Question

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: $$ \begin{array}{rl} \hline \text { Time }(\mathrm{min}) & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M)} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?(\mathbf{b})\) What is the rate constant? (c) Using this rate constant, calculate the concentration of sucrose at 39,80,140 , and 210 min if the initial sucrose concentration was \(0.316 \mathrm{M}\) and the reaction was zero order in sucrose.

Short answer

(a) The reaction is first-order with respect to sucrose concentration. (b) The rate constant is \(3.131 \times 10^{-3}\) min^{-1}. (c) The concentration of sucrose at the given times: 39 min: 0.274 M, 80 min: 0.238 M, 140 min: 0.190 M, 210 min: 0.146 M.

Step-by-step solution

Identify the rate law for the reaction

The rate law indicates the relationship between the rate of a chemical reaction and the concentration of the reactants. We need to determine the order of the reaction with respect to the sucrose concentration. Two potential rate laws can be considered: 1) First-order reaction: Rate = k [Sucrose] 2) Second-order reaction: Rate = k [Sucrose]^2 In both cases, k is the rate constant. We will analyze the data and determine which rate law best matches the given data.

Calculate the change in concentration over time

We need to calculate the change in sucrose concentration over time. Using the given data, we can create a table that displays the time, sucrose concentration, change in sucrose concentration, and ratio of change in concentration to initial concentration. $$ \begin{array}{|c|c|c|c|} \hline \text{Time(min)} & [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}](\mathrm{M}) & \Delta [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}] & \frac{\Delta [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}]}{[C_{12}H_{22}O_{11}]_0} \\ \hline 0 & 0.316 & 0 & \\ 39 & 0.274 & -0.042 & 0.133 \\ 80 & 0.238 & -0.078 & 0.247 \\ 140 & 0.190 & -0.126 & 0.398 \\ 210 & 0.146 & -0.170 & 0.538 \\ \hline \end{array} $$

Test for first-order reaction

First, we will test if the reaction is of first order with respect to sucrose concentration. For first-order reactions, we have the equation: \[ t = \frac{1}{k} \ln \frac{[C_{12}H_{22}O_{11}]_0}{[C_{12}H_{22}O_{11}]_t} \] We can determine the average value of k for different time periods and see if it remains consistent: 1) From 0 to 39 minutes: \[ k_1 = \frac{\ln(\frac{0.316}{0.274})}{39} = 3.139 \times 10^{-3}\ \text{min}^{-1} \] 2) From 39 to 80 minutes: \[ k_2 = \frac{\ln(\frac{0.274}{0.238})}{80 - 39} = 3.144 \times 10^{-3}\ \text{min}^{-1} \] 3) From 80 to 140 minutes: \[ k_3 = \frac{\ln(\frac{0.238}{0.190})}{140 - 80} = 3.111 \times 10^{-3}\ \text{min}^{-1} \] It can be observed that the rate constant, k, remains reasonably consistent over different time periods. Thus, we can affirm that the reaction is first-order with respect to sucrose concentration.

Calculate the average rate constant

We can now calculate the overall average rate constant, k, for the first-order reaction: \[ k_{\text{avg}} = \frac{k_1 + k_2 + k_3}{3} = \frac{3.139 \times 10^{-3} + 3.144 \times 10^{-3} + 3.111 \times 10^{-3}}{3} = 3.131 \times 10^{-3}\ \text{min}^{-1} \] The reaction thus has the rate law: Rate = \(k_{\text{avg}} \times\) [Sucrose] Answer: (a) The reaction is first-order with respect to sucrose concentration. (b) The rate constant is \(3.131 \times 10^{-3}\) min^{-1}.

Calculate the concentration of sucrose at given times

Since we know that reaction's rate constant and the initial sucrose concentration, we can use the first-order reaction equation to calculate the concentration of sucrose at given times: (a) t = 39 min: \[ [C_{12}H_{22}O_{11}]_{39} = [C_{12}H_{22}O_{11}]_0 \times e^{-k \times t} = 0.316 \times e^{-3.131 \times 10^{-3} \times 39} = 0.274\ \mathrm{M} \] (b) t = 80 min: \[ [C_{12}H_{22}O_{11}]_{80} = 0.316 \times e^{-3.131 \times 10^{-3}\times 80} = 0.238\ \mathrm{M} \] (c) t = 140 min: \[ [C_{12}H_{22}O_{11}]_{140} = 0.316 \times e^{-3.131 \times 10^{-3}\times 140} = 0.190\ \mathrm{M} \] (d) t = 210 min: \[ [C_{12}H_{22}O_{11}]_{210} = 0.316 \times e^{-3.131 \times 10^{-3}\times 210} = 0.146\ \mathrm{M} \] Answer: (c) The concentration of sucrose at the given times can be calculated as follows: 39 min: 0.274 M, 80 min: 0.238 M, 140 min: 0.190 M, 210 min: 0.146 M.

Key concepts

First-Order Reaction

A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one reactant. This means that if you double the concentration of the reactant, the rate of the reaction will also double. The formula for a first-order reaction is: \[ \text{Rate} = k [\text{Reactant}] \] where:

• Rate is the speed of the reaction.
• k is the rate constant.
• [Reactant] is the concentration of the reactant.

First-order reactions are common in chemical kinetics, as many unimolecular reactions fall into this category. They are characterized by a constant half-life, which is the time required for the reactant concentration to decrease to half its initial value. For example, in the hydrolysis of sucrose, we observe that the reaction is first-order, as the decomposition of sucrose into glucose and fructose follows this principle. This is verified by checking the consistency of the rate constant over time with different concentrations.

Rate Constant

The rate constant, denoted as \(k\), is a crucial part of the reaction kinetics equation. It essentially gives you an idea of the speed of a reaction at a given temperature and is specific to each reaction. It appears in almost all rate laws, reflecting the nature and conditions of the reaction. The units of the rate constant vary depending on the order of the reaction:

• - For first-order reactions, the unit is \(\text{time}^{-1}\), such as \(\text{min}^{-1}\).

In the context of sucrose hydrolysis, knowing the rate constant helps predict how quickly sucrose decomposes under certain conditions, such as in an acidic solution at a specified temperature. Understanding how to calculate and utilize the rate constant is fundamental for predicting reaction behavior and timing, which is particularly important in industrial and laboratory settings. For instance, for the sucrose hydrolysis reaction discussed, the calculated average rate constant is \(3.131 \times 10^{-3}\ \text{min}^{-1}\), providing a stable measure of the reaction rate across varying sucrose concentrations.

Sucrose Hydrolysis

Sucrose hydrolysis is a specific reaction where sucrose, a disaccharide sugar, is broken down into its monosaccharide components, glucose and fructose. This reaction occurs in the presence of an acid, such as hydrochloric acid, and involves the addition of a water molecule. The chemical reaction can be represented simply as: \[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \] where sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) reacts with water (\( \text{H}_2\text{O} \)) to form glucose and fructose, both with the formula \(\text{C}_6\text{H}_{12}\text{O}_6\). This reaction is important in many biological and industrial processes. It is a classical example of bimolecular nucleophilic substitution. However, its behavior is tracked using reaction kinetics concepts, making it a first-order reaction in terms of the sucrose concentration. The monitoring of sucrose hydrolysis is important as it can affect sweetness and affect the end product in food processing and preparation. By understanding the kinetics, including factors like the rate constant, industries can effectively control and optimize conditions for desired results.