Question

Consider the following reaction:

A (aq) <==> B (aq) ΔG^o_rxn= 3.93 kJ

A reaction vessel was charged with compound A (4.31 M) and compound B (6.11 M) and allowed to react at 345 K. After a period of 6.1 hours, ΔG was determined to be 2.71 kJ. What was the concentration of compound A at that time?

Please enter your answer to two decimal places.​

Answer 1

we know that

dG = dGo + RT lnK

given

dG = 2.71 x 1000 J

dGo = 3.93 x 1000 J

Temperature (T) = 345

R = 8.314

so

using those values

we get

2.71 x 1000 = (3.93 x 1000 ) + ( 8.314 x 345 x ln K )

K = 0.65355

now

consider the given reaction

A ---> B

the value of K is given by

K = [B] / [A]

now

A ---> B

initially

[A] = 4.31

[B] = 6.11

now

using ICE table

finally

[A] = 4.31 - x

[B] = 6.11 + x

K = [B] / [A]

0.65355 = [6.11 + x] / [4.31 - x]

2.816805 - 0.65355x = 6.11 + x

x = -1.99

so

[A] = 4.31 + 1.99 = 6.3

[B] = 6.11 - 1.99 = 4.12

so

at that time the concentration of [A] is 6.30 M