Respuesta :
Answer : The concentrations of A, B, and C at equilibrium is, 0.11 M, 0.21 M and 0.19 M respectively.
Explanation :
The relation between the equilibrium constant and standard Gibbs, free energy is:
[tex]\Delta G^o=-RT\times \ln K[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs, free energy  = -5.43 kJ/mol = -5430 J/mol
R = gas constant = 8.314 J/mol.K
T = temperature  = [tex]25^oC=273+25=298K[/tex]
K = equilibrium constant = ?
Now put all the given values in the above relation, we get:
[tex]-5430J/mol=-(8.314J/mol.K)\times (298K)\times \ln K[/tex]
[tex]K=8.95[/tex]
Now we have to calculate the concentrations of A, B, and C at equilibrium.
The balanced equilibrium chemical reaction is:
             [tex]A(aq)+B(aq)\rightleftharpoons C(aq)[/tex]
Initial conc. Â Â Â 0.30 Â Â Â 0.40 Â Â Â Â 0
At eqm. Â Â Â Â (0.30-x) Â Â (0.40-x) Â Â x
The expression for equilibrium constant for this reaction is,
[tex]K=\frac{[C]}{[A][B]}[/tex]
Now put all the given values in this expression, we get:
[tex]8.95=\frac{x}{(0.30-x)\times (0.40-x)}[/tex]
[tex]x=0.19\text{ and }0.62M[/tex]
As we know that the concentration at equilibrium can not be more than initial concentration. So, neglecting the value of x = 0.62 M
The value of 'x' will be, 0.19 M
Thus, the concentrations of A at equilibrium = (0.30-x) = (0.30-0.19) = 0.11 M
The concentrations of B at equilibrium = (0.40-x) = (0.40-0.19) = 0.21 M
The concentrations of C at equilibrium = x = 0.19 M
