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Chemistry Help - Application of Hess's Law [entries|archive|friends|userinfo]
Chemistry Help

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Application of Hess's Law [Oct. 12th, 2010|07:49 pm]
Chemistry Help
Sorry if this posted twice or something weird happened, my internet crashed at the most inopportune moment!

One of my homework questions is:

"Calcium metal will react in water to form calcium hydroxide, Ca(OH)2. Use the thermochemical data that follow and Hess's Law to calculate the value of ΔH° for the reaction

Ca(s) + 2H2O(l) = Ca(OH)2(s) + H2 (g)

a. H2(g) + ½O2(g)= H2O(l) ΔH°= -286kJ
b. CaO(s) +H2O (l) = Ca(OH)2(s) ΔH° = -64kJ
c. Ca(s) + ½O2(g)= CaO(s) ΔH°= -635kJ "

I know to flip line a and change the sign to a + 286 kJ. I was wondering why I didn't have to multiply line b by 2 because in the reaction H2O has a coefficient of 2.

Any help will be appreciated! I seem to be incompetent when it comes to Hess's law.

[User Picture]From: uberjason
2010-10-13 12:07 am (UTC)
It's because you ultimately add up all three equations to get the overall equation. When you flipped line a, you put H2O on the left. Line b also has H2O on the left. Line c has no H2O in it, so in total (adding all three equations together) you have 2 H2O on the left - just as in the overall reaction equation.
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From: jinx24
2010-10-13 11:10 am (UTC)
ah, yes, thank you so much :)!
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[User Picture]From: felixdiecat
2010-10-26 08:25 pm (UTC)
After flipping line a, you'd have another H20 on the left sift of the equation, so you'd have two total. Thus, no need to multiply. :D
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