Respuesta :
Answer : The concentration of [tex]A,B\text{ and }C[/tex] at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.
Explanation :
The given chemical reaction is,
[tex]A(aq)+B(aq)\rightleftharpoons C(aq)[/tex]
First we have to calculate the equilibrium constant for the reaction.
The relation between the equilibrium constant and standard free‑energy is:
[tex]\Delta G^o=-RT \ln k[/tex]
where,
[tex]\Delta G^o[/tex] = standard free‑energy change = -4.20 kJ/mole
R = universal gas constant = 8.314 J/mole.K
k = equilibrium constant = ?
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in the above relation, we get:
[tex]-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k[/tex]
[tex]k=5.45[/tex]
Now we have to calculate the concentrations of A, B, and C at equilibrium.
The given equilibrium reaction is,
[tex]A(aq)+B(aq)\rightleftharpoons C(aq)[/tex]
Initially 0.30 0.40 0
At equilibrium (0.30-x) (0.40-x) x
The expression of equilibrium constant will be,
[tex]k=\frac{[C]}{[A][B]}[/tex]
[tex]5.45=\frac{x}{(0.30-x)\times (0.40-x)}[/tex]
By solving the term x, we get
[tex]x=0.168\text{ and }0.716[/tex]
From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.
The value of x will be, 0.168 M
The concentration of [tex]A[/tex] at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M
The concentration of [tex]B[/tex] at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M
The concentration of [tex]C[/tex] at equilibrium = x = 0.168 M
The equilibrium concentrations of A, B and C are; 0.2 M, 0.30 M, 0.1 M.
Given that;
ΔG = -RTlnK
ΔG = change in free energy = −4.20 kJ/mol or 4.20 × 10^3J/mol
R = molar gas constant = 8.314 J/K/mol
T = absolute temperature = 25 °C + 273 = 298 K
K = equilibrium constant = ??
K = ΔG/ -RT
Substituting values;
K = −(4.20 × 10^3J/mol)/ -(8.314 J/K/mol × 298 K)
K = 1.695
Now we must set up an ICE table as follows;
A(aq) + B(aq) ⇄ C(aq)
I 0.30 0.40 0
C -x -x +x
E 0.30 - x 0.40 - x +x
K = [C]/[A] [B]
1.695 = [x]/[ 0.30 - x] [0.40 - x]
1.695 [ 0.30 - x] [0.40 - x] = [x]
We now have the quadratic equation;
0.23 - 2.187x + 1.695x^2 = 0
x = 1.2 or 0.1
Since x can not exceed the initial concentration of reactants, hence, x = 0.1
[C] = 0.1 M
[B] = 0.40 M - 0.1 M = 0.30 M
[A] = 0.30 M - 0.1 M = 0.2 M
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