Problem 77 - Chemistry: An Atoms-Focused Approach

Question

The metabolism of table sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ) begins with the hydrolysis of the disaccharide to glucose and fructose (both \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) ): $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$$ The kinetics of the reaction were studied at \(24^{\circ} \mathrm{C}\) in a reaction system with a large excess of water, so the reaction was pseudo first order in sucrose. Determine the rate law and the pseudo-first-order rate constant for the reaction from the following data: $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ 0 & 0.562 \\ \hline 612 & 0.541 \\ \hline 1600 & 0.509 \\ \hline 2420 & 0.484 \\ \hline 3160 & 0.462 \\ \hline 4800 & 0.442 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Question: Determine the rate law and the pseudo-first-order rate constant for the hydrolysis reaction of sucrose in this experiment. Answer: The rate law for this pseudo-first-order reaction is given by: $$\textrm{rate} = k[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]$$ To find the rate constant k, plot the natural logarithm ln of sucrose concentration against time (t). The slope of the linear regression line represents the negative value of the pseudo-first-order rate constant \(-k\). Determine the rate constant k from the slope and include it in the rate law.

Step 1: Identify the reaction order

Since it is given that the reaction is pseudo-first order in sucrose, the rate law will take the form: $$\textrm{rate} = k[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]$$ where k is the pseudo-first-order rate constant that needs to be determined.

Step 2: Apply the integrated rate law

For a pseudo-first-order reaction, the integrated rate law is given by: $$\ln[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]=-\textrm{k}\textrm{t}+\ln[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]_0$$ where \([\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]_0\) is the initial concentration of sucrose and t is the time.

Step 3: Calculate the natural logarithm ln of sucrose concentrations

In order to apply the integrated rate law, let's compute the natural logarithm ln of sucrose concentration for each time point given in the data: $$\begin{array}{ccc} \text { Time (s) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} & {\ln[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]} \\ 0 & 0.562 & 0 \\ \hline 612 & 0.541 & -0.036 \\ \hline 1600 & 0.509 & -0.086 \\ \hline 2420 & 0.484 & -0.127 \\ \hline 3160 & 0.462 & -0.161 \\ \hline 4800 & 0.442 & -0.207 \\ \hline \end{array}$$

Step 4: Plot ln[sucrose concentration] against time

Now, plot the natural logarithm ln of sucrose concentration against time (t). The slope of the linear regression line of this graph will represent the negative value of the pseudo-first-order rate constant \(-k\). $$\textrm{Slope} (\textrm{of ln}[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}] \textrm{vs. t}) = -k$$

Step 5: Determine the rate constant k and the rate law

After analyzing the graph or using data analysis tools to determine the slope, you will get the negative value of the rate constant \(-k\) and then calculate the actual rate constant k by changing the sign. Finally, the rate law will take the form: $$\textrm{rate} = k[\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}]$$ using the value of k calculated.

Key Concepts

Rate Laws

In chemical kinetics, rate laws describe how the rate of a chemical reaction depends on the concentration of the reactants. Knowing the rate law is fundamental to understanding how a reaction progresses over time. In our exercise, the reaction for the hydrolysis of sucrose is analyzed to determine its rate law. The goal is to understand how the concentration of sucrose affects the overall rate of reaction. The rate law for a reaction is typically expressed as:

• \( \text{rate} = k[A]^m[B]^n \)
• Here, \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants \( A \) and \( B \).

In the case of sucrose hydrolysis studied here, you deal with a situation where one reactant is in large excess (water), simplifying the rate law to focus primarily on sucrose. Thus, the rate law is expressed as:

• \( \text{rate} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}] \)
• This indicates a pseudo-first-order reaction where the rate depends linearly on sucrose concentration.

Reaction Order

The reaction order is a key component of reaction rate laws and indicates the power to which the concentration of a reactant is raised. It defines how changes in concentration affect the rate of the reaction. For each reactant, the order is denoted by its exponent in the rate law expression.

• For the sucrose hydrolysis, the reaction is specified as pseudo-first order.
• This pseudo-order is derived from the large excess of water, rendering the concentration of water effectively constant.

In conventional terms, first-order reactions have rate laws of the form:

• \( \text{rate} = k[A] \)
• This implies that the reaction rate changes directly with the concentration of the reactant \( A \).

In our exercise, the focus is placed on the sucrose concentration as the determining factor for reaction kinetics, despite the presence of other reactants. This approach simplifies analysis and measurements.

Pseudo-first-order Reaction

Pseudo-first-order reactions occur when one reactant is in a large excess compared to others, causing its concentration to remain almost constant throughout the reaction. This simplifies rate calculations and data analysis by allowing the reaction to be treated as though it is first-order with respect to the reactant of interest.

• In our exercise, the hydrolysis of sucrose in a large amount of water is studied as a pseudo-first-order reaction.
• Since the concentration of water does not change appreciably, it simplifies the situation by transforming a potentially more complex reaction order into a simpler first-order scenario relative to sucrose.

This means that reaction rate is primarily influenced by changes in sucrose concentration, even though water is a reactant but is present in such excess it doesn't appear in the rate law explicitly. Understanding this concept ensures accurate data interpretation and helps streamline process kinetic studies.

Integrated Rate Law

The integrated rate law provides a mathematical expression that relates reactant concentration to time for a simplified understanding of reaction progress. For first-order reactions, the integrated rate law is particularly useful as it allows determination of the rate constant \( k \) more directly through linear analysis.The integrated rate law for a first-order reaction is:

• \( \ln[A] = -kt + \ln[A]_0 \)
• Here, \([A]_0\) is the initial concentration of the reactant, \( t \) is time, and \( \ln[A] \) is the natural logarithm of concentration at time \( t \).

In the exercise, this equation helps us determine the rate constant by plotting \( \ln[\text{C}_{12}\text{H}_{22}\text{O}_{11}] \) against time \( t \). The slope of this plot equals \(-k\), which leads us to calculate the rate constant directly by finding the slope's value. Such graphical and algebraic approaches enable a deeper understanding of reaction dynamics over time.