We’re being asked to determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C.

We’re given the rate constant at another temperature and the activation energy of the reaction. This means we need to use the ** two-point form of the Arrhenius Equation**:

$\overline{){\mathbf{ln}}{\mathbf{}}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}}{\mathbf{[}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{]}}}$

where **k _{1}** = rate constant at T

**Let's first calculate the activation energy:**

**T _{1} **= 27°C + 273.15

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C_{12}H_{22}O_{11} + H_{2}O ⟶ C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6} follows a first-order rate equation for the disappearance of sucrose: Rate = k[C_{12}H_{22}O_{11}] (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 × 10^{−11} s ^{−1} at 27 °C and 8.5 × 10^{−11} s ^{−1} at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).

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