Question

Hi, I have a lab question and I'm kind of confused on how to approach it.

Evaluation of molar enthalpy of dissociation of formic acid, deltaHdiss.

Given:
H1 = -58010 J/mol; H2 = -59758.4J/mol

They tell use to use Hess's Law to determine the enthalpy of dissociation

Thanks!

Answer 1

Hess's law states that the heat change in a chemical reaction that is carried out at a constant pressure is the same whether it occurs one one or several steps.

so

\[\Delta H_1 = \Delta H_2 + \Delta H_3 + \Delta H_4\]

Answer 2

so if the dissociation of formic acid is a two step process you simply just need to add the enthalpies to figure out the total enthalpy of the reaction

Answer 3

Ohhhh.. okay, makes sense as I did add both of the equations that they gave us which is
1. H3O+ + OH- > 2H2O H1= -58010 J/mol
2. HCOOH + OH- > H2O + HCOO- H2 = -59758 J/mol

2. HCOOH + OH- > H2O + HCOO- H2 = -59758 J/mol
1. [x-1] = 2H2O > OH + H3O H1 = 580101 J/mol
--------------------------------------------------------
HCOOH + H2O > H3O + HCOO H= -1748 J/mol

Is that correct to find the molar enthalpy of dissociation of formic acid?

Answer 4

Cause my friend did a totally difference process and I just wanted to make sure if this was correct. Thanks!

Answer 5

yeah that seems correct

Answer 6

Here is an example to reassure you of its correctitude

Answer 7

Thank you very much!!!!!

Answer 8

I'm sorry I couldn't explain it better I'm still learning this stuff to be honest and my brain feels like it is bleeding atm ha

Answer 9

No worries! thanks again and hope you feel better!