Asked by Jerry
When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced.
O2M(s) ----> M(s) +O2(g) delta G= 288.9 Kj/mol
When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s).
I got an answer of :
O2M(s)+ C(s) ----> M(s) + CO2(g)
This is correct. I need help on this part of the question:
What is the thermodynamic equilbrium constant for the coupled reaction?
I attempted this twice and got
K=1.22 and 0.89 and they are both wrong. Please help.
O2M(s) ----> M(s) +O2(g) delta G= 288.9 Kj/mol
When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s).
I got an answer of :
O2M(s)+ C(s) ----> M(s) + CO2(g)
This is correct. I need help on this part of the question:
What is the thermodynamic equilbrium constant for the coupled reaction?
I attempted this twice and got
K=1.22 and 0.89 and they are both wrong. Please help.
Answers
Answered by
DrBob222
I would add 288.9 to dGof for CO2 of -394.4 for -105.5 kJ/mol for the reaction you show. Then
(105,500/8.314*298) = lnK and i get a huge number for K. Somethng like 3.1E18
Check that carefully.
(105,500/8.314*298) = lnK and i get a huge number for K. Somethng like 3.1E18
Check that carefully.
Answered by
Jerry
yes it is correct! Thanks so much! You are the best!!!!!
Answered by
Coolguy
K=e raised to -(-106.4 kj/mol)/(8.314e-3kj/mol*298k)
K=4.476e18 is what I got.
K=4.476e18 is what I got.
Answered by
Andrew
I keep getting the equation wrong!
Answered by
J
my deltaGf was dGf=290.4kJ/mol
i got K=1.69e18 as my answer and it was correct.
still getting the equation wrong though :/
i got K=1.69e18 as my answer and it was correct.
still getting the equation wrong though :/
Answered by
Janavi
1: Set all the delta g's for solids equal to 0
2:Find thr delta g for the gases.
3:Find the delta g for the second equation.
4:Add the delta g's for the 2 questions together.
5: Solve for the value of Q using the equation
-Delta G not =RTLnQ
2:Find thr delta g for the gases.
3:Find the delta g for the second equation.
4:Add the delta g's for the 2 questions together.
5: Solve for the value of Q using the equation
-Delta G not =RTLnQ
Answered by
HitGirl
COMBINE THE 2 EQUATIONS TOGETHER PEOPLE!
MO2 + C <-> CO2 + M
MO2 + C <-> CO2 + M
Answered by
Chris
Original deltaG=288.9
deltaG(system)=deltaG[CO2]+deltaG[M]-deltaG[MO2]-deltaG[C]
look up deltaG value from table for CO2=-394.4
DeltaG for a solid is almost always zero (C)
So -394.4+288.9=deltaGsystem
then
K=e^((deltaGsystem)/((8.314/1000)*298)))
deltaG(system)=deltaG[CO2]+deltaG[M]-deltaG[MO2]-deltaG[C]
look up deltaG value from table for CO2=-394.4
DeltaG for a solid is almost always zero (C)
So -394.4+288.9=deltaGsystem
then
K=e^((deltaGsystem)/((8.314/1000)*298)))
Answered by
sdtfer
ert5y547
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