Question

A reaction
A + B <==> C
has a standard free-energy change of -3.27 kJ/mol at 25^oC

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

Answer 1

T= 25.0 oC

= (25.0+273) K

= 298 K

deltaG = -3.27 KJ/mol

deltaG = -3270 J/mol

we have below equation to be used:

deltaG = -R*T*ln Kc

-3270 = - 8.314*298.0* ln(Kc)

ln Kc = 1.3198

Kc = 3.743

Let's prepare the ICE table

[A] [B] [C]

initial 0.3 0.4 0

change -1x -1x +1x

equilibrium 0.3-1x 0.4-1x +1x

Equilibrium constant expression is

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

3.743 = (1*x)/((0.3-1*x)(0.4-1*x))

3.743 = (1*x)/(0.12-0.7*x + 1*x^2)

0.44916-2.6201*x + 3.743*x^2 = 1*x

0.44916-3.6201*x + 3.743*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 3.743

b = -3.62

c = 0.4492

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 6.38

putting value of d, solution can be written as:

x = {3.62 + √(6.38)}/7.486

x = {3.62 - √(6.38)}/7.486

solutions are :

x = 0.821 and x = 0.1462

x can't be 0.821 as this will make the concentration negative.so,

x = 0.1462

At equilibrium:

[A] = 0.3-1x = 0.3-1*0.14616 = 0.154 M

[B] = 0.4-1x = 0.4-1*0.14616 = 0.254 M

[C] = 0+1x = 0+1*0.14616 = 0.146 M

[A] = 0.154 M

[B] = 0.254 M

[C] = 0.146 M