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
Lisa Howell asked 7 months ago
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.

9 Answers
Lisa Howell answered 7 months ago
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.

Lisa Howell answered 7 months ago
yes it is correct! Thanks so much! You are the best!!!!!

Lisa Howell answered 7 months ago
K=e raised to -(-106.4 kj/mol)/(8.314e-3kj/mol*298k)

K=4.476e18 is what I got.

Lisa Howell answered 7 months ago
I keep getting the equation wrong!

Lisa Howell answered 7 months ago
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 :/

Lisa Howell answered 7 months ago
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

Lisa Howell answered 7 months ago
COMBINE THE 2 EQUATIONS TOGETHER PEOPLE!
MO2 + C CO2 + M

Lisa Howell answered 7 months ago
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)))

Lisa Howell answered 7 months ago
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