Q65 E The hydrolysis of the sugar sucr... [FREE SOLUTION]

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Chapter 12: Q65 E (page 710)

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \({{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}{\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{ + }}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\) follows a first-order rate equation for the disappearance of sucrose: \({\bf{Rate = k}}\left( {{{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}} \right)\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
In neutral solution, \({\bf{k = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 27 掳C and \({\bf{8}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 37 掳C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 掳C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is\({\bf{1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\). How long will it take the solution to reach equilibrium at 27 掳C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Short Answer

Expert verified

The energy of activation for the reaction is 108.123kJ/mol.
The time required for the reaction to reach equilibrium is \(7.56 \times {10^6}\) days
The assumption that the reaction is irreversible simplifies calculation as we do not have to worry about the product being converted back to the reactants.

Step by step solution

Step 1: Calculation of Activation Energy

From the Arrhenius Equation, the rate of reaction at two different temperatures is given as

\({\bf{log}}\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = }}\frac{{{{\bf{E}}_{\bf{a}}}}}{{{\bf{2}}{\bf{.303*R}}}}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}{\bf{ - }}\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right)\)

where \({{\bf{k}}_{\bf{1}}}\) and \({{\bf{k}}_{\bf{2}}}\) are the rate constants at \({{\bf{T}}_{\bf{1}}}\) and \({{\bf{T}}_{\bf{2}}}\) where \({{\bf{T}}_{\bf{1}}}{\bf{ < }}{{\bf{T}}_{\bf{2}}}\). \({{\bf{E}}_{\bf{a}}}\)is the activation energy (in J) and R is the gas constant.

From the question, the ratio of \(\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = 1}}{\bf{.47}}\)

Replacing the values in the Arrhenius equation,

\(\begin{align}\log \frac{{8.5*{{10}^{ - 11}}}}{{2.1*{{10}^{ - 11}}}} &= \frac{{{E_a}}}{{2.303*8.314}}\left( {\frac{1}{{300}} - \frac{1}{{310}}} \right)\\{E_a} &= 0.607*2.303*8.314*\left( {\frac{{300*310}}{{310 - 300}}} \right)\\{E_a} &= 108123J/mol\\{E_a} &= 108.12kJ/mol\end{align}\)

Calculation of Frequency factor

The Arrhenius equation is given as \({\bf{k = A}}{e^{\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{{\bf{RT}}}}}}\) ,where A is the frequency factor, k is the rate constant, \({{\bf{E}}_{\bf{a}}}\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The frequency factor can be calculated by considering either rate constant value given. Considering the rate constant at 37^oC and replacing the values in the equation.

\({\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ = A}}{{\bf{e}}^{\frac{{{\bf{ - 108123}}}}{{{\bf{8}}{\bf{.314*310}}}}}}\)

\( \Rightarrow {\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ = A}}{{\bf{e}}^{{\bf{ - 41}}{\bf{.95}}}}\)

\( \Rightarrow {\bf{A = }}\frac{{{\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}}}{{{\bf{1}}{\bf{.65*1}}{{\bf{0}}^{{\bf{ - 19}}}}}}\)

\( \Rightarrow {\bf{A = 1}}{\bf{.408*1}}{{\bf{0}}^{\bf{8}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Calculation of Rate constant

Again, from the Arrhenius equation

\({\bf{k = A}}{e^{\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{{\bf{RT}}}}}}\)

where A is the frequency factor (calculated previously), k is the rate constant, \({{\bf{E}}_{\bf{a}}}\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

At 47^oC, the rate can be calculated as

\(k = 1.408*{10^8}*{e^{\frac{{ - 108123}}{{8.314*320}}}}\)

\(\begin{align}\Rightarrow k &= 1.408*{10^8}*{e^{ - 40.6}}\\ \Rightarrow k &= 3.15*{10^{ - 10}}\end{align}\)

Integrated First law rate equation

The integral rate equation for the first-order reaction is given as follows.

\({\bf{k = }}\frac{{{\bf{2}}{\bf{.303}}}}{{\bf{t}}}{\bf{log}}\frac{{{{\left( {\bf{A}} \right)}_{\bf{o}}}}}{{\left( {\bf{A}} \right)}}\)

Where (A)₀ is the initial concentration and (A) is the concentration at time t.

Calculation of required time

Here, \({\left( {\bf{A}} \right)_{\bf{0}}}{\bf{ = 0}}{\bf{.150M}}\) and \(\left( {\bf{A}} \right){\bf{ = 1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}\)^.Replacing the values in the above equation,

At k=\({\bf{2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}\) at T=27⁰C

\(\begin{align}\Rightarrow t &= \frac{{2.303}}{k}\log \frac{{{{\left( A \right)}_o}}}{{\left( A \right)}}\\ \Rightarrow t &= \frac{{2.303}}{{2.1*{{10}^{ - 11}}}}\log \frac{{0.150}}{{1.65*{{10}^{ - 7}}}}\\ \Rightarrow t &= 6.53*{10^{11}}{s^{}}\\ \Rightarrow t &= \frac{{6.53*{{10}^{11}}}}{{60*60*24}}days\\ \Rightarrow t &= 7.56*{10^6}days\end{align}\)

The assumption that the reaction is irreversible simplifies the calculation as we do not have to worry about the product being converted back to the reactants.

Step 6: Assumption

The assumption that the reaction is irreversible simplifies the calculation as we do not have to worry about the product being converted back to the reactants.