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Chapter 14: Problem 50

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$

Short Answer

Expert verified
To determine the rate law and rate constant of the sucrose decomposition reaction, first analyze the given concentration data and calculate the initial rate of the reaction using the formula \(Initial\:rate = \frac{\Delta[C12H22O_{11}]}{\Delta t}\). Then, determine the order of the reaction by trying different values of n in the rate law equation, Rate = \(k[C12H22O_{11}]^n\). Finally, use the initial rate and the initial concentration of sucrose to calculate the rate constant k using the rearranged rate law equation \(k = \frac{Rate}{[C12H22O_{11}]^n}\).

Step by step solution

01

Analyze the given data

First, let's analyze the given data for how the concentration of sucrose changes over time. We have: Time (min) | [C12H22O11] (M) -----------|--------------- 0 | 0.316 39 | 0.274 80 | 0.238 140 | 0.190 210 | 0.146
02

Calculate the initial rate of the reaction

To find the initial rate of the reaction, let's use the first two data points. We have: Initial concentration of sucrose: \([C12H22O11]_0 = 0.316\) M Concentration of sucrose at 39 minutes: \([C12H22O11]_{39} = 0.274\) M Then, we can calculate the initial rate of the reaction using the formula: Initial rate = \(\frac{\Delta [C12H22O11]}{\Delta t}\) Initial rate = \(\frac{0.274 - 0.316}{39 - 0}\) = \(-0.00108\) M/min The negative sign indicates that the concentration of sucrose decreases as time progresses.
03

Determine the order of the reaction

The rate law for a single reactant A would be in the form: Rate = \(k [A]^n\) In this case, the reactant A is sucrose. To find the order of the reaction (n), we can try different values of n and analyze how well they correlate with the given concentration data. For example, we can try first-order (n=1) and second-order (n=2) reactions. For a first-order reaction, the rate law becomes: Rate = \(k [C12H22O11]\) For a second-order reaction, the rate law becomes: Rate = \(k [C12H22O11]^2\) After trying different values of n, we will find the one that best describes the given data.
04

Calculate the rate constant k

Once the order of the reaction (n) is determined, we can use the initial rate and the initial concentration of sucrose to calculate the rate constant k. We can rearrange the rate law equation to solve for k: k = \(\frac{Rate}{[C12H22O11]^n}\) By plugging in the initial rate of the reaction and the initial concentration of sucrose, we can calculate the value of k for the sucrose decomposition reaction.
05

Conclusion

In order to determine the rate law and rate constant of the sucrose decomposition reaction, we analyzed the given concentration data and calculated the initial rate of the reaction. Then, we determined the order of the reaction and used the rate law equation to calculate the rate constant k. With this information, we have a complete understanding of the kinetics of this reaction under the given conditions.

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Most popular questions from this chapter

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with \(\mathrm{ICl}\) : $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(A\) and \(B\) ? What is the overall reaction order? (c) What are the units of the rate constant?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C}\) ? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?
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