Determination of the enthalpy of combustion of Magnesium using Hess's law


by Sekminara
Tags: combustion, determination, enthalpy, hess, magnesium
Sekminara
Sekminara is offline
#1
Nov23-08, 02:45 AM
P: 2
1. The problem statement, all variables and given/known data

ok so i'm trying to find the change in enthalpy for this reaction:
Mg(s) + (1/2)O2--> MgO(s)

I've done the lab to find the temperatures of certain reactions, these were:
9.70 Celsius for the reaction Mg(s) + 2HCl(aq)--> MgCl2(aq) + H2(l)

7.60 Celsius for the reaction MgO(s) + 2HCl --> MgCl2(aq) + H2O(l) reaction

(and we know the enthalpy of change for the reaction
H2(g) + (1/2)O2(g) --> H2O(l) is 285.8 kJ)


2. Relevant equations

change in enthalpy = mass (100ml/1000ml) * specific heat (4.18) * change in temp (above)


3. The attempt at a solution

using the (above) equation i found that the enthalpy of change for the first reaction was -405 kJ per mol and the second was -318 kJ per mol.

so i built a Hess's law chart and found that [Mg(s) + 2HCl(aq)--> MgCl2(aq) + H2(l)] + [H2(g) + (1/2)O2(g) --> H2] = [MgO(s) + 2HCl --> MgCl2(aq) + H2O(l)] + [Mg(s) + (1/2)O2--> MgO(s)]

so basically the equation look like this: -405 + (-285.8) = -318 + x

therefore x should be -372.8 kJ per mol

but all the answers i've found on the net so far say that the enthalpy of combustion is approximately 602 kJ per mol

help?
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Borek
Borek is offline
#2
Nov23-08, 03:49 AM
Admin
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P: 22,671
-372.8 kJ/mol agrees with your experimental data, so if it doesn't agree with tables - you have to look for experimental error.
Sekminara
Sekminara is offline
#3
Nov23-08, 03:56 AM
P: 2
dang...ouch 38% error

well...i guess thats what you get when you use Styrofoam insulators...


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