[ada]

I was wondering if anyone could please help me. I am doing a write up on an experiment of a Bunsen ice calorimeter and have to calculate the standard heat of formation of Mg+2, the value i have calculated is significantly different from the expected value of -462.0 kJ/mol and i feel like i am missing a step in my calculations.
here is the information:
Mg(s) + 2HCl(aq) > MgCl2(aq) + H2(g)
V(HCl) = 0.005L
c(HCl) = 2.00 M
n(Mg) = 0.005 mol

questions are:
1. determine the difference in volume
2. calculate the heat of the reaction
3. calculate the standard heat of formation of Mg+2

EQNs given:
ΔV = V(i) - V(w)
 so my change in volume = 0.36 mL decrease
then calculated the mass of the ice melted/water formed to be 3.98g, this was using m = ΔV (D(i) x D(w) / D(w) - D(i) ) where D(i) is density of ice (0.9168 g/mL) and D(w) was density of water (0.9998 g/mL)

to find enthalpy of reaction they have given the enthalpy fusion of ice (334J/g)

Also need help on how i should calculate errors, is it acceptable to just do it at the end of the calculation by percentage error = (accepted value-experimental value)/ accepted value x 100?

[Dan]

the value i have calculated is significantly different from the expected value of -462.0 kJ/mol and i feel like i am missing a step in my calculations.

Please show your calculations.

[ada]

ΔV = Vi - Vw = 0.47 - 0.11 = 0.36mL decrease
ΔV = m (1/Di - 1/Dw)
m = ΔV (Di.Dw / Dw - Di)
where Di - density of ice is 0.9168g/mL and Dw - density of water is 0.9998g/mL
m = 0.36 x (0.9168.0.9998 / 0.9998 - 0.9168) = 3.98 g of ice melted/water formed
then ..
ΔrxnH = mL
ΔrxnH = ΔV(Di.Dw / Dw - Di) ΔfusH[H2O(s)]    = 3.98 x 334j/g = 1327.88J/g
then standard heat of formation
ΔfH° = ΔrxnH/n = 1327.88/0.005       ;because n(Mg) = 0.005mol
= 241 432.61J/mol
convert to kJ = 241.43kJ/mol and because it is exothermic -241.43kJ/mol

thats how i came to my answer, but the literature value is -462.0 kJ/mol

[ada]

appreciate any help thank you

[mjc123]

How did you measure the volume?
Have you calibrated the calorimeter?

[Enthalpy]

What I see in your message from 03:17:39 PM is an enthalpy of reaction. It is a sum and difference of all involved enthalpies of formation, not just Mg²⁺ (aq). At least one other species has reacted too.

As a sidenote, I'd suggest to mention the condition of Mg²⁺, here dissolved in water. If it were a plasma you'd get a completely different value, in an other solvent too (is there any?).

[ada]

The volume was measured from a graduated pipette, at 30 second intervals I would note down the volume in the pipette onto a graph until the rate was less than 0.01mL/min over a period of five minutes. I then worked out the change in volume from the graph I had drawn.

Yes the calorimeter was calibrated

Ok how do I then calculate the standard heat of formation of Mg+2 (aq) on its own?
Thank you, and no there were no other solvents

[mjc123]

The reaction (ignoring the Cl^- spectator ion) is
 Mg(s) + 2H⁺(aq)  Mg²⁺(aq) + H₂(g)
for which ΔH = ΔH_fMg²⁺(aq) - 2ΔH_fH⁺(aq)
Now ΔH°_fH⁺(aq) is assigned a value of 0 by definition (since you can't get absolute values for individual ions). So the above reaction should give you ΔH_fMg²⁺(aq). (Apart from the heat of dilution of 2M HCl, as the standard state is 1M, but I'm guessing that's not high.) Without knowing more about the details of the procedure, I'm at a loss to explain the discrepancy.