Question

A 65.0 g sample of 0.600 M HI at 18.46 degrees celsius is added to 84.0 g of solution containing excess KOH, also at 18.46 degrees celsius. The final temperature is 21.96 degrees celsius. Calculate the enthalpy of H for the reaction.

Assume the specific heat of the solution is 4.184 J/g(degrees celsius), and the calorimeter has negligible heat capacity .

HI(aq) + KOH(aq) --> KI(aq) + H2O(l)

enthalpy of H ?

Answer 1

You have 65.0 g soln from HI and 84.0 g H2O with the KOH so total mass soln is 149 g.
q = mass H2O x specific heat water x (delta T).

USUALLY delta H for a reaction is done in kJ/mol. This problem has no easy way to get to moles but q/mol = J/mol and you can convert that to kJ/mol if needed. If you assume the density of the HI solution is 1.00 g/mL, that can be used to solve for mols HI; e.g., 65 mL of 0.600 M HI would be moles = M x L = 0.600 x 0.065 = ? and you can use that for mols. I doubt that the density is 1. g/mL.

Answer 2

Awesome

I am defiantly starting to understand this stuff better.

.065L x (.600 mol/L) = 0.0390 mol

q=MC(delta T)
= 149g x 4.184 J/g(C) x 3.5 celsius
= 2181.9 J
=2.2 kJ

Then I did
2.2 kJ / 0.0390 mol
= 56 kJ/mol

for my final answer I got 56 kJ/mol