Ok, I don't know were to start. Help please!!!
DG° = -210.155 kJ for the reaction 1A + 2B ==> 1C + 2D. What is DG (in kJ) for this reaction for the concentrations listed below? Assume 25.00°C.
Concentrations (mol/L)given:
A is 4.447
B is 2.281
C is 2.581
D is 4.196
Please Help Me!!!!!!!!!
2007-09-24 08:23:00 · 3 answers · asked by Nickname 1 in Science & Mathematics ➔ Chemistry
the above solution is not quite complete.
DG°= -RT ln Kc only applies for standard condition hence the subscript ° in DG. to calculate the gibbs free enthalpy at non standard condition and assuming those reaction that you mention are gas phase reaction and follows the ideal equation of state the DG(298 K or 25 °C) = DG° + kT ln Kc
where Kc = [C] [D]^2/([A] [B]^2)
k is the Boltzmann constant = 1.38 10^-23
T = 25°C = 298 K
the rest I believe you'd manage to calculate
2007-09-24 08:48:54 · answer #1 · answered by IonicLiquids 2 · 0⤊ 0⤋
Find the equilibrium constant K, assuming your concentrations are in equilibrium. Once you do that, DG=-RTlnK
2007-09-24 15:50:31 · answer #2 · answered by vasilios s 2 · 0⤊ 0⤋
Work out Kc from delta G = -RT ln Kc
Then apply the new figures.
2007-09-24 15:29:37 · answer #3 · answered by Gervald F 7 · 0⤊ 0⤋