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Why is the answer to this question D? [Enthalpy]
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Topic: Why is the answer to this question D? [Enthalpy] (Read 1371 times)
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Roddy3
Regular Member
Posts: 10
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Why is the answer to this question D? [Enthalpy]
«
on:
December 11, 2019, 10:21:14 PM »
Here’s a link to the problem:
https://imgur.com/a/aVzm4Gd
Using the enthalpy equation (sum of products minus sum of reactants), I keep getting answer C, -722. But the book says the answer is -994. If I do sum of products *plus* the sum of reactants, I get the right answer. I am confused, so any help would be appreciated.
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Roddy3
Regular Member
Posts: 10
Mole Snacks: +0/-0
Re: Why is the answer to this question D? [Enthalpy]
«
Reply #1 on:
December 11, 2019, 10:33:24 PM »
I might have figured it out, please correct me if I’m wrong:
If it’s asking for just delta H, then I simply add the two energies together.
If it’s asking for delta Hf, then I do what I thought I was supposed to do.
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...
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Re: Why is the answer to this question D? [Enthalpy]
«
Reply #2 on:
December 12, 2019, 02:37:29 AM »
Check the second chemical equation
3 O
2
(g)
2 O
3
(g)
Divide the related reaction enthalpy by the coefficient "2" and try to calculate your standard reaction enthalpy again...
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MNIO
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Re: Why is the answer to this question D? [Enthalpy]
«
Reply #3 on:
December 26, 2019, 05:07:03 PM »
starting with
(1) 1 H2(g) + ½ O2(g) ----> 1 H2O(l) dH = -286 kJ
(2) 3 O2(g) ----> 2 O3(g) dH = +271 kJ
multiplying (1) by 3 to get (3)
inverting (2) to get (4)
(3) 3 H2(g) + 3/2 O2(g) ----> 3 H2O(l) dH = -858 kJ
(4) 2 O3(g) -----> 3 O2(g) dH = -271 kJ
dividing (4) by 2
(3) 3 H2(g) + 3/2 O2(g) ----> 3 H2O(l) dH = -858 kJ
(5) 1 O3(g) -----> 3/2 O2(g) dH = -135.5 kJ
adding (3) + (5)
(6) 3 H2(g) + 3/2 O2(g) + 1 O3(g) ----> 3 H2O(l) + 3/2 O2(g) dH = -994
canceling
(7) 3 H2(g) + 1 O3(g) ----> 3 H2O(l) dH = -994
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Why is the answer to this question D? [Enthalpy]