Question

Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your answer numerically using two significant figures. hints

Answer 1

The equilibrium constant for isomerization reaction is $\boxed{9.615}$

Further Explanation:

The standard Gibbs free energy change in a reaction $\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)$ is the difference of sum of the standard free energies of formation of product molecules and sum of standard free energies of formation of reactant molecules at the standard conditions. The formula used to calculate the value of standard Gibbs free energy  change for a reaction $\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)$ is as follows:

$\Delta\text{G}_{\text{rxn}}^{\circ}=\sum\text{n}\Delta\text{G}_{\text{f}(\text{products})}^{\circ}-\sum\text{m}\Delta\text{G}_{\text{f}(\text{reactants})}^{\circ}$

Here, n is the stoichiometric coefficients of products, and m are the stoichiometric coefficients of reactants in a balanced chemical equation.

The formula to determine the relationship between change in standard Gibbs free energy $\left( \Delta{\text{G}^{\circ}} \right)$ and equilibrium constant $\left({\text{K}}\right)$ is given as follows:

${\Delta }}{{\text{G}}^{{^\circ }}} = - {\text{RTlnK}}$       ......(1)

Here,

$\Delta{\text{G}^{\circ}$ is the standard Gibbs free energy change.

${\text{R}$ is the gas constant.

${\text{T}}$ is the temperature in Kelvin.

${\text{K}}$ is the equilibrium constant.

The isomerization of glucose-1-phosphate to fructose-6-phosphate occurs in 2 steps:  

The reaction of step 1 is as follows:

${\text{glucose - 1 - phosphate}} \to {\text{glucose - 6 - phosphate}}$

                                       ......(2)

$\Delta{\text{G}^{\circ}_{1}$ for equation (2) is  $- 7.28\;{\text{kJ/mol}}$

The reaction of step 2 is as follows:

${\text{fructose - 6 - phosphate}} \to {\text{glucose - 6 - phosphate}}$

                                                   ......(3)

$\Delta{\text{G}^{\circ}_{2}$ for equation (3) is  $- 1.67\;{\text{kJ/mol}}$

Reverse the reaction of step 2.

${\text{glucose - 6 - phosphate}} \to {\text{frutcose - 6 - phosphate}}$

                                                ......(4)

$\Delta{\text{G}^{\circ}_{3}$ for equation (4) is $+ 1.67\;{\text{kJ/mol}}$

Add equation (1) and (3) to get the final equation.

${\text{glucose - 1 - phosphate}} \to {\text{frutcose - 6 - phosphate}}$

To calculate $\Delta {\text{G}}_{{\text{rxn}}}^{^\circ }}$, add $\Delta{\text{G}^{\circ}_{1}$ and $\Delta{\text{G}^{\circ}_{3}$ as follows:  

$\Delta{\text{G}^{\circ}_{\text{rxn}}=\Delta{\text{G}^{\circ}_{1}+\Delta{\text{G}^{\circ}_{3}$                       ......(5)

Substitute $- 7.28\;{\text{kJ/mol}}$ for $\Delta{\text{G}^{\circ}_{1}$ and $+ 1.67\;{\text{kJ/mol}}$ for $\Delta{\text{G}^{\circ}_{3}$ in equation (5).

$\begin{aligned}\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }} &= - 7.28\;{\text{kJ/mol + }} + 1.67\;{\text{kJ/mol}}\\{\text{}}&= - 5.61\;{\text{kJ/mol}}\\\end{aligned}$

For equilibrium constant (K), rearrange equation (1)

${\text{K}}={\text{e}}\frac{-\Delta{\text{G}}^{\circ}}{\text{RT}}$    ......(6)

Substitute $- 5.61\;{\text{kJ/mol}}$ for $\Delta{\text{G}^{\circ}$,$8.314\;{\text{J/mol}} \cdot {\text{K}}$ for R and $298\;{\text{K}}$ for T in equation (6)

$\begin{aligned} {\text{K}}&= {{\text{e}}^{\frac{{ - \left( { - 5.61\;{\text{kJ/mol}}} \right)}}{{\left( {8.314\;{\text{J/mol}} \cdot {\text{K}}} \right)\left( {\frac{{{\text{1J}}}}{{1000{\text{kJ}}}}} \right)\left( {298\;{\text{K}}} \right)}}}}\\&= {{\text{e}}^{2.2634}}\\&= 9.615\\\end{aligned}$

The equilibrium constant for the reaction is 9.615.

Learn more:

1. The change in standard gibbs free is for a reaction: brainly.com/question/10838453

2. Determination of the equilibrium constant for pure water: brainly.com/question/3467841

Answer details:

Grade: Senior Secondary School

Subject: Chemistry

Chapter: Chemical Equilibrium

Keywords: Standard Gibbs free energy, equilibrium, constant, glucose-1-phosphate and fructose-6-phosphate.

Answer 2

We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol

Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62