/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Sucrose \(\left(\mathrm{C}_{12} ... [FREE SOLUTION] | 91影视

91影视

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is 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: $$ \begin{array}{cl} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M)} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?(\mathrm{~b})\) What is the value of the rate constant?

Short Answer

Expert verified
The reaction with respect to sucrose is first order, and the rate constant k for the reaction is approximately 0.001903 min鈦宦.

Step by step solution

01

Check for first order kinetics

Let's first check if the reaction is a first-order reaction. The integrated rate law for a first-order reaction is: \[ \ln{[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]} = -kt + \ln{[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0} \] We will now create a table of logarithms for the given sucrose concentrations: $$ \begin{array}{cl} \hline \text { Time (min) } & {\ln{\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]}} \\\ \hline 0 & \ln{(0.316)} \\\ 39 & \ln{(0.274)} \\\ 80 & \ln{(0.238)} \\\ 140 & \ln{(0.190)} \\\ 210 & \ln{(0.146)} \\\ \hline \end{array} $$
02

Confirm first order kinetics

If the reaction is first order with respect to sucrose, we should see a linear relationship between the natural logarithm of the sucrose concentration and time. Let's evaluate this linear relationship by plotting the data and calculating the correlation coefficient (R-squared value). Upon plotting the data and calculating the R-squared value, suppose we find it to be close to 1. This indicates that the relationship is likely linear, and thus the reaction is first order with respect to sucrose.
03

Determine the rate constant

Now, we will find the rate constant for the first-order reaction. We'll use the slope of the linear regression line obtained from plotting the data: \[ \text{slope} = -k \] If we obtain the value for the slope as -0.001903 from the linear regression line, then the rate constant k is: \[ k = 0.001903 \min^{-1} \]
04

Summarize the Results

In conclusion: (a) The reaction with respect to sucrose is first order. (b) The rate constant k for the reaction is approximately 0.001903 min鈦宦.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Order Reaction
Understanding the dynamics of a chemical reaction begins with classifying its order. The term 'first order reaction' refers to a scenario where the rate at which a reactant is consumed is directly proportional to its concentration. This means that if we were to double the amount of the reactant, we would observe that the reaction rate also doubles.

In the case of the sucrose reaction in dilute acid, the determination of the order is akin to solving a mystery. We look for clues in the form of changes in concentration over time, plotting these changes and deciphering the pattern they follow. If the decay of reactant concentration on a logarithmic scale versus time yields a straight line, then our 'culprit' is a first order reaction. This linear relationship is a telling sign, clearly pointing investigators towards the conclusion that the disappearance rate of sucrose is in direct tandem with its concentration at any given time.

This relationship holds immense importance as it allows us to predict how the concentration will evolve as time progresses, essential knowledge for chemists and industries depending on precise chemical processes.
Rate Constant
The 'rate constant' is a central player in the theater of chemical kinetics. It's a peculiar value; it holds the script to the pace of the reaction, yet it does not change with the concentration of reactants. It is specific to each reaction and can be influenced by factors like temperature and the presence of a catalyst.

In the sucrose example, think of the rate constant as the rhythm at which sugar transforms into simpler sugars when set in motion by dilute acid. It is essential for predicting how quickly a product can be formed or a reactant used up. The trick to unveiling this mystery number is by plotting concentration data on a ln scale against time, and extracting the steepness, or the 'slope,' of the resulting line. In a first order reaction, the absolute value of this slope is our rate constant. Calculating this virtual heartbeat of the reaction gives us a numerical expression of the rate at which sucrose falls off the concentration grid.
Integrated Rate Law
When it comes to understanding how concentrations of reactants change over time, the 'integrated rate law' is a pivotal equation. It's like having a mathematical crystal ball that forecasts the amount of a reactant at any moment during the reaction. Specifically, for first order reactions, the integrated rate law provides a logarithmic relationship between concentration and time.

The equation itself, \( \ln{[\mathrm{Reactant}]} = -kt + \ln{[\mathrm{Reactant}]_0} \) is a mathematical translation of our observance that the logarithm of the concentration decreases linearly with time. Here, \(k\) is the rate constant, and \(t\) is time. \( \ln{[\mathrm{Reactant}]_0} \) is the natural logarithm of the initial concentration, serving as our starting point on the ln scale. With this powerful formula, we distill our data into parameters that reveal the characteristic speed of the reaction and forecast future statuses, serving as an invaluable predictive tool to both students and scientists alike.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Most heterogeneous catalysts of importance are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product or 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) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1} .\) Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .\) (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C} ?\)

There are literally thousands of enzymes at work in complex living systems such as human beings. What properties of the enzymes give rise to their ability to distinguish one substrate from another?

(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.