Question

What is the value of K for an aqueous reaction at 298 K when the Delta G = 16.17

Answer 1

https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm

Answer 2

https://www.khanacademy.org/science/physics/thermodynamics/v/gibbs-free-energy-and-spontaneity

Answer 3

delta G = -RT ln K

R=8.314 J /mol K

Answer 4

that K okay

Answer 5

so it'd be

16.17=-(8.314)(298k) ln K

Answer 6

yea the equilibrium constant

Answer 7

which is 0.00652= ln k

Answer 8

then do you have to take the e^-(0.00652) ??

Answer 9

be careful it is your delta G in Joules or kJoules

Answer 10

Kc = Qc

Answer 11

it's in kj

Answer 12

R it is in Joules

Answer 13

Kc is at the equilibrium, Qc is the rate at any other position different than at equilibrium. If Qc = Kc the reaction it is at equilibrium

Answer 14

ye

Answer 15

soooo... I got 6.5 x 10^-6

Answer 16

then do I do e^(-6.5 x 10^-6)

Answer 17

sorry, if the delta G is in kJ you have to multiply that value for 10^3 before do the calculation

Answer 18

Oh. okay! No worries :)

Answer 19

so its just 6.53 and then is right to do e^-6.53?

Answer 20

K= 1.46 x 10^(-3)

Answer 21

Perfect!!!!

Answer 22

AHHH You're the best! Thank you so much! :) I really appreciate it!