How Do You Calculate the Molar Enthalpy of Formic Acid?

[JameB]

Homework Statement

I'm doing report for a Thermochemistry lab and I need to calculate the molar enthalpy of formic acid, ΔH°₂.

Two reactions are taking place

1) H₃O + OH <-> 2H₂O with ΔH°₁ = -58400J/mol and some q1
2) HCOOH + OH <-> H₂O + HCOO and we need to calculate ΔH°₂.

I also know the pH is 1.96 before the neutralization took place.

Homework Equations

q = nΔH°
q_system = q₁ + q₂ = -q_surroundings = -(C_calΔT_cal + C_vV_acidΔT_acid +C_vV_baseΔT_base)...[2]

The Attempt at a Solution

So my general approach is:::

First calcualate what q₁ is because I know the pH.

[H3O] = 10^-1.96
[H3O] = 0.011 M
=> moles of H3O = Vacid*[H3O] = 0.05L * 0.011M = 0.00055 mol H3O {Is this right?}

Then, q = nΔH° = 0.00055mol * 58400 J/mol = 32.12 {does this make sense?}

Now I calculate q₂ using equation [2]
Ccal = 76 J/°C
ΔTcal = 6.2°C

Cv = 4.16
Vacid = 50mL
ΔTacid = 6.2°C

Vbase = 50mL
ΔTbase = 5.2°C

q₂ = 2842.4J {correct?} Now, ΔH°₂ = q₂/moles of HCOOHHow do I find the moles of HCOOH? I'm told that "some of the formic acid dissociated and some of the OH reacted with H3O, so you need to determine the mole sof HCOOH left over after the reaction of H3O and OH"

How do I do that? and am I right so far?

 

[Borek]

It won't hurt if you will tell us what you really did, what are q₁ and q₂, and what are you expected to determine. We don't know how to read your mind, and what you wrote so far is not enough, it is chaotic and leaves a lot ambiguity. Some things you did are wrong for sure, no matter what you were trying to calculate.

 

[JameB]

Ignore that whole question, I have a whole new question:

if you have a galvanic cell set up like:

Pt | 0.1M Fe³⁺, 0.1M Fe²⁺ || 0.1M AgNO₃ | Ag

What does it mean to write a balanced cell reaction equation and how do I do it?

I know that the + electrode was Ag_(s) so that means that the cathode was Ag meaning Ag was reduced from Ag²⁺. So is the answer Pt_(s) + Ag²⁺_(aq) -> Pt²⁺_(aq) + Ag_(s)?

 

[Borek]

Please don't pot new questions in old threads, start a new one.

Pt is just an electrode material, it doesn't take part in the reaction.