91Ó°ÊÓ

ΔG°is the standard free energy change and it indicates the amount of energy released when the reactants are converted to products under standard conditions.The standard free energy which are calculated according to the biochemistry convention are valid at pH 7 only, so under biochemistry conventionΔG° is customarily symbolized byΔ³Ò°′.

In a reaction A → B, an equilibrium point is reached where no further net chemical change occurs, which means A is being converted to B at the same rate as B is being converted to A.In equilibrium state, the ratio of [B] to [A] is constant and is denoted by K_eq(equilibrium constant)_.

K_eq =$\frac{\left[B\right]eq}{\left[A\right]eq}$

Here, [B]_eqis the concentration of B at equilibrium

[A]_eqis the concentration of A at equilibrium

If a reaction A →B is in equilibrium at constant temperature and pressure, the standard free energy change is given by:

Δ³Ò°′= - RT ln$\frac{\left[B\right]}{\left[A\right]}$

So,Δ³Ò°′= - RT ln K_eq

Here, R = gas constant

T = absolute temperature

K_eq= equilibrium constant

Thus, equilibrium constant can be calculated by:

K_eq== e^{-Δ³Ò°′/ RT}

It is given,

Δ³Ò°′= -7.1 kJ. mol^-1= -7100J mol^-1

T = 25°C= 25 + 273 = 298 K

R = 8.315 JK^-1 mol^-1

The equation for equilibrium constant is,

K_eq== e^{-Δ³Ò°′/ RT}

Substituting the values in the equation,

$\frac{\left[G6P\right]}{\left[G1P\right]}$= e^ {-7100 J mol^-1 / -(8.315 JK^-1 mol^-1× 298 K)}

$\frac{\left[G6P\right]}{\left[G1P\right]}$= e^2.86

^{$\frac{\left[G6P\right]}{\left[G1P\right]}$}= 17.56

$\frac{\left[G6P\right]}{\left[G1P\right]}$= 0.057

Thus, the equilibrium ratio of [G1P] to [G6P] is 0.057.