Quiz 1 prep- 2014 #9






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Liat Bainvoll 3G
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Joined: Fri Sep 26, 2014 2:02 pm

Quiz 1 prep- 2014 #9

Postby Liat Bainvoll 3G » Mon Jan 26, 2015 6:42 pm

Hi!
The question is asking for the standard deltaG of reaction for the decomposition of mercury (II) oxide @ 298K. It gives the standard delta H of formation and he standard molar entropies. How do I solve this if I do not have the deltaS values? The answer should be +117.1kJ/mol
Thank you!

Olivia_1C
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Joined: Sun Nov 23, 2014 3:00 am

Re: Quiz 1 prep- 2014 #9

Postby Olivia_1C » Mon Jan 26, 2015 7:34 pm

If you flip forward to page 10 in the workbook, you'll find a set a delta S values. They should all be there.

claytonkersh2I
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Re: Quiz 1 prep- 2014 #9

Postby claytonkersh2I » Mon Jan 26, 2015 8:00 pm

I'm still stuck on this question. OP, did you get the answer?

Liat Bainvoll 3G
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Joined: Fri Sep 26, 2014 2:02 pm

Re: Quiz 1 prep- 2014 #9

Postby Liat Bainvoll 3G » Mon Jan 26, 2015 8:33 pm

but those are all standard molar S values like is given in the problem, don't we need delta S?

Niharika Reddy 1D
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Joined: Fri Sep 26, 2014 2:02 pm

Re: Quiz 1 prep- 2014 #9

Postby Niharika Reddy 1D » Mon Jan 26, 2015 10:36 pm

To solve for ΔS°r, or the standard reaction entropy, you use the standard molar entropies of the products and reactants in the balanced chemical equation.

ΔS°r=(sum of nS°m(products))-(sum of nS°m(reactants))
Where n is the stoichiometric coefficient of the product/reactant from the balanced chemical equation and S°m is the standard molar entropy, which is given for all the products and reactants in the question itself.

Then you will have all the information to find ΔG°r:
ΔG°r=ΔH°r-TΔS°r

claytonkersh2I
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Joined: Tue Nov 25, 2014 3:00 am

Re: Quiz 1 prep- 2014 #9

Postby claytonkersh2I » Tue Jan 27, 2015 4:26 pm

Found my mistake. Forgot to multiply the delta H by 2.

Justin Le 2I
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Re: Quiz 1 prep- 2014 #9

Postby Justin Le 2I » Wed Jan 28, 2015 2:59 pm

For this problem, shouldn't the answer be 58.56 kj.mol-1, half of the answer given in the back? 58.56kj.mol-1 would be the delta G for the decomposition of one mole of HgO whereas 117.1 kj.mol-1 would be the answer for one mole of the reaction where the equation is 2 HgO(s) --> 2 Hg(l) + O2(g).

Emilie Flores 2G
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Re: Quiz 1 prep- 2014 #9

Postby Emilie Flores 2G » Wed Jan 27, 2016 1:43 pm

Justin Le. The question asks to calculate (delta) G for the BALANCED REACTION showing... Etc. Because the balanced reaction requires a coefficient of 2 in front of the HgO, the question is asking for the (delta) G of 2 moles of Mercury (II) Oxide. So you do not need to divide the change in free energy by 2 because the question is looking to find the value for 2 moles. Does this make sense? I can see that the question WAS worded kind of weird, I hope this helps.

604468944
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Joined: Mon Jan 26, 2015 2:17 pm

Re: Quiz 1 prep- 2014 #9

Postby 604468944 » Sun Jan 31, 2016 4:51 pm

Can anyone tell me why when we plug 2*90.83 to "delta G=delta H-T(delta S)" it is a positive? (If i plug it in as a negative my answer comes our wrong but as a positive it is right) I'm having trouble understanding that concept.

Emilie Flores 2G
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Joined: Fri Sep 25, 2015 3:00 am

Re: Quiz 1 prep- 2014 #9

Postby Emilie Flores 2G » Sun Jan 31, 2016 5:50 pm

Remember, Delta H = (The Sum of Products H) - (The sum of the reactants H). The two blank spaces in indicate that the delta H values for both products are zero. So, delta H = 0 - 2(90.83) = -181.66 kJ/mol


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