Asked by Anonymous
The standard internal energy change for a reaction can be symbolized as ΔU°rxn or ΔE°rxn.
For the following reaction equations, calculate the energy change of the reaction at 25 °C and 1.00 bar.
Sn(s) + 2Cl2(g) --> SnCl4(l)
ΔH°rxn = -511.3 kJ/mol
ΔU°rxn = ?
H2(g) + Cl2(g) --> 2HCl(g)
ΔH°rxn = -184.6 kJ/mol
ΔU°rxn = ?
I know ΔU°rxn = q + work
q = ΔH°rxn
How do I find work for each reaction?
For the following reaction equations, calculate the energy change of the reaction at 25 °C and 1.00 bar.
Sn(s) + 2Cl2(g) --> SnCl4(l)
ΔH°rxn = -511.3 kJ/mol
ΔU°rxn = ?
H2(g) + Cl2(g) --> 2HCl(g)
ΔH°rxn = -184.6 kJ/mol
ΔU°rxn = ?
I know ΔU°rxn = q + work
q = ΔH°rxn
How do I find work for each reaction?
Answers
Answered by
DrBob222
For #2.
dUrxn = q + w
q = dH = -184.6 for 1 mol or x 9 for 9 mols = -? kJ.
w = -pdV
volume = nRT/P = 9mol*0.08206 L*atm*K*298K/0.9869 atm = about 223L
w = -0.9869*446 = about -44 kJ
Then dU = -? kJ - ?w = ?
dUrxn = q + w
q = dH = -184.6 for 1 mol or x 9 for 9 mols = -? kJ.
w = -pdV
volume = nRT/P = 9mol*0.08206 L*atm*K*298K/0.9869 atm = about 223L
w = -0.9869*446 = about -44 kJ
Then dU = -? kJ - ?w = ?
Answered by
Anonymous
How come 223L was doubled?
And do I need to use 9 mols
so -184.6 x 9 = -1661.4 kJ/mol?
Can I do this:
(1 mol*0.08206*298K)/0.9869 = 24.8L
then
w = -0.9869 * 24.8 = -2.45 kJ
then
dU = -184.6 - 2.45 = 209.1 kJ/mol ?
And do I need to use 9 mols
so -184.6 x 9 = -1661.4 kJ/mol?
Can I do this:
(1 mol*0.08206*298K)/0.9869 = 24.8L
then
w = -0.9869 * 24.8 = -2.45 kJ
then
dU = -184.6 - 2.45 = 209.1 kJ/mol ?
Answered by
DrBob222
<b>Yes and no.
I goofed on the 9 mols; don't ask where I picked that up because I don't see it anywhere in the problem. However, what is n for H2 and n for Cl2. That isn't clear from the problem.</b>
(1 mol*0.08206*298K)/0.9869 = 24.8L
<b>Yes you can use that but subtitute mols HCl for n. If you used 1 mol H2 and 1 mol Cl2 it produces 2 mols HCl and that's what goes in for n.</b>
then
w = -0.9869 * 24.8 = -2.45 kJ
<b> You should note here that the is p in atm and v in L(but probably not 24.8L) so the units are atm*L and not joules. You must multiply that number by 101.325 to change L*atm to joules.</b>
then
dU = -184.6 - 2.45 = 209.1 kJ/mol ?
<b>Remember that the -184.6 kJ/mol must be multiplied by mols HCl also to find total kJ for dH. Of course the 2.45 you have here isn't the final number either due to the above changes.
Hope this looks ok. I'll check it and make corrections if necessary.
</b>
I goofed on the 9 mols; don't ask where I picked that up because I don't see it anywhere in the problem. However, what is n for H2 and n for Cl2. That isn't clear from the problem.</b>
(1 mol*0.08206*298K)/0.9869 = 24.8L
<b>Yes you can use that but subtitute mols HCl for n. If you used 1 mol H2 and 1 mol Cl2 it produces 2 mols HCl and that's what goes in for n.</b>
then
w = -0.9869 * 24.8 = -2.45 kJ
<b> You should note here that the is p in atm and v in L(but probably not 24.8L) so the units are atm*L and not joules. You must multiply that number by 101.325 to change L*atm to joules.</b>
then
dU = -184.6 - 2.45 = 209.1 kJ/mol ?
<b>Remember that the -184.6 kJ/mol must be multiplied by mols HCl also to find total kJ for dH. Of course the 2.45 you have here isn't the final number either due to the above changes.
Hope this looks ok. I'll check it and make corrections if necessary.
</b>
Answered by
DrBob222
It looks ok to me. Let me know if you have other questions about the solution. Do you have clarification on n H2 and nCl2.
Answered by
Anonymous
That is incorrect
Answered by
Anonymous
This answer is wrong
Answered by
Anonymous
The second reaction has no change in the number of moles of gas, so w = 0, and ΔU = ΔH.
Answered by
Soli
w=Δ°nRT
so just find the change in mols from the equation.
so just find the change in mols from the equation.
Answered by
Ders
Equation #1
The change in moles is -2
dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = 506.3 kJ/mol
The change in moles is -2
dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = 506.3 kJ/mol
Answered by
Ders
Equation #1
The change in moles is -2
dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = -506.3 kJ/mol
The change in moles is -2
dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = -506.3 kJ/mol
Answered by
windal
1) -511.3
and
2)-184.6
there is no change in moles. Just completed this homework and the above answers are correct. delta-H=delta-U
and
2)-184.6
there is no change in moles. Just completed this homework and the above answers are correct. delta-H=delta-U
Answered by
WINDAL FAN
WINDAL HAS RIGHT ANSWER!!!!
Answered by
WINDAL FAN 2
WINDAL IS GOD
Answered by
welp
the answer are
equation #1 = -506.3
equation #2 = -184.6
equation #1 = -506.3
equation #2 = -184.6
Answered by
WelpisGod
Welp has the right answer. Not Windal.
Answered by
MarcoPolo
I concur that welp is correct, with the answers being
equation #1 = -506.3
equation #2 = -184.6
because windal forgot to convert into K for equation 1
equation #1 = -506.3
equation #2 = -184.6
because windal forgot to convert into K for equation 1