Question

$\alpha$- D glucose undergoes muta rotation to $\beta$-D glucose in aqueous solution.If at 298 K there is 60% conversion. Calculate $△G^0$ for the reaction.

Answer 1

Solution:-

$\underset{\text{At equillibrium}}{\underset{\text{Initially}}{}}\underset{1-\alpha }{\underset{1}{\alpha -\text{Glucose}}}\rightleftharpoons \underset{\alpha }{\underset{0}{\beta -\text{Glucose}}}$

Given that at $298K$, the conversion is $60\%$.

$\therefore \alpha =\frac{60}{100}=0.6$

Therefore,

$[\alpha -\text{Glucose}]=1-\alpha =0.4$

$[\beta -\text{Glucose}]=\alpha =0.6$

$\therefore K_c=\frac{[\alpha -\text{Glucose}]}{[\beta -\text{Glucose}]}$

$\Rightarrow K_c=\frac{0.4}{0.6}=\frac{2}{3}$

Now, as we know that,

$\Delta G^0=-2.303RT\log K_c$

$\Rightarrow \Delta G^0=-2.303\times 8.314\times 298\times \log \frac{2}{3}J$

$\Rightarrow \Delta G^0=5.7\times {10}^{-3}\times (\log 2-\log 3)J$

$\Rightarrow \Delta G^0=5.7\times {10}^{-3}\times (-0.176)=1.0032kJ$

Hence the value of $\Delta G^0$ for the reaction will be $1.0032kJ$