Answer:
The maximum amount of work that can be done by this system is -2.71 kJ/mol
Explanation:
Maximum amount of work denoted change in gibbs free energy [tex](\Delta G)[/tex] during the reaction.
Equilibrium concentration of B = 0.357 M
So equilibrium concentration of A = (1-0.357) M = 0.643 M
So equilibrium constant at 253 K, [tex]K_{eq}= \frac{[B]}{[A]}[/tex]
[A] and [B] represent equilibrium concentrations
[tex]K_{eq}=\frac{0.357}{0.643}=0.555[/tex]
When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M
So reaction quotient at this situation, [tex]Q=\frac{0.133}{0.867}=0.153[/tex]
We know, [tex]\Delta G=RTln(\frac{Q}{K_{eq}})[/tex]
where R is gas constant and T is temperature in kelvin
Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and [tex]K_{eq}[/tex] is 0.555
So, [tex]\Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K[/tex]
= -2710 J/mol
= -2.71 kJ/mol
The maximum amount of workdone is mathematically given as
dG= -2.71 kJ/mol
Question Parameter(s):
A (aq) ⇌ B (aq) at 253 K
initial concentration of A = 1.00 M
initial concentration of B = 0.000 M.
the concentration of B = 0.357 M.
Equilibrum conc= (1-0.357) M = 0.643 M
Generally, the equation for the Equilbrum constant is mathematically given as
[tex]K_{eq}= \frac{[B]}{[A]}[/tex]
Therefore
[tex]K_{eq}=\frac{0.357}{0.643}[/tex]
Keq=0.555
The reaction quotient
Q=0.133/0.867
Q=0.153
In conclusion
[tex]d G=RTln(\frac{Q}{K_{eq}})[/tex]
Hence
[tex]d G=8.314\times 253* ln(\frac{0.153}{0.555})[/tex]
dG= -2.71 kJ/mol
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