Question

Reaction A(aq) + B(aq) $\rightleftharpoons$ C(aq) has the standard free-energy change of -3.51 kJ/mol at 25 ^$\circ$C.

What are the concentrations of A, B and C at equilibrium if their their initial concentrations are 0.30 M, 0.40 M, and 0 M?

Answer 1

T= 25.0 oC

= (25.0+273) K

= 298 K

ΔGo = -3.51 KJ/mol

ΔGo = -3510 J/mol

use:

ΔGo = -R*T*ln Kc

-3510 = - 8.314*298.0* ln(Kc)

ln Kc = 1.4167

Kc = 4.124

ICE Table:

CA] CB] [C7 initial change equilibrium 0.3 -1x 0.3-1x 0.4 -1x 0.4-1x +1x +1x

Equilibrium constant expression is

Kc = [C]/[A]*[B]

4.124 = (1*x)/((0.3-1*x)(0.4-1*x))

4.124 = (1*x)/(0.12-0.7*x + 1*x^2)

0.4949-2.887*x + 4.124*x^2 = 1*x

0.4949-3.887*x + 4.124*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 4.124

b = -3.887

c = 0.4949

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 6.944

roots are :

x = 0.7907 and x = 0.1518

x can't be 0.7907 as this will make the concentration negative.so,

x = 0.1518

At equilibrium:

[A] = 0.3-1x = 0.3-1*0.1518 = 0.1482 M

[B] = 0.4-1x = 0.4-1*0.1518 = 0.2482 M

[C] = +1x = +1*0.1518 = 0.1518 M

Answer:

[A] = 0.148 M

[B] = 0.148 M

[C] = 0.152 M