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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: TF2 on March 16, 2009, 03:56:56 PM

Title: Standard enthalpy of formation?
Post by: TF2 on March 16, 2009, 03:56:56 PM

for CuSO4.5H2O(s) is -2278 jk/mole
What is the chemical equation to whihc this applies
Is it

1)Cu(s) + S(s) +5H2O(g) +2O2(g) ---->CuSo4.5H2O(S)

2) 2Cu(s) + 2SO2(g) +5H2O(g) ---->2CuSo4.5H2O(S)

3) 2Cu(s) + 2SO2(g) +5H2O(g) ---->2CuSo4.5H2O(S)

4) Cu(s) + SO4(g) +5H2O(g) ---->CuSo4.5H2O(S)

5) Cu(s) + S(s) + 5/9 O2(g)+5H2(g) ---->CuSo4.5H2O(S)

6) Cu(s) + S(s) + 9/2 O2(g)+5H2(g) ---->CuSo4.5H2O(S)

Title: Re: Standard enthalpy of formation?
Post by: Borek on March 16, 2009, 04:10:24 PM

Simply apply the definition.

Title: Re: Standard enthalpy of formation?
Post by: TF2 on March 16, 2009, 04:22:40 PM

im getting number 4....is this correct?

Title: Re: Standard enthalpy of formation?
Post by: Borek on March 16, 2009, 04:45:58 PM

What is the definition of the standard enthalpy of formation?

Title: Re: Standard enthalpy of formation?
Post by: TF2 on March 16, 2009, 11:19:19 PM

are you asking or do you know?

Title: Re: Standard enthalpy of formation?
Post by: Borek on March 17, 2009, 03:58:33 AM

I know, but seems to me you have no idea.

Title: Re: Standard enthalpy of formation?
Post by: TF2 on March 17, 2009, 09:37:33 AM

well when you start a reply with what is....and ended it with a question im guessing you are asking, instead of helping. And i asked what is the chemical formula not the definition. Anyway ill go ask someone else who actually knows and stop wasting my time with your nonsense

Title: Re: Standard enthalpy of formation?
Post by: sjb on March 17, 2009, 09:49:21 AM

OK then, what do you believe the definition of the standard enthalpy of formation is?

Why have you chosen 4, rather than say 5?

S

Title: Re: Standard enthalpy of formation?
Post by: Borek on March 17, 2009, 10:21:05 AM

Quote from: TF2 on March 17, 2009, 09:37:33 AM

well when you start a reply with what is....and ended it with a question im guessing you are asking, instead of helping. And i asked what is the chemical formula not the definition. Anyway ill go ask someone else who actually knows and stop wasting my time with your nonsense

I was trying to help you find out the correct answer on your own. But if you don't want to be helped - be it.