227 g of graphite and partial pressures of 2.00 bar CO2 and 125 bar CO are added to a sealed 1.00 L sealed container at 298 K. In which direction is the reaction spontaneous?
Do I use deltaG= -RTInK equation to solve this question?
C(graphite) + CO2 (g) = 2 CO (g)
K= (2.00)^2/ 125
K= 0.032
deltaG= -RTInK
= -(8.214)(298) In(0.032)
= 8527.85 J
= 8.527 kJ
I am not sure if this is the correct answer, please correct me if I am wrong.
I am not positive about the question. I don't think your K is correct because no where in the problem is a K listed AND the values given in the problem are not equilibrium values. I assume the problem is this but correct me if I'm wrong. I think you added 2 bar CO2 and 125 bar CO to an empty 1.00 L container, sealed it, let the reaction reach equilibrium and the questin is which way does the reaction shift; i.e., t the left or right. You need a Kp and you don't have one. Here is a way to get it.,
Look up dGo values and calculate dGo for the reactioin.
dGo rxn = (n*dGo products) - (dGo reactants), then calculate Kp from dGo = -RT*lnKp.
Next, using the values in the probem, calculate Qp, then compare Qp with Kp.
Post your work if you get stuck and/or correct me if the problem is not the same as I've assumed.
This is the full question:
C(graphite) + CO2 (g) ⇌ 2 CO (g) ΔH°= 173.5 kj ΔG°= 120.0 kj
a) Provide the expression for the reaction quotient Q, for this reaction:
My answer: Q= [CO] ^2 / [CO2]
b) 227 g of graphite and partial pressures of 2.00 bar CO2 and 125 bar CO are added to a sealed 1.00 L sealed container at 298 K. In which direction is the reaction spontaneous?
My answer: would the equation be ΔG= ΔG°+RTInK
Correction for b) ΔG= ΔG°+RTIn(Q)
To be honest about it, it sure would have been nice to have the entire question instead of just part of it initially.
a is correct.
b. I don't think the equation you have in mind is the correct one. This is why. The problem gives you dGo. The value of dGo is understood to be at 25 C or 298 K. Part b is asking for dG at 298 which actually is dGo. So dG = dGo = -RT*lnQ
So you do as I suggested above. Calculate Kp and Qp and compare. I think there is another way to approach the problem since the dHo is given as well as dGo. Having dGo and dHo and T of 298 gives you the opportunity to calculate dSo and since both products and reactants are gases (good job on not trying to use C(graphite) and recognizing it is a solid and doesn't enter into any of the calculations) use dS to determine which way is spontaneous.
I used what you have suggested ΔG= ΔH-TΔS
120.0 kj= 173.5 kj- 298K ΔS
ΔS= -0.96386 kj x1000J
ΔS= -963.86 J
so it will be spontaneous, more products than reactants.
correct me if i made any mistakes, thank you for the guidance.
To determine the direction in which the reaction is spontaneous, you can use the concept of Gibbs free energy (ΔG) and the associated equation ΔG = ΔG° + RT ln(Q), where ΔG is the change in Gibbs free energy, ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.
To use this equation, you need to know the standard Gibbs free energy change (ΔG°) for the reaction. ΔG° can be found using standard thermodynamic data tables or by calculating it from the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) using the equation ΔG° = ΔH° - TΔS°.
Once you have ΔG°, you can substitute it into the ΔG equation along with the values of R (8.314 J/(mol K)) and T (298 K). Then, calculate the value of Q by using the given partial pressures of CO2 and CO.
If the calculated value of ΔG is negative, it means that the reaction is spontaneous in the forward direction. If the calculated value is positive, it means that the reaction is spontaneous in the reverse direction.
Therefore, by using the ΔG equation, you can determine the direction in which the reaction is spontaneous.