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Topic: Hess's Law - is it always HrxnP-HrxnR?  (Read 1861 times)

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Offline castromatic

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Hess's Law - is it always HrxnP-HrxnR?
« on: July 18, 2018, 11:58:36 AM »
My brain is really trying to make this one work.

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion
of graphite is -393.5 kJ/mol and that of diamond is -395.4 kJ/mol:

C(graphite) + O2{g) -----> C02{g)  DeltaH1 = -393.5 kJ
C(diamond) + O2(g) -----> CO2 (g) DeltaH2 = -395.4 kJ

Calculate DeltaH for the conversion of graphite to diamond
C(graphite) ----> C(diamond)

The answer in the workbook is positive 1.9 kilojoules.

But using Hess's equation:
reverse diamond so it's in the product side
-DeltaH2=395.4

395.4-(-393.5) = 788.9 kiloJoules! what???

if we don't reverse (which is the method I've been using for the 2 practice problems I solved)
-395.4-(-393.5) =-1.9 kilojoules.

How does it become positive?
Do I also need to reverse the first equation?
(395.4-393.5=1.9)
Or is it
-393.5-(-395.4)=1.9?
*delete me*

Also, saw in some manuals that you just add the two (395.4+(-353.5)) = 1.9

But that's not what I was taught!

Can someone educate me lol

Offline mjc123

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Re: Hess's Law - is it always HrxnP-HrxnR?
« Reply #1 on: July 18, 2018, 12:41:03 PM »
Don't try to blindly apply formulas you don't understand. Think about the situation. Draw the Hess's Law cycle for this scenario. What path is equivalent to graphite  :rarrow: diamond?

Be careful about using heats of combustion. You may be familiar with the equation
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
But in the equation for the formation of X, X is the product. In the equation for the combustion of X, X is a reactant. How then would you write ΔHrxn in terms of the enthalpies of combustion?

Offline Enthalpy

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Re: Hess's Law - is it always HrxnP-HrxnR?
« Reply #2 on: July 19, 2018, 05:18:11 AM »
Some reasons to think instead of applying formulas:
  • You are a human, not a software.
  • In 20 years, you'll have forgotten all formulas.
  • Occasionally, conventions differ. This must not fool you.
  • Some data sources contain sign mistakes.
  • If you have 10% sign error probability on 10 individual steps, the result is random.
Here the heats of combustion are nearly identical, so the reactants have nearly the same heat of formation.

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