September 29, 2020, 10:37:50 PM
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### Topic: Why is the answer to this question D? [Enthalpy]  (Read 657 times)

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#### Roddy3

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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.

#### Roddy3

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##### 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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##### 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 O2(g) 2 O3(g)

Divide the related reaction enthalpy by the coefficient "2" and try to calculate your standard reaction enthalpy again...

#### 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

(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