Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

Correction for b) ΔG= ΔG°+RTIn(Q)

Answer 5

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.

Answer 6

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.