Question

A reaction

A (aq) + B (aq) <--> C (aq)

has a standard free-energy change of –3.85 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?

[A]= M

[B]= M

[C]= M

Answer 1

Ans:

∆G^o = -RTlnK

where ∆G = standard free energy change = -3.85 kJ/mol = -3.85 x 10³ J mol^-1

         R = universal gas constant = 8.314 J K^-1 mol^-1

         T = temperature in kelvin = 298 K (25+273)

         K = equilibrium constant

∆G^o = -RTlnK

-3.85 x 10³ J mol^-1 = -(8.314 J K^-1 mol^-1 x 298 K)lnK

lnK = 1.554

K = e^(1.554) = 4.73

Reaction: A (aq) + B (aq) ⇋ C (aq)

Let ‘xM’ of A react with ‘xM’ B to give ‘xM’ C at equilibrium

	A (aq) +	B (aq) ⇋	C (aq)
Initial	0.30 M	0.40 M	0 M
Change	-x M	-x M	+x M
Equilibrium	(0.30-x) M	(0.40-x) M	x M

According to law of chemical equilibrium, equilibrium constant K is given by

4.73x² − 4.311x + 0.5676 = 0

For the quadratic equation ax² + bx + c

x = 0.7518 M & x = 0.1596 M

Since x = 0.7518 M is greater than the initial concentration, it is ignored for this situation.

So, x = 0.1596 M

Equilibrium concentration:

[A] = (0.30-x) = (0.30-0.1596) = 0.1404 M

[B] = (0.40-x) = (0.40-0.1596) = 0.2404 M

[C] = x M = 0.1596 M