Question

The key step in the metabolism of glucose for energy is the isomerization of glucose−6−phosphate (G6P) to fructose−6−phosphate (F6P): G6P ⇌ F6P K = 0.510, at 298K. A. Calculate ΔG when Q, the [F6P]/[G6P] ratio, equals 10.0.B. Calculate ΔG when Q = 0.100.C. Calculate Q if ΔG = 2.50 kj/mol.

Answer 1

Answer :

(A) The value of $\Delta G_{rxn}$ is, 7.37 kJ/mol

(B) The value of $\Delta G_{rxn}$ is, -4.03 kJ/mol

(C) The value of Q for the reaction is, 1.39

Explanation :

The given balanced chemical reaction is,

$G6P\rightarrow F6P$

The expression for reaction quotient will be :

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

Given:

$Q=\frac{[F6P]}{[G6P]}=10.0$

k = 0.510

First we have to calculate the $\Delta G^o$ for the reaction.

$\Delta G^o=-RT\ln k$

$\Delta G^o=-(8.314J/mol.K)\times (298K)\times \ln (0.510)$

$\Delta G^o=1668.259J/mol=1.67kJ/mol$

Part A:

Now we have to calculate the value of $\Delta G_{rxn}$.

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$    ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction  = ?

$\Delta G_^o$ =  standard Gibbs free energy  = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 10.0

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

$\Delta G_{rxn}=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (10.0)$

$\Delta G_{rxn}=7.37kJ/mol$

Thus, the value of $\Delta G_{rxn}$ is, 7.37 kJ/mol

Part B:

Now we have to calculate the value of $\Delta G_{rxn}$.

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$    ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction  = ?

$\Delta G_^o$ =  standard Gibbs free energy  = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 0.100

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

$\Delta G_{rxn}=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (0.100)$

$\Delta G_{rxn}=-4.03kJ/mol$

Thus, the value of $\Delta G_{rxn}$ is, -4.03 kJ/mol

Part C:

Now we have to calculate the value of Q.

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$    ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction  = 2.50 kJ/mol

$\Delta G_^o$ =  standard Gibbs free energy  = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = ?

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

$2.50kJ/mol=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (Q)$

$Q=1.39$

Thus, the value of Q for the reaction is, 1.39