Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of rate determining.

Short Answer

Expert verified
The rate-determining step is the slowest step of a reaction, limiting the overall rate.

Step by step solution

01

Understanding the Reaction Steps

In a multi-step chemical reaction, several individual steps occur in sequence to convert reactants into products. These steps can have different reaction rates, which is how quickly they proceed.
02

Identifying the Slowest Step

The rate-determining step (RDS) is the slowest step in a reaction mechanism. Even if other steps are faster, the overall rate of the reaction cannot exceed the rate of this slowest step because subsequent steps must wait for it to complete before proceeding.
03

Everyday Analogy: The Traffic Example

Imagine driving through a series of traffic lights along a road. If one light is particularly slow to change (taking a long period before allowing traffic to pass), it will dictate the overall time it takes for a convoy of cars to get through the series of lights. This slow light is similar to the rate-determining step in a chemical reaction.

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 Vaia!

Key Concepts

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

Understanding Reaction Mechanism
When we talk about the reaction mechanism, we're referring to the step-by-step journey that reactants undergo to become products. Think of it like a play, with each part having its own scene in the larger narrative. Each "scene" or step involves the breaking and forming of chemical bonds. This sequence of steps maps out how the transformation takes place.

In a multi-step reaction, the mechanism isn't just one move, but rather a series of small moves. Here are some key points to remember about reaction mechanisms:
  • Each step involves a different chemical process, such as bond cleavage or electron transfer.
  • Intermediate species might form and dissipate quickly.
  • Understanding these steps helps in predicting the behavior of the reaction.
Like solving a puzzle, knowing the reaction mechanism helps chemists understand how different factors affect the overall reaction.
Defining Reaction Rate
The reaction rate tells us how quickly or slowly a reaction proceeds. It's like the speed of a car in a race – some reactions speed along quickly, while others lumber along like a slow-moving vehicle. This speed depends on various factors:
  • The concentration of reactants: More reactant molecules mean more frequent collisions, increasing the rate.
  • Temperature: Higher temperature usually speeds up reactions, as molecules move faster and collide more intensely.
  • The presence of catalysts: Catalysts can provide alternative pathways, lowering the activation energy required, thus increasing the rate.
  • The rate-determining step: Regardless of other steps, this slowest step ultimately limits how fast the reaction can go.
Knowing the reaction rate helps in controlling industrial processes and can be crucial in designing efficient chemical syntheses.
Exploring Multi-Step Reactions
Multi-step reactions are like a relay race, where each runner (or step) must do their part to complete the race efficiently. These reactions involve a sequence of steps, each contributing to the overall transformation from reactants to products. Some key ideas to grasp about multi-step reactions include:
  • Each step has its own rate, and they can differ significantly from one another.
  • The rate-determining step (RDS) acts like the baton handover area in a relay race – if it's slow, it affects the whole race.
  • Understanding each step can help identify which parts may be optimized or improved to speed up the overall reaction.
In practice, multi-step reactions require careful study and manipulation to ensure that the process runs smoothly and efficiently. Identifying and improving the rate-determining step can lead to significant time and cost savings in chemical production.

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

Many reactions involving heterogeneous catalysts are zeroth order; that is, rate \(=k\). An example is the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) over tungsten \((\mathrm{W})\) $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ It is found that the reaction is independent of \(\left[\mathrm{PH}_{3}\right]\) as long as phosphine's pressure is sufficiently high \((\geq 1\) atm \()\). Explain.

On which of the following properties does the rate constant of a reaction depend: (a) reactant concentrations, (b) nature of reactants, (c) temperature?

Consider the reaction: $$ \mathrm{A} \longrightarrow \mathrm{B} $$ The rate of the reaction is \(1.6 \times 10^{-2} \mathrm{M} / \mathrm{s}\) when the concentration of A is \(0.15 M\). Calculate the rate constant if the reaction is (a) first order in \(\mathrm{A}\) and \((\mathrm{b})\) second order in \(\mathrm{A}\).

The following data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$ $$ \begin{array}{cccc} \text { Experiment } & {\left[\mathrm{H}_{2}\right](M)} & {[\mathrm{NO}](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate.)

The rate law for the decomposition of ozone to molecular oxygen: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) $$ is $$ \text { rate }=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]} $$ The mechanism proposed for this process is: $$ \begin{array}{l} \mathrm{O}_{3} \stackrel{k_{1}}{\rightleftarrows} \mathrm{O}+\mathrm{O}_{2} \\\ \mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2} \end{array} $$ Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

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.

Sign-up for free