and 160nl of 0.75 mol/l hno3 are mixed in an insulated cup. initially, the solutiona are at the same temperature. If the highest temp during mixing was 24.5C, what was the iniyial temp of the solutions? The density is 1.00g/ml and the specific heat capacity is 4.19J/gC.
Vt=280mL m=280g \[DeltaH=-53.1kJ\] n=0.12mol Ti=(Q/mc) -Tf This is all I got so far.
LiOH+HNO3-------> LiNO3 + H2O+ 53.1kJ
Is that the whole question? It seems a bit vague. :( How did you get the delta H by the way?
Yeah that is the question. This equation was given LiOH+HNO3-------> LiNO3 + H2O+ 53.1kJ.
Oh, okay, how about the mass 280 g?
so i calculated the total volume and then i used the density formula to find the total mass.
Hmm, I feel like the problem is a bit more complicated than this, but why don't you just try substituting what you?\[T _{i}^{} = \frac {Q}{mc} - T _{f}^{} = \frac {-53 kJ}{280 g * 0.00419 J/gC} - 24.5 C\]
What you have*
0.00419 kJ/gC*
my teach told me to use deltaHr =deltaH/n
so I first found the n=mv=1.00mol/L * 0.12L=0.12mol
Ah, right, 53.0 kJ of energy is released for every 1 mole of LiNO3 formed. So the actual delta H for the sample given would be 53.0 kJ/mol * 0.12 mol = 6.36 kJ
but it is in kJ not kJ/mol :S I'm confused.
Well, according to the reaction, when 1 mole is formed when 53.0 kJ is released. So the delta H rxn = 53.0 kJ/mol.
Hmm, am I explaining that or just repeating? Don't really know how to say it...
Oh okay, now it makes sense :) Thank you soooo much. Let me calculate the final answer and show it to you.
i got -19.1C
The Q should be negative right?
I would guess so, since energy is released. right? then the answer would be -29.9
I really don't like this question :( It'd be great if someone else could also do this question to verify the answer.
Join our real-time social learning platform and learn together with your friends!