Posted by Jerry on .
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.

Chemistry 
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. 
Chemistry 
Jerry,
yes it is correct! Thanks so much! You are the best!!!!!

Chemistry2 
Coolguy,
K=e raised to (106.4 kj/mol)/(8.314e3kj/mol*298k)
K=4.476e18 is what I got. 
Chemistry 
Andrew,
I keep getting the equation wrong!

Chemistry 
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 :/ 
Chemistry 
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 
Chemistry 
HitGirl,
COMBINE THE 2 EQUATIONS TOGETHER PEOPLE!
MO2 + C <> CO2 + M 
Chemistry 
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)))