1. Hydrolysis of sucrose give Sucrose+H_2O +Fructose Equilibrium … - Vidyadip

Question

Hydrolysis of sucrose give

$\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$

Equilibrium constant $K_c$ for the reaction is $2 \times 10^{13}$ at $300K$. Calculate $\Delta\text{G}^\circ\text{ at }300\text{K.}(\log 2=0.3010)$

The concentration of hydrogen in two sample of soft drinks A and B $4.0 \times 10^{-7}$ and $3.2 \times 10^{-6}$ respectively. Which of these two soft drinks has higher pH?

✓

Answer

$\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$

$\text{K}_{\text{c}}2\times10^{13},\text{T}=300\text{K},\Delta\text{G}^\circ=?$

$\Delta\text{G}^\circ=-2.30\text{ RT }\log\text{K}_{\text{c}}$

$=-2.303\times8.314\times300\log2\times10^{13}$

$=-19.147\times300(\log2+\log^{13})$

$=\frac{-19.147\times300\times13.3010\text{J}}{1000}$

$=-76.402\text{kJ mol}^{-1}$

For A,

$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log4.0\times10^{-7}$

$=-\log4.0-\log10^{-7}$

$=-0.6021+7.000=6.3979$

For B,

$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log3.2\times10^{-6}$

$=-\log3.2-\log10^{-6}$

$=-0.5050+6.000=5.4950$

pH of 'A' is higher than 'B'.

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