## Related questions with answers

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$ follows a first-order rate law for the disappearance of sucrose: rate $=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$ (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, $k=2.1 \times 10^{-11} \mathrm{~s}^{-1}$ at $27^{\circ} \mathrm{C}$ and $8.5 \times 10^{-11} \mathrm{~s}^{-1}$ at $37^{\circ} \mathrm{C}$. Determine the activation energy, the frequency factor, and the rate constant for this equation at $47^{\circ} \mathrm{C}$ (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of $0.150 \mathrm{M}$ reaches equilibrium, the concentration of sucrose is $1.65 \times 10^{-7} \mathrm{M}$. How long will it take the solution to reach equilibrium at $27^{\circ} \mathrm{C}$ in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Solution

VerifiedTo calculate for the activation energy of the reaction, we can use the Arrhenius equation:

$\begin{aligned}k = Ae^{\frac{-E_{a}}{RT}}\end{aligned}$

where $k$ stands for the rate constant, $A$ stands for the frequency factor, $E_{a}$ stands for the activation energy, $R$ stands for the gas constant and $T$ stands for the temperature.

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