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April 20, 2014

April 20, 2014

Posted by **Rose Bud** on Thursday, February 16, 2012 at 1:06am.

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 ?

- Chemistry-Thermochemistry (grade 12) -
**DrBob222**, Thursday, February 16, 2012 at 1:36amYou 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.

- Chemistry-Thermochemistry (grade 12) -
**Rose Bud**, Thursday, February 16, 2012 at 2:00amAwesome

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

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