## anonymous 4 years ago 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.

1. anonymous

Vt=280mL m=280g $DeltaH=-53.1kJ$ n=0.12mol Ti=(Q/mc) -Tf This is all I got so far.

2. anonymous

LiOH+HNO3-------> LiNO3 + H2O+ 53.1kJ

3. Rogue

Is that the whole question? It seems a bit vague. :( How did you get the delta H by the way?

4. anonymous

Yeah that is the question. This equation was given LiOH+HNO3-------> LiNO3 + H2O+ 53.1kJ.

5. Rogue

Oh, okay, how about the mass 280 g?

6. anonymous

so i calculated the total volume and then i used the density formula to find the total mass.

7. Rogue

Hmm, I feel like the problem is a bit more complicated than this, but why don't you just try substituting what you?$T _{i}^{} = \frac {Q}{mc} - T _{f}^{} = \frac {-53 kJ}{280 g * 0.00419 J/gC} - 24.5 C$

8. Rogue

What you have*

9. Rogue

0.00419 kJ/gC*

10. anonymous

my teach told me to use deltaHr =deltaH/n

11. anonymous

so I first found the n=mv=1.00mol/L * 0.12L=0.12mol

12. Rogue

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

13. anonymous

but it is in kJ not kJ/mol :S I'm confused.

14. Rogue

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.

15. Rogue

Hmm, am I explaining that or just repeating? Don't really know how to say it...

16. anonymous

Oh okay, now it makes sense :) Thank you soooo much. Let me calculate the final answer and show it to you.

17. anonymous

i got -19.1C

18. Rogue

The Q should be negative right?

19. anonymous

I would guess so, since energy is released. right? then the answer would be -29.9

20. Rogue

I really don't like this question :( It'd be great if someone else could also do this question to verify the answer.