Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin?

Question

(Use the reaction free energy given below.)


2SO3 = 2SO2 + O2 in the gaseous state


ΔGo = 140.0 kJ/mol



Explaining how the answer was derived would be greatly appreciated!!!

Answer 1

There is a single equation which relates delta Go to Keq. The equation is:

DGo=-RTlnKeq.

Plug in the temperature, the gas constant (in units of j/molK) and solve for K.

Answer 2

Lancenigo di Villorba (TV), Italy

I will apply the "van't Hoff's Isothermal Equation", that is

DeltaG = DeltaG° + R * T * SUM[nu,i * LN(a,i)]

where nu,i means the Stoichiometric Coefficient of the i-th Chemical Stuff while a,i means the Chemical Activity of the i-th Chemical Stuff.
As you know, the Chemical Equilibrium is a condition where it verify the following facts :

-) SUM[nu,i * LN(a,i)] = LN(Keq)
-) DeltaG = 0.0

I may derive that

0.0 = DeltaG = DeltaG° + R * T * LN(Keq)

that is

Keq = EXP(0 - DeltaG° / (R * T))

CALCULATIONs
By applying the obtained relation

Keq = EXP(0 - DeltaG° / (R * T)) =
= EXP(0 - (-1.4E+5) / (8.3 * 298)) = 3.8E+24

I hope this helps you.

Answer 3

2.889e-25

G=-RTlnKeq
140=-8.314*298*lnKeq
Keq=2.889e-25