I also need help with this textbook question, I'm just confused how to get to internal energy?
Oxygen difluoride is a colorless, very poisonous gas that reacts rapidly and exothermically with water vapor to produce O2 and HF:
OF2(g) + H2O(g) --> O2(g) + 2 HF(g), delta H = 2318 kJ
What is the change in internal energy for the reaction of 1.00 mol OF2?
Thanks :)
Textbook Question 4D.7
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Re: Textbook Question 4D.7
Use the first law of thermodynamics: delta U=q+w. Use the ideal gas law pv=nrt to find-p*deltav (note that temperature is 298k and you would manipulate it so that -p*delta v=-delta n*r*t). Work is equal to -p*delta v and q is equal delta h at constant pressure.
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Re: Textbook Question 4D.7
Ryan Laureano 3I wrote:Use the first law of thermodynamics: delta U=q+w. Use the ideal gas law pv=nrt to find-p*deltav (note that temperature is 298k and you would manipulate it so that -p*delta v=-delta n*r*t). Work is equal to -p*delta v and q is equal delta h at constant pressure.
what would you do here for the change in moles?
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Re: Textbook Question 4D.7
The change in moles is delta n = 1 mol.
Using PV = nRT, you change it to P(deltaV) = (delta n)RT
Then use P(delta V) to get the work done by expansion.
Also thank you for the explanation, this makes much more sense now!
Using PV = nRT, you change it to P(deltaV) = (delta n)RT
Then use P(delta V) to get the work done by expansion.
Also thank you for the explanation, this makes much more sense now!
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Re: Textbook Question 4D.7
Ryan Laureano 3I wrote:Use the first law of thermodynamics: delta U=q+w. Use the ideal gas law pv=nrt to find-p*deltav (note that temperature is 298k and you would manipulate it so that -p*delta v=-delta n*r*t). Work is equal to -p*delta v and q is equal delta h at constant pressure.
Wait one more questions, how do you know what P and T are since they aren't given? Can I assume 1 atm and 298K?
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Re: Textbook Question 4D.7
I was wondering this too, why do we assume that the temperature is 298 K for this problem?
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Re: Textbook Question 4D.7
I was also wondering why the temperature is 298k when its not written in the problem, and if we can assume it, in what other situations would we assume the temperature is 298k?
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Re: Textbook Question 4D.7
Seraphina Joseph 1C wrote:I was also wondering why the temperature is 298k when its not written in the problem, and if we can assume it, in what other situations would we assume the temperature is 298k?
Something in the problem seems to be missing because we can't assume the temperature to be 298K unless we are told that. The only way we can assume a temp is if the problem says STP, meaning the temp would be 273 K.
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Re: Textbook Question 4D.7
I think sapling is missing some information here but I just assumed SATP (298K and 1 atm).
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Re: Textbook Question 4D.7
Seraphina Joseph 1C wrote:I was also wondering why the temperature is 298k when its not written in the problem, and if we can assume it, in what other situations would we assume the temperature is 298k?
I think that, if nothing is stated explicitly in the problem, you can just assume that the reaction occurs at SATP (25 degrees C and 1 atm) because most reactions occur at room temperature.
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