Hi all,
So if you calculate #9 on Quiz 1 2014 using the balanced equation 2HgO -> 2Hg+O2, you get 117.1 KJ/mol (the answer in the book).
If you solve the problem in terms of the decomposition of 1 mol HgO, HgO->Hg+0.5O2, you get 58.56 KJ/mol, (half the answer in the book).
So if you solve the problem using the first method, shouldn't you have to divide the final answer by 2, since you multiply the quantities HgO by 2, and all subsequent relationships?
Thanks!
Renee Crippen 2I
Why don't we divide Quiz 1, 2014 #9 by 2?
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Re: Why don't we divide Quiz 1, 2014 #9 by 2?
Hi can you please post the entire question so that we can help you.
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Re: Why don't we divide Quiz 1, 2014 #9 by 2?
You don't divide by 2 because the problem is asking for delta G for the BALANCED reaction.
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Re: Why don't we divide Quiz 1, 2014 #9 by 2?
But by balanced, does it mean for the least whole integer?
For example if the balanced reaction for the combustion of molecule "A" is:
2A+4B->C
Do we use the stated equation above, or do we use
A+2B-> 0.5C
Because it is referring to the combustion of molecule A, which implies it should be calculated for 1 mol?
Thanks,
Renee Crippen 2I
For example if the balanced reaction for the combustion of molecule "A" is:
2A+4B->C
Do we use the stated equation above, or do we use
A+2B-> 0.5C
Because it is referring to the combustion of molecule A, which implies it should be calculated for 1 mol?
Thanks,
Renee Crippen 2I
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