Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$

Step by step solution

01

Determining the order of reaction

Let's denote the rate of reaction with \(k\). Then the rate law will have the form: \( -\frac{d[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]}{dt} = k[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]^{n} \), where \(n\) is the order of the reaction. Observing the change in concentration over time in the table, it's possible to calculate the rate constant \(k\) for each interval. For a first order reaction, plotting \(ln ([\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}] )\) against time will give a straight line. If such relationship holds, the order of the reaction is 1.

02

Calculating the reaction time

For a first order reaction, the time taken for a certain percentage of a reactant to be consumed can be calculated by using the formula \( t = \frac{1}{k} \ln \left( \frac{[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0}{[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}] } \right) \), where \(t\) is the time, \(k\) is the rate constant and \( [\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}] \) is the concentration of sucrose at any time, and \( [\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0 \) is the initial concentration. For 95% hydrolysis, we need to replace \([\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]\) with \(0.05[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0\), then solve for \(t\).

03

The role of Water

Water is present in large excess in this reaction. The concentration of water essentially remains constant during the reaction and consequently does not influence the rate of the reaction. Thus, the concentration of water does not appear in the rate law.

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