Question

A key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose-6- phosphate (F6P):

$\mathbf{\text{G6P}}\mathbf{\rightleftharpoons }\mathit{F}\mathbf{6}\mathit{P}\mathbf{;}\mathbf{}\mathbf{}\mathit{K}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{510}\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathbf{298}\mathit{K}$

(a) Calculate $\mathbf{∆}{\mathit{G}}^{\mathbf{°}}$at $\mathbf{298}\mathit{K}$.

(b) Calculate $\mathbf{∆}\mathit{G}$when Q, the [F6P]/[G6P] ratio, equals$\mathbf{10}\mathbf{.}\mathbf{0}$.

(c) Calculate $\mathbf{∆}\mathit{G}$when $\mathit{Q}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{100}$.

(d) Calculate Q if$\mathbf{}\mathit{G}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$.

Short answer

a) Because the concentration equilibrium constant equals $50$, the reaction is product-favorable, and$K_c>1$ .

b) For the specified reaction and conditions,$K_p=K_c$ .

c)$\text{- 16.78kJ}$ is the$∆G_{rxn}^{°}$ .

d) Gibb's energy is unaffected since the equilibrium constant remains unchanged.

Step-by-step solution

Definition of Isomerization

The process of transforming a molecule, ion, or molecular fragment into an isomer having a distinct chemical structure is known as isomerization. Tautomerization and enolization are both examples of isomerization.

Determining∆G°  at 298K

(a) Gibb's energy change$298K$can be estimated using the K.

$$\begin{gathered} ∆G^{°}=-R\times T\times InK \\ ∆G^{°}=-8.314\frac{J}{mol\times K}\times 298K\times In\left(0.510\right) \\ ∆G^{°}=1668.26\frac{J}{mol}=1.668\frac{kJ}{mol} \end{gathered}$$

The $∆G^{°}$has a value of $\text{1.668kJ/mol}$.

Determining ∆G when Q = [F6P][G6P] = 10.00

(b) The response quotient Q is calculated as follows:

$\text{Q =}\frac{\text{[F6P]}}{\text{[G6P]}}\text{= 10.00}$

• At$298K$ , Gibb's energy change may be estimated.

$$\begin{gathered} ∆G=∆G^{°}+R\times T\times InQ \\ ∆G=1.668\times {10}^3\frac{J}{mol}+8.314\frac{J}{mol\times K}\times 298K\times In\left(10.00\right) \\ ∆G=7372.820\frac{J}{mol}=7.373\frac{kJ}{mol} \end{gathered}$$

$\text{7.373kJ/mol}$is the$∆G$ .

Determining ∆Gwhen Q=0.100

(c )For$\text{Q = 0.100}$ , $∆G$is recalculated in the same way as in b).

$$\begin{gathered} ∆G=1.668\times {10}^3\frac{J}{mol}+8.314\frac{J}{mol\times K}\times 298K\times In\left(10.00\right) \\ ∆G=-4036.82\frac{J}{mol}=-4.0368\frac{kJ}{mol} \end{gathered}$$

$\text{- 4.0368kJ/mol}$is the$∆G$ .

Determining Q when ∆G = - 2.50kJ/mol

(d) Now, given that$∆G=-2.50kJ/mol$ , calculate Q in the other direction.

$$\begin{gathered} ∆G=∆G^{°}+R\times T\times InQ \\ InQ=\frac{∆G-∆G^{°}}{R\times T} \\ InQ=e^{InQ} \end{gathered}$$

- Using the given and derived values,

$$\begin{gathered} InQ=\frac{\left(\left(-2.50kJ/mol-\left(1.668kJ/mol\right)\right)\right)\times {10}^{-3}J/kJ}{8.314{\displaystyle \frac{J}{mol\times K}}\times 298K} \\ InQ=-1.683 \\ Q=0.1859=0.19 \end{gathered}$$

In these circumstances, the reaction quotient Q is $0.19$