can somebody help me with chemistry? I have to reaction equations which I'm supposed to find the enthalpy of so that I can use Hess's law in order the enthalpy of a third reaction equation
I have the temp of the hydrochloric acids before and after dropping their respective solid (Zinc and Zinc Oxide I also have the amount of everything
Can you guys help me detemine the enthalpy of both reactions so that I can find out the enthalpy of Zn + (1/2)O2(g) -> ZnO(s)
pls halp, guide me, im a noobie
Joshua Smith
Just drop the class and switch to sociology. You're simply wasting time furrowing your brow in a class that's completely over your head (intro chem lol)
Zachary Peterson
I have to take it for CS and yes I know im a pleb
Adrian Hill
help, you fucking niggers!
Blake Edwards
I got a fucking A in chemistry, but I don't remember for the love of god
How about you google it, dumb shit?
Tyler Green
there's a time limit. right now I'm reading the chapter, but it'd be nice if some anons did homework for me
Jayden Peterson
You need the volumes of the water solution as well. If you have the temps and volumes, then you can use the specific heat of water to determine your change in enthalpy and, given conservation of energy, your change in entropy.
To determine specific enthalpy changes, just figure out what your volumes or masses were to figure out how man moles you're dealing with. Then you can simply find the ratio of enthalpy change to moles.
It's been three years since I've taken any chem, though, so what the hell do I know.
Nathaniel Sullivan
Cute cat, when does it transform to the girl? Is it like only under a full moon?
Nicholas Roberts
thats the thing, the hcl was the liquid, no water unless you mean I do it for water
Jace Miller
*I do it for HCl
Gavin Gomez
Oh yeah, you need to determine the limiting reactant, too (probably the zinc (oxide)) and the molarity of the HCl.
The Cl- isn't really part of the equation, being aqueous on both sides of the equation. All you're doing in the first reaction is stripping two electrons from zinc and giving one to each hydrogen ion, leading to a bond between the hydrogen atoms. So it will take energy to remove those electrons, but energy will be released from the electrons being added and the bond being formed. Similar thing is happening in the second reaction, but now an oxygen ion is having two electrons stripped from it, which are transferred to hydrogen. Knowing the energy it takes for these two processes to happen should tell you everything you need to know about the last reaction, where zinc is losing two electrons, which are given to oxygen. It's the last leg on the triangle, as it were, so you know enough to figure it out.
Thomas Miller
HCl is water. Just with HCl dissolved in it.
Just use water's specific heat, unless you can find a better value somewhere.
Evan White
yeah, I found a better value give me a sec, I will come back if I have more questions thanks for your input
Elijah Butler
Can't promise I'll be around, but gl and shit.
Noah Roberts
np, thanks again, famalam
Jayden Collins
ok, familia, I have the molar heat of reaction (or enthalpy, right?) for both reaction formula
now, how do I find the enthalpy of the third reaction formula in
Robert Mitchell
You need the enthalpies of formation for the reactions. You can literally look them up. Then basic arithmetic
Alexander Cox
is molar heat of reaction the same as enthalpy? because I have molar heat of reaction
Sebastian Smith
Wouldn't that basically be cheating in this case, though? I figured OP had to determine the enthalpies of formation, considering the experiment is rendered moot if he just uses tables.
You can add the inverse of the second equation to the first. You'll get H20+Zn-->ZnO+H2+(h1-h2)
Now what you do from there, I have no fucking clue nowadays. Might have to use a table, like said.
Caleb Moore
By the way, for the table, if you use it, you just need to find the energy associated with a hydrogen-oxygen single bond, and a hydrogen-hydrogen single bond. You have two of the former on the left side, one of the latter on the right side. Oh, and I guess you need the enthalpy for an oxygen-oxygen double bond, as well, to account for breaking up the oxygen molecule in the combustion.
Then you add them up appropriately. Should end up being something like
hreac=(h1-h2)-[h(h-h)-2*h(h-o)]-h(o--o)
Hope that makes some sense. I know the notation is crazy. You're basically just trying to make this reaction look like your desired reaction.
Owen Jones
Halve that h(o-o), as well, so the equation is
hreac=(h1-h2)-[h(h-h)-2*h(h-o)]-.5*h(o--o) Pretty sure redox equations would have made this easier.
John Rodriguez
ok guys, I got -375 kJ/mol the real value is -348 kJ/mol percent error is 8% thats an ok percent error, right? thanks anyways
Jace Green
Yes.
Joshua Brown
If you don't know, don't chirp. No, this value isn't something you can find any way other rhan empirical measurement. And the Molar values won't change, but are the product of experiment.
Adrian Smith
>If you don't know, don't chirp
Guess I shouldn't have chirped.
Hudson Long
I did and got a close answer pic related
Jordan Ross
It's not terrible for an experiment that's temperature sensitive. Some of the heat is lost to the environment. Add in issues with the scale and thermometer, and you're fine.