Gibbs free energy from partial pressures

[kmwolf92]

Homework Statement

Consider the following reaction:
CH₃OH(g) <-> CO(g)+2H₂(g)

Calculate ΔG for this reaction at 298 K under the following conditions:
P_CH3OH=0.895atm
P_CO=0.115atm
P_H2=0.200atm

Homework Equations

ΔG=-R*T*ln(K)
where R is the gas constant 8.314 J/molK, T is 298 in this case, and K is determined from the partial pressures.

The Attempt at a Solution

I calculated K by (0.200²)(0.115)/0.895 and found it to be 0.00514.
This produced deltaG=-8.314*298*ln(0.00514)=-13.1 kJ, but this is incorrect according to the website. I'm really not sure what is going wrong; this is my last attempt on any of the pressure problems on this homework, so I'd like to get this figured out. My guess right now is that it's related to the # of moles not being used (as R suggests), but I'd appreciate input before I use my last attempt, especially so it's drilled into my brain after this assignment.

Thanks for any help!

 

[linhbear]

The consistency in units is often the case

Since the problem doesn't give Delta G naught, we calculate Delta G reaction assuming the reaction is at equilibrium (ΔG = 0)

ΔG° reaction (kJ) = - R (J/°K.mol) x T(°K) x ln (Pp/Pr)​

In this case, convert R = 8.314x10^-3 (kJ/°K.mol)

 

[mjc123]

You neglected the minus sign. (Clue: if K < 1, ΔG⁰ must be positive.) Is +13.1 kJ/mol correct, according to your website?

 

[DrDu]

The consistency in units is often the case

Since the problem doesn't give Delta G naught, we calculate Delta G reaction assuming the reaction is at equilibrium (ΔG = 0)

ΔG° reaction (kJ) = - R (J/°K.mol) x T(°K) x ln (Pp/Pr)​

In this case, convert R = 8.314x10^-3 (kJ/°K.mol)

The use of °K instead of K was discouraged in 1967.