Question

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):

\({\text{G}}6{\text{P}} \rightleftharpoons {\text{F}}6{\text{P}}\;\;\;\Delta G_{298}^\circ = 1.7\;{\text{kJ}}\)

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be \(1M\), however, in a typical cell, they are not even close to these values. Calculate \(\Delta G\)when the concentrations of G6P and F6P are \(120\mu M\) and \(28\mu M\)respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37°C

Short answer

(a)The reaction is non –spontaneous.

(b)The change of free energy is \(\Delta G = - 2.09\frac{{kJ}}{{mol}}\) and the reaction is spontaneous.

Step-by-step solution

Definition of Gibbs free energy

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the topmost quantum of work done in a thermodynamic system when temperature and pressure remain constant.

Determine the Change of free energy.

a)

• In order to determine whether the reaction is spontaneous or non-spontaneous, we need to look at the standard free energy change.
• Since the standard free energy change is positive, the reaction is not spontaneous under standard thermodynamic conditions.

Hence, the reaction is non–spontaneous.

b)

To calculate the free energy change under these conditions, we can use the formula:

\(\Delta G = \Delta G({\rm{ standard }}) + RT\ln Q\)

\(\begin{array}{l}Q = \frac{{c(F6P)}}{{c(G6P)}}\\Q = \frac{{28}}{{120}}\\Q = 0.23\end{array}\)

The temperature is\(37C\), that is\(310\;{\rm{K}}\). We can convert any temperature in degrees Celsius to Kelvin by simply adding \(273\) to the temperature in degrees Celsius.

We can convert the free energy change from \(\frac{{kJ}}{{mol}}\) to \(\frac{J}{{mol}}\) by multiplying the energy change by \(1000(1kJ = 1000J)\).

\(\Delta G = 1700{\rm{Jmo}}{{\rm{l}}^{ - 1}} + 8.314J\;{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \cdot 310\;{\rm{K}} \cdot \ln (0.23)\)

\(\Delta G = - 2.09\frac{{kJ}}{{mol}}\)

We can convert the free energy change from \(\frac{J}{{mol}}\) to \(\frac{{kJ}}{{mol}}\) by dividing the energy change by\(1000(1kJ = 1000J)\).

The reaction, under these conditions, is spontaneous because the free energy change is negative.

Hence, the change of free energy is \(\Delta G = - 2.09\frac{{kJ}}{{mol}}\) and the reaction is spontaneous.