Respuesta :
Answer :
(a) The value of equilibrium constant is, 262.163
(b) The value of equilibrium constant is, 611.807
(c) The value of equilibrium constant is, 0.286
Solution :
The relation between the standard Gibbs free energy and equilibrium constant are,
[tex]\Delta G^o=-2.303\times RT\times \log K_{eq}[/tex] ...........(1)
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy
R = universal gas constant = 8.314 J/K/mole
T = temperature = [tex]25^oC=273+25=298K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant
Now we have to determine the value of [tex]K_{eq}[/tex] for the following reaction.
(a) [tex]\text{Glucose}-6-\text{phosphate}+H_2O\rightarrow \text{Glucose}+\text{Phosphate}[/tex] [tex]\Delta G^o=-13.8kJ/mol[/tex]
Now put all the given values in the above formula 1, we get:
[tex]-13.8kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]-13.8\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]K_{eq}=262.163[/tex]
(b) [tex]\text{Lactose}+H_2O\rightarrow \text{Glucose}+\text{galactose}[/tex] [tex]\Delta G^o=-15.9kJ/mol[/tex]
Now put all the given values in the above formula 1, we get:
[tex]-15.9kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]-15.9\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]K_{eq}=611.807[/tex]
(c) [tex]\text{Malate}\rightarrow \text{fumarate}+H_2O[/tex] [tex]\Delta G^o=+3.1kJ/mol[/tex]
Now put all the given values in the above formula 1, we get:
[tex]+3.1kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]+3.1\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}[/tex]
[tex]K_{eq}=0.286[/tex]
