# When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount

QuestionsCategory: ChemistryWhen the oxide of generic metal M is heated at 25.0 °C, only a negligible amount
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

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.

yes it is correct! Thanks so much! You are the best!!!!!

K=e raised to -(-106.4 kj/mol)/(8.314e-3kj/mol*298k)

K=4.476e18 is what I got.

I keep getting the equation wrong!

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 :/

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

COMBINE THE 2 EQUATIONS TOGETHER PEOPLE!
MO2 + C CO2 + M

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)))