If ∆G°' for a biochemical reaction is -17.5 kJ mol-1 what is the equilibrium constant (Keq)?

Question

Faraday constant: 96485 J V-1 mol-1
Molar gas constant: 8.314 J mol-1 K-1
Standard temperature: 298 K

The relationship between kJ and J needs to be considered I believe.. It has kept throwing me off..

Cain- Where did 2.303 come from?

Answer 1

K = e(-delta G/RT) = e (17500/ (8.314 * 298)) = 1.17 * 10^3

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 - (+3.16E+4) / (8.3 * 298)) = 2.8E-6 I hope this helps you.

Answer 3

(delta)G = -2.303RT*logK or -RTlnK

You have R (gas constant). You have temperature in K.
Plug n chug, dude....

(delta)G = -RT*lnK
-17,500 J/mol = (-8.314 J/mol*K)(298K)*lnK

*Note the need to change the kJ to J for G expression so the units were compatible.

Now it's your turn...do the math...solve for ln K and then take the anti-log (natural) to get K.