/*! 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 71 One proposed mechanism for the f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.

Short Answer

Expert verified
The rate of reaction expression for the formation of the DNA double helix is Rate = k[S1][S2].

Step by step solution

01

Identify the Rate Determining Step

The rate determining step is the slowest step in a reaction mechanism, which here is given as the second step: \( (S_{1}: S_{2})^{*} \longrightarrow S_{1}: S_{2} \). The rate of reaction would therefore be initially defined as, Rate = k[(S1: S2)*] where 'k' is the rate constant.
02

Recognize and Substitute the Intermediate

Note that \((S_{1}: S_{2})^{*}\) is an intermediate for it appears in the mechanism but not in the overall reaction. By using the fast step, \( (S_{1}+S_{2})=\left(S_{1}: S_{2}\right)^{*} \), we can express \((S_{1}: S_{2})^{*}\) in terms of S1 and S2 since intermediates must not appear in the rate law. Thus, replace \((S_{1}: S_{2})^{*}\) with [S1][S2], assuming the fast step is at equilibrium.
03

Write Rate Law

Substitute the intermediate into our rate law expression to obtain the final rate law: Rate = k[S1][S2]. This rate law indicates that the rate of helix formation is proportional to the concentration of strand S1 and strand S2.

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 91Ó°ÊÓ!

Key Concepts

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

Rate Determining Step
Understanding the 'rate determining step' is crucial when studying chemical kinetics. It's akin to the slowest runner in a relay race — no matter how fast the other runners are, the overall speed of the team is limited by the slowest member. Similarly, in chemistry, the rate of the entire reaction is controlled by the slowest step in the reaction mechanism, often referred to as the rate determining step.

In the context of the given exercise, the DNA helix formation involves a fast step followed by a slower one. The slow step, the transformation of the unstable helix \( (S_{1}: S_{2})^{*} \) into a stable double helix \( S_{1}: S_{2} \) is the rate determining step. Therefore, when writing the rate expression for the reaction, we focus on the concentrations of the reactants participating in this step. It is important to realize that a reaction can comprise multiple steps, but the overall reaction rate is dictated by this single bottleneck — the rate determining step.
Reaction Intermediate
Another interesting aspect of reaction mechanisms is the 'reaction intermediate'. In the storyline of a reaction, intermediates are like characters that play a role in the plot but don't appear in the beginning or the end of the story. They are produced and then consumed during the course of the reaction sequence.

In the given DNA helix formation scenario, the unstable helix \( (S_{1}: S_{2})^{*} \) acts as a reaction intermediate. It is formed in the first, fast step and then used up in the second, slower step. When it comes to writing the rate expression, intermediates should not be included directly. Instead, we express them in terms of the reactants that form them. Here, the assumption that the fast step is at equilibrium allows us to substitute the intermediate \( (S_{1}: S_{2})^{*} \) with the concentrations of \( S_{1} \) and \( S_{2} \) — the reactants that combine to form it. This ensures our rate law includes only species that are present at the beginning of the reaction.
Rate Constant
Delving into the reaction kinetics, the 'rate constant' emerges as a pivotal figure. It is the factor that quantifies the speed of a reaction at a given temperature and is denoted by the symbol 'k'. Think of it as the chef's secret ingredient that influences the pace at which a recipe is prepared. In our exercise, the rate constant 'k' tells us how rapidly the DNA double helix is formed once the reactants \( S_{1} \) and \( S_{2} \) are combined.

The value of the rate constant depends on various factors, such as temperature and the presence of a catalyst, but it is independent of the concentration of reactants. When we write out the rate law, \( Rate = k[S_{1}][S_{2}] \), 'k' multiplies the concentration terms, signifying its direct impact on the rate. It's a unique value specific to each reaction under particular conditions, characterizing the intrinsic reactivity of the process.

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

In the first-order reaction \(A \longrightarrow\) products, \([\mathrm{A}]=0.816 \mathrm{M}\) initially and \(0.632 \mathrm{M}\) after \(16.0 \mathrm{min}.\) (a) What is the value of the rate constant, \(k ?\) (b) What is the half-life of this reaction? (c) At what time will \([\mathrm{A}]=0.235 \mathrm{M} ?\) (d) What will [A] be after 2.5 h?

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$

Acetoacetic acid, \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH},\) a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\mathrm{CO}_{2}(\mathrm{g}).\) $$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}(\mathrm{aq}) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})$$ This first-order decomposition has a half-life of 144 min. (a) How long will it take for a sample of acetoacetic acid to be \(65 \%\) decomposed? (b) How many liters of \(\mathrm{CO}_{2}(\mathrm{g}),\) measured at \(24.5^{\circ} \mathrm{C}\) and 748 Torr, are produced as a \(10.0 \mathrm{g}\) sample of \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}\) decomposes for 575 min? [Ignore the aqueous solubility of \(\mathrm{CO}_{2}(\mathrm{g}) \cdot \mathrm{l}.\)

The following data were obtained for the dimerization of 1,3 -butadiene, \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g}),\) at 600 K: \(t=0 \min ,\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0169 \mathrm{M} ; 12.18 \mathrm{min}, 0.0144 \mathrm{M} ; 24.55 \mathrm{min}, 0.0124 \mathrm{M} ; 42.50 \mathrm{min}, 0.0103 \mathrm{M}, 68.05 \min , 0.00845 \mathrm{M}.\) (a) What is the order of this reaction? (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.00423 \mathrm{M} ?\) (d) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0050 \mathrm{M} ?\)

Briefly describe each of the following ideas, phenomena, or methods: (a) the method of initial rates; (b) activated complex; (c) reaction mechanism; (d) heterogeneous catalysis; (e) rate-determining step.

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.