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.
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.
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)))
I keep getting the equation wrong!
COMBINE THE 2 EQUATIONS TOGETHER PEOPLE!
MO2 + C <-> CO2 + M
yes it is correct! Thanks so much! You are the best!!!!!
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
ert5y547
To determine the thermodynamic equilibrium constant for the coupled reaction, you need to use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K).
The standard Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation:
ΔG° = -RTln(K)
Where:
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (K)
- ln represents the natural logarithm
In this case, the given standard Gibbs free energy change (ΔG°) is 288.9 kJ/mol, but we need to convert it to joules per mole:
ΔG° = 288.9 kJ/mol × 1000 J/1 kJ = 288,900 J/mol
Now, let's assume the equilibrium constant for the coupled reaction is K.
Substituting the values into the equation, we have:
288,900 J/mol = -8.314 J/(mol·K) × T × ln(K)
To find the equilibrium constant K, you'll need to know the temperature (T) at which the reaction is taking place. Once you have the temperature, you can rearrange the equation to solve for K:
K = e^(-ΔG° / (RT))
where e is the base of the natural logarithm (approximately 2.71828).
Make sure you use the appropriate units for temperature (Kelvin) and gas constant (J/(mol·K)).
If you provide the temperature (in Kelvin) at which the reaction is taking place, I can help you calculate the equilibrium constant (K) correctly.
K=e raised to -(-106.4 kj/mol)/(8.314e-3kj/mol*298k)
K=4.476e18 is what I got.