Question

. Methanol, CH3OH, is used as fuel in race cars. This liquid fuel can be synthesized using the following process

C(s) + 1⁄2O2(g) + 2H2(g) CH3OH(l)
Calculate ÆGo using the ÆGof of the reaction.

Calculate Kp.

What is the relation between the magnitude of ÆGo and Kp?

Answer 1

the reaction:

C(s) + 1⁄2O2(g) + 2H2(g)---> CH3OH(l)

is already balanced

Calculate Kp

Kp = 1/ [(P-H2)^2 * (P-O2)^0.5]

Note that we can relate Kp to Kc and dG via:

dG = Gproducts - Greactants

dG = G-Methanol - (G-C + 0.5*G-O2 + 2*G-H2)

dG = -166.6kJ/mol - (0)

dG = -166.6 kJ/mol

dG = -166600 J/mol

Now,

dG = -RT*ln(K)

dG = -8.314 J/molK * 298K * ln(Keq)

ln(Keq) = -dG/(RT)

Keq = exp(-dG/(RT)) = exp(166600 /(8.314*298)) = 1.59725*10^29

Note that

Keq is not Kp!

Keq is based on concentrations (mol per liter)

so

relate Kp to Keq:

Kp = Kc*(RT)^dn

dn = moles of gas porducts - moles of gas reactants = 0 - (0.5+2) = -2.5

sustitute

Kp = Kc*(RT)^dn

Kp = (1.59725*10^29 )*((0.082*298)^-2.5)

Kp = 5.411*10^25

Relationship is:

Kp > 1 if dG is negative

and Kp < 1 id dG is positive

which is accordingly what we will expect in equilibrium favouiring prodcts/reactnats