Question

Calculate the equilibrium constants K’eq for each of the following reactions at pH 7.0 and 25oC, using the ∆Go’ values given: (a) Glucose-6-phosphate + H2O → glucose + PI ∆Go’= -13.8 kJ/mol (b) Lactose + H2O → glucose + galactose ∆Go’= -15.9 kJ/mol (c) Malate → fumarate + H2O ∆Go’= +3.1 kJ/mol

Answer 1

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,

$\Delta G^o=-2.303\times RT\times \log K_{eq}$      ...........(1)

where,

$\Delta G^o$ = standard Gibbs free energy

R = universal gas constant = 8.314 J/K/mole

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant

Now we have to determine the value of $K_{eq}$  for the following reaction.

(a) $\text{Glucose}-6-\text{phosphate}+H_2O\rightarrow \text{Glucose}+\text{Phosphate}$     $\Delta G^o=-13.8kJ/mol$

Now put all the given values in the above formula 1, we get:

$-13.8kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$-13.8\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$K_{eq}=262.163$

(b) $\text{Lactose}+H_2O\rightarrow \text{Glucose}+\text{galactose}$     $\Delta G^o=-15.9kJ/mol$

Now put all the given values in the above formula 1, we get:

$-15.9kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$-15.9\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$K_{eq}=611.807$

(c) $\text{Malate}\rightarrow \text{fumarate}+H_2O$     $\Delta G^o=+3.1kJ/mol$

Now put all the given values in the above formula 1, we get:

$+3.1kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$+3.1\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}$

$K_{eq}=0.286$