Question

$\alpha$-D-glucose undergoes mutation to $\beta$-D-glucose in aqueous solution. If 298K there is 60% conversion, Calculate $\vartriangle G$for the reaction.

Answer 1

The free energy of a system can be defined as the maximum number of energy available to the system during a process which can be converted into useful work. Hence, we can also say that free energy is a measure of capacity to do useful work. $\vartriangle G = - {W_{max}}$

Complete step by step solution:
We know that the free energy is the capacity to do work. So, here we assume that the reaction happens in an equilibrium. Hence, we need a relation between the equilibrium constant and free energy.
We know that the free energy change $\left( {\vartriangle G} \right)$ is related to standard free energy $\left( {\vartriangle ^\circ {\text{G}}} \right)$ as follows,
$\vartriangle G = \vartriangle ^\circ G + 2.303RT\;{\text{log}}\left( Q \right)$
Where $R$is the ideal gas constant,
And $T$ is the temperature
$Q$ is the reaction quotient
Also we consider the equation is at equilibrium, then
$Q = K\;and\;\vartriangle G = 0$
Where K is the equilibrium constant. Applying this in the equation we get,
$\vartriangle ^\circ G = 2.303RTlog\left( K \right)$
Now, we have to find the $K$from the reaction.
In question it is given that $\alpha$-D-glucose undergoes mutation to $\beta$-D-glucose. Hence the reaction becomes,
$\alpha - D - glu{\text{cos}}e\; \rightleftharpoons {{\beta }} - {\text{D}} - {\text{glucose}}$
Let us assume that some, $x$ concentration of $\alpha - D - glu{\text{cos}}e$ is reacted and at equilibrium it is given that 60% of them has been converted to ${{\beta }} - {\text{D}} - {\text{glucose}}$. Hence we can say
$\alpha - D - glu{\text{cos}}e\; \rightleftharpoons {{\beta }} - {\text{D}} - {\text{glucose}}$
$x - 0.6x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.6x$

The concentration of $\alpha - D - glu{\text{cos}}e$ reduces by its 60%, i.e., $x - 0.6x$ and the concentration of ${{\beta }} - {\text{D}} - {\text{glucose}}$ increases by $0.6x$.
Now, we can find the equilibrium constant, K
It is the ratio of product of molar concentration of the products to product of molar concentration of the reactants, each raised to the power equal to its stoichiometric coefficient at constant temperature.
${\text{Equilibrium}}\;{\text{constant}},\;{\text{K}} = \dfrac{{0.6{\text{x}}}}{{{\text{x}} - 0.6{\text{x}}}} = \dfrac{{0.6{\text{x}}}}{{{\text{x}}\left( {1 - 0.6} \right)}}$
Cancelling x in numerator and denominator, we get
${{Equilibrium\;constant}},{{\;K}} = \dfrac{{0.6}}{{0.4}} = 1.5$
We also the value of $R = 8.314\;J/mol\;K$and $T = 298$, substituting this in the equation of free energy we get,
$\vartriangle ^\circ G = 2.303 \times 8.314 \times 298\;log\left( {1.5} \right)$
$\vartriangle ^\circ G = 1004.7\;J/mol$

Hence the free energy for the reaction is 1004.7 J/mol

Note: While choosing the value of R we have to be careful as it has many values depending upon the unit we are choosing. If we had chosen some other unit, we should use the respecting units in the solution also. Here we took it as 8.314 J/molK hence the answer is also in J.